Proof. $$(cf)(r\vb x)=cf(r\vb x)=crf(\vb x)=r(cf)(\vb x)$$ $$(cf)(\vb x+\vb y)=cf(\vb x+\vb y)=c(f(\vb x)+f(\vb y))=(cf)(\vb x)+(cf)(\vb y)$$ $$(f\pm g)(r\vb x)=f(r\vb x)\pm g(r\vb x)=rf(\vb x)\pm rg(\vb x)=r(f\pm g)(\vb x)$$ $$(f\pm g)(\vb x+\vb y)=f(\vb x+\vb y)\pm g(\vb x+\vb y)=(f(\vb x)+f(\vb y))\pm (g(\vb x)+g(\vb y))=(f\pm g)(\vb x)+(f\pm g)(\vb y)$$ $$(h\circ f)(r\vb x)=h(rf(\vb x))=r(h\circ f)(\vb x)$$ $$(h\circ f)(\vb x+\vb y)=h(f(\vb x)+f(\vb y))=(h\circ f)(\vb x)+(h\circ f)(\vb y)$$ $\blacksquare$

Proof. $$f(\vb 0)=f(0\vb0)=0f(\vb0)=\vb0$$ $\blacksquare$

Proof. Let $f:V\to W$ be a bijective linear function with inverse $f^{-1}:W\to V$. Since for all $\vb x,\vb y\in W$ and $c\in F$, $$f(cf^{-1}(\vb x))=cf(f^{-1}(\vb x))=c\vb x=f(f^{-1}(c\vb x))$$ and $$f(f^{-1}(\vb x)+f^{-1}(\vb y))=f(f^{-1}(\vb x))+f(f^{-1}(\vb y))=\vb x+\vb y=f(f^{-1}(\vb x+\vb y))$$ By injectivity of $f$, we conclude that $$cf^{-1}(\vb x)=f^{-1}(c\vb x)$$ and $$f^{-1}(\vb x)+f^{-1}(\vb y)=f^{-1}(\vb x+\vb y)$$ implying that $f^{-1}$ is linear. $\blacksquare$

Proof. Since $f$ is linear, it is a homomorphism. Since $f$ is bijective and linear, $f^{-1}$ is linear and thus homomorphic. Hence $f$ is an isomorphism. $\blacksquare$

Proof. $-\vb u=(-1)\vb u\in S$.

Since $S$ is non-empty, let $\vb u\in S$, then $-\vb u\in S$, hence $\vb0=\vb u+(-\vb u)\in S$. $\blacksquare$

Proof. Let $X$ be a vector space. Let $U$ be an independent subset of $X$. Let $\mathcal C$ be the set of independent subsets of $X$ that contains $U$, then $U\in\mathcal C$, and $(\mathcal C,\subseteq)$ is a non-empty partially ordered set. Let $C$ be a chain in $\mathcal C$ and let $C'=C\cup\{U\}$, then $C'$ is also a chain in $\mathcal C$. Let $D$ be a finite subset of $\bigcup C'$, indexed into $\vb x_1,\ldots,\vb x_k$ for some $k\in N$. Then for each $i$, there exists $C_i\in C'$ such that $\vb x_i\in C_i$. Define $C_0=U$, then by induction, there exists $i\in\{0,\ldots,k\}$ such that $\vb x_j\in C_i$ for all $j\in\{1,\ldots,k\}$, implying $D$ is independent. Thus $\bigcup C'$ is independent, implying it is an upper bound of $C$ in $\mathcal C$. By Zorn's lemma, $\mathcal C$ has a maximal element $M$. Suppose for contradiction that $M$ does not span $X$, then there exists $\vb x\in X$ that is not spanned by $M$.

• Suppose for all $k\in N$, pairwise distinct $\vb x_1,\ldots,\vb x_k\in M$, and $c,c_1,\ldots,c_k\in F$ such that $\sum_{i=1}^kc_i\vb x_i+c\vb x=\vb0$, we have $c=0$. Let $D$ be a finite subset of $M\cup\{\vb x\}$. If $\vb x\notin D$, then $D$ is independent. If $\vb x\in D$, index $D\setminus\{\vb x\}$ into $\vb x_1,\ldots,\vb x_k$ for some $k\in N$ and define $\vb x_0=\vb x$, then $\vb x_1,\ldots,\vb x_k\in M$ and they are pairwise distinct. Let $c_0,\ldots,c_k\in F$ such that $\sum_{i=0}^kc_i\vb x_i=\vb0$, then $c_0=0$, implying $c_j=0$ for all $j\in\{0,\dots,k\}$ since $(\vb x_1,\ldots,\vb x_k)$ is independent. Thus $(\vb x_0,\ldots,\vb x_k)$, and hence $D$, is independent. We have shown that $M\cup\{\vb x\}\in\mathcal C$, contradicting that $M$ is a maximal element of $\mathcal C$.
• Suppose there exist $k\in N$, pairwise distinct $\vb x_1,\ldots,\vb x_k\in M$, and $c,c_1,\ldots,c_k\in F$ such that $\sum_{i=1}^kc_i\vb x_i+c\vb x=\vb0$ and $c\neq0$. Then $\vb x=\frac{1}{-c}\sum_{i=1}^kc_i\vb x_i$, which is spanned by $M$, a contradiction.

Therefore, $M$ spans $X$, and is a basis of $X$. $\blacksquare$

Proof. Apply the above lemma with the independent subset being $\emptyset$. $\blacksquare$

Proof. Let $\kappa,\mu$ denote $\abs{A},\abs{B}$. Reindex $(\vb v_a)_{a\in A},(\vb w_b)_{b\in B}$ into $(\vb v_a)_{a\in\kappa},(\vb w_b)_{b\in\mu}$, then $(\vb v_a)_{a\in\kappa}$ is independent in $X$ and $(\vb w_b)_{b\in\mu}$ spans $X$. Suppose for contradiction that $\kappa\gt\mu$.

Suppose $\mu\lt\omega$. Let $k\lt\mu$ and suppose for inductive hypothesis that $(\vb v_0,\ldots,\vb v_{k-1},\vb w_{\sigma_k},\ldots,\vb w_{\sigma_{\mu-1}})$ spans $X$ for some permutation $\sigma$ of $\mu$. Then $\vb v_k=\sum_{i=0}^{k-1}c_i\vb v_i+\sum_{i=k}^{\mu-1} c_i\vb w_{\sigma_i}$ for some $c_0,\ldots,c_{\mu-1}\in F$. Since $(\vb v_0,\ldots,\vb v_k)$ is independent, for some $j\in\{k,\ldots,\mu-1\}$, $c_j\neq0$, implying $\vb w_{\sigma_j}$ is a linear combination of $(\vb v_0,\ldots,\vb v_k,\vb w_{\sigma_k},\ldots,\vb w_{\sigma_{j-1}},\vb w_{\sigma_{j+1}},\ldots,\vb w_{\sigma_{\mu-1}})$, denoted $\mathcal A$. Then each of $(\vb v_0,\ldots,\vb v_{k-1},\vb w_{\sigma_k},\ldots,\vb w_{\sigma_{\mu-1}})$ is a linear combination of $\mathcal A$, implying that $\mathcal A$ spans $X$. Let $\tau$ be the permutation of $\mu$ such that $\tau(0)=j$, $\tau(i)=i-1$ for all $i\in\{1,\ldots,j\}$, and $\tau(i)=i$ for all $i\in\{j+1,\ldots,\mu-1\}$. Then $\mathcal A$ is $(\vb v_0,\ldots,\vb v_k,\vb w_{(\sigma\tau)_k+1},\ldots,\vb w_{(\sigma\tau)_{\mu-1}})$. With the trivial base case, we conclude that $(\vb v_0,\ldots,\vb v_{\mu-1})$ spans $X$. Since $\kappa\gt\mu$, $\vb v_\mu$ is spanned by $(\vb v_0,\ldots,\vb v_{\mu-1})$, implying $(\vb v_0,\ldots,\vb v_\mu)$ is not independent, a contradiction.

Suppose $\mu\ge\omega$. Note that $(\vb v_a)_{a\in\kappa}$ can be extended into a basis $(\vb x_a)_{a\in A}$ of $X$, and $\abs{A}\ge\kappa\gt\mu$. For all $b\in\mu$, there exists finite $B_b\subseteq A$ such that for some scalars $(c_a)_{a\in B_b}$, $\vb w_b=\sum_{a\in B_b}c_a\vb x_a$. Note that $\bigcup_{b\in\mu}B_b\subseteq A$ and $\abs{\bigcup_{b\in\mu}B_b}\le\omega\mu=\mu\lt\abs{A}$, implying for some $a^*\in A$, $a^*\notin\bigcup_{b\in\mu}B_b$. Since $\vb x_{a^*}$ is a linear combination of $(\vb w_b)_{b\in\mu}$, it is also a linear combination of $(\vb x_a)_{a\in\bigcup_{b\in\mu}B_b}$, contradicting that $A$ is independent.

Therefore, $\kappa\le\mu$, implying $\abs{A}\le\abs{B}$. $\blacksquare$

Proof. Let $X$ be a vector space, then it has a basis $(\vb v_a)_{a\in A}$ whose cardinality is $\abs{A}$. Let $(\vb w_b)_{b\in B}$ be any basis of $X$, since $(\vb v_a)_{a\in A}$ and $(\vb w_b)_{b\in B}$ are both independent and spanning $X$, we have $\abs{B}=\abs{A}$, meaning that the cardinality of $(\vb w_b)_{b\in B}$ is $\abs{A}$. $\blacksquare$

Proof. Let $(\vb x_a)_{a\in A}$ be a basis of $X$, then $(\vb x_a)_{a\in A}$ spans $S$, implying $\dim S\le\abs{A}=\dim X$. $\blacksquare$

Proof. Clearly, $\{\vb 0\}$ is a subspace of $X$. Note that $\emptyset$ indexed by itself is a basis of $\{\vb0\}$. Thus $\dim\{\vb0\}=\abs{\emptyset}=0$.

Suppose $S$ is a subspace of $X$ with dimensionality $0$. Then there exists a basis $(\vb x_a)_{a\in A}$ of $S$ with $\abs{A}=0$, implying $A=\emptyset$. Thus the span of $(\vb x_a)_{a\in A}$ is $\{\vb0\}$, implying $S=\{\vb0\}$. $\blacksquare$

Proof. Let $n$ denote $\dim X$. Trivially, $X$ is a subspace of $X$ with dimensionality $n$. Now suppose we have a subspace $S$ of $X$ such that $\dim S=n$, then $S$ contains independent vectors $(\vb v_1,\ldots,\vb v_n)$. Let $\vb x\in X$, suppose for contradiction that $\vb x$ is not spanned by $(\vb v_1,\ldots,\vb v_n)$. If $\sum_ic_i\vb v_i+c\vb x=\vb0$ then $\sum_ic_i\vb v_i=-c\vb x$, so $c=0$ because otherwise $\vb x$ is spanned by $(\vb v_1,\ldots,\vb v_n)$, hence $\sum_ic_i\vb v_i=\vb0$, and by independence of $(\vb v_1,\ldots,\vb v_n)$, we have $c_i=0$ for all $i$. Thus $(\vb v_1,\ldots,\vb v_n,\vb x)$ is independent, a contradiction to $X$ having dimensionality $n$. Therefore, $\vb x$ is spanned by $(\vb v_1,\ldots,\vb v_n)$, and hence $\vb x\in S$. Since $\vb x$ is chosen arbitrarily from $X$, we have $X\subseteq S$, and hence $S=X$. $\blacksquare$

Proof. Let $B$ be a finite subset of $A$. Let $(c_b)_{b\in B}$ be scalars such that $\sum_{b\in B}c_b\Phi(\vb x_b)=\vb0$. Then $\Phi\p{\sum_{b\in B}c_b\vb x_b}=\vb0=\Phi(\vb0)$, thus $\sum_{b\in B}c_b\vb x_b=\vb0$, implying $c_b=0$ for all $b\in B$. We have shown that $(\Phi(\vb x_a))_{a\in A}$ is independent.

Let $\vb y\in Y$, then $\Phi^{-1}(\vb y)\in X$, thus there exists finite $B\subseteq A$, such that for some scalars $(c_b)_{b\in B}$, $\sum_{b\in B}c_b\vb x_b=\Phi^{-1}(\vb y)$. Then $\sum_{b\in B}c_b\Phi(\vb x_b)=\Phi\p{\sum_{b\in B}c_b\vb x_b}=\vb y$. We have shown that $(\Phi(\vb x_a))_{a\in A}$ spans $Y$. $\blacksquare$

Proof. There exist a basis $(\vb x_a)_{a\in A}$ of $X$ and an isomorphism $\Phi:X\to Y$. Then $(\Phi(\vb x_a))_{a\in A}$ is a basis of $Y$. Thus $\dim X=\abs{A}=\dim Y$. $\blacksquare$

Proof. Suppose $(c_1,\ldots,c_k)$ and $(d_1,\ldots,d_k)$ are both coordinates of $\vb x$ with respect to the basis $(\vb x_1,\ldots,\vb x_k)$. Then $\vb x=\sum_ic_i\vb x_i$ and $\vb x=\sum_id_i\vb x_i$, hence $\sum_i(c_i-d_i)\vb x_i=\sum_i(c_i\vb x_i-d_i\vb x_i)=\sum_ic_i\vb x_i-\sum_id_i\vb x_i=\vb x-\vb x=\vb 0$. Since $(\vb x_1,\ldots,\vb x_k)$ is a basis, it is independent, so $c_i-d_i=0$ for all $i$, implying $(c_1,\ldots,c_k)=(d_1,\ldots,d_k)$. $\blacksquare$

Proof. Suppose we have $X$ with basis $B=(\vb x_1,\ldots,\vb x_n)$. Let $\psi$ denote the corresponding coordinate map.

Suppose $\psi(\vb a)=(a_1,\ldots,a_n)$ and $\psi(\vb b)=(b_1,\ldots,b_n)$. Then $\vb a=\sum_ia_i\vb x_i$ and $\vb b=\sum_ib_i\vb x_i$. So given $c\in F$, we have $c\vb a=\sum_i(ca_i)\vb x_i$, and we also have $\vb a+\vb b=\sum_i(a_i+b_i)\vb x_i$. Hence, $\psi(c\vb a)=(ca_1,\ldots,ca_n)=c\psi(\vb a)$ and $\psi(\vb a+\vb b)=(a_1+b_1,\ldots,a_n+b_n)=\psi(\vb a)+\psi(\vb b)$, implying $\psi$ is linear.

If $\psi(\vb a)=\psi(\vb b)$, then $\vb a=\sum_i\psi(\vb a)_i\vb x_i=\sum_i\psi(\vb b)_i\vb x_i=\vb b$, so $\psi$ is injective.

Given $(c_1,\ldots,c_n)\in F^n$, we have $\sum_ic_i\vb x_i\in X$ and $\psi(\sum_ic_i\vb x_i)=(c_1,\ldots,c_n)$, so $\psi$ is surjective. $\blacksquare$

Proof. Suppose $(\vb x_1,\ldots,\vb x_n)$ is independent. Let $\vb x\in X$ and define $\vb x_0=\vb x$. Since $\dim X=n$, $(\vb x_0,\ldots,\vb x_n)$ is not independent, implying for some $c_0,\ldots,c_n\in F$ with $\sum_{i=0}^nc_i\vb x_i=\vb0$, not all $c_i$ are $0$. Suppose for contradiction that $c_0=0$, then $\sum_{i=1}^nc_i\vb x_i=\vb0$, implying all $c_i$ are $0$, a contradiction. Thus $c_0\neq0$, implying $\vb x_0$, and thus $\vb x$, is a linear combination of $(\vb x_1,\ldots,\vb x_n)$. We have shown that $(\vb x_1,\ldots,\vb x_n)$ spans $X$.

Suppose $(\vb x_1,\ldots,\vb x_n)$ is not independent. Then for some $c_1,\ldots,c_n\in F$ with $\sum_{i=1}^nc_i\vb x_i=\vb0$, not all $c_i$ are $0$. Thus for some $j\in\{1,\ldots,n\}$, $\vb x_j$ is a linear combination of $(\vb x_1,\ldots,\vb x_{j-1},\vb x_{j+1},\ldots,\vb x_n)$, which then has the same span as $(\vb x_1,\ldots,\vb x_n)$. Since $\dim X=n$, $(\vb x_1,\ldots,\vb x_{j-1},\vb x_{j+1},\ldots,\vb x_n)$ does not span $X$, thus $(\vb x_1,\ldots,\vb x_n)$ does not span $X$. $\blacksquare$

Proof. Suppose $\dim V=n$, then $\dim\ker f=k$ for some $k\le n$. Let $(\vb x_1,\ldots,\vb x_k)$ be a basis of $\ker f$, then it can be extended into a basis $(\vb x_1,\ldots,\vb x_n)$ of $V$. Let $\vb y_i=f(\vb x_{k+i})$ for all $i$ from $1$ to $n-k$.

Suppose $\sum_{i=1}^{n-k}c_i\vb y_i=\vb0$ for some $c_1,\ldots,c_{n-k}\in F$. Then $$f\p{\sum_{i=1}^{n-k}c_i\vb x_{k+i}}=\sum_{i=1}^{n-k}c_if(\vb x_{k+i})=\vb0$$ implying $\sum_{i=1}^{n-k}c_i\vb x_{k+i}\in\ker f$. Let $(d_1,\ldots,d_k)$ be the coordinates of $\sum_{i=1}^{n-k}c_i\vb x_{k+i}$ with respect to the basis $(\vb x_1,\ldots,\vb x_k)$, then $\sum_{i=1}^kd_i\vb x_i=\sum_{i=1}^{n-k}c_i\vb x_{k+i}$, or $$\sum_{i=1}^kd_i\vb x_i+\sum_{i=1}^{n-k}-c_i\vb x_{k+i}=\vb0$$ which implies $c_i=0$ for all $i$ since $(\vb x_1,\ldots,\vb x_n)$ is independent. We have shown that $(\vb y_1,\ldots,\vb y_{n-k})$ is independent.

Let $\vb u\in\Im f$, then there exists $\vb v$ such that $f(\vb v)=\vb u$. Let $(c_1,\ldots,c_n)$ be the coordinates of $\vb v$ with respect to the basis $(\vb x_1,\ldots,\vb x_n)$. Note that $f(\sum_{i=1}^kc_{i}\vb x_{i})=\vb0$. Thus $$\sum_{i=1}^{n-k}c_{k+i}\vb y_i=\sum_{i=1}^{n-k}c_{k+i}f(\vb x_{k+i}) =f\p{\sum_{i=k+1}^nc_{i}\vb x_{i}}+f\p{\sum_{i=1}^kc_{i}\vb x_{i}}=f(\vb v)=\vb u$$ We have shown that $(\vb y_1,\ldots,\vb y_{n-k})$ spans $\Im f$.

Therefore, $(\vb y_1,\ldots,\vb y_{n-k})$ is a basis of $\Im f$, implying $\dim\Im f=n-k$. Hence $$\dim\ker f+\dim\Im f=k+(n-k)=n=\dim V$$ $\blacksquare$

Proof. If $f$ is injective, then $\dim\ker f=0$, thus $\dim\Im f=\dim V=\dim U$. Since $\Im f$ is a subspace of $U$ with the same finite dimensionality, $\Im f=U$. Hence $f$ is surjective.

If $f$ is surjective, then $\dim\Im f=\dim U=\dim V$. Thus $\dim\ker f=0$, implying $\ker f=\{\vb0\}$. Hence $f$ is injective. $\blacksquare$

Proof. Note that $u_1+u_2=(\vb v_{u_1}+\vb v_{u_2})+W$, where $\vb v_1+W=u_1=\vb v_{u_1}+W$ and $\vb v_2+W=u_2=\vb v_{u_2}+W$, implying $\vb v_1-\vb v_{u_1}\in W$ and $\vb v_2-\vb v_{u_2}\in W$. Then $(\vb v_1+\vb v_2)-(\vb v_{u_1}+\vb v_{u_2})\in W$, implying $(\vb v_1+\vb v_2)+W=(\vb v_{u_1}+\vb v_{u_2})+W=u_1+u_2$.

Note that $cu=(c\vb v_u)+W$, where $\vb v+W=u=\vb v_u+W$, implying $\vb v-\vb v_u\in W$. Then $c\vb v-c\vb v_u\in W$, implying $(c\vb v)+W=(c\vb v_u)+W=cu$. $\blacksquare$

Proof. Define a function $f:V\to V/W$ by $f(\vb v)=\vb v+W$, then $f$ is linear. Given $\vb v\in V$,

• if $\vb v\in W$, then $f(\vb v)=\vb v+W=\vb 0+W$, hence $\vb v\in\ker f$;
• if $\vb v\notin W$, $\vb v+W\cap\vb0+W=\emptyset$, hence $f(\vb v)=\vb v+W\neq\vb 0+W$, implying $\vb v\notin\ker f$.

Thus $\ker f=W$. Let $W^*\in V/W$, then there exists $\vb v\in V$ such that $f(\vb v)=\vb v+W=W^*$, hence $\Im f=V/W$. Therefore, $$\dim V/W=\dim\Im f=\dim V-\dim\ker f=\dim V-\dim W$$ $\blacksquare$

Proof. Suppose $\sum_ic_i\vb x_i=\vb0$. then $$c_j=c_j\norm{\vb x_j}^2=\sum_ic_i\braket{\vb x_j}{\vb x_i}=\braket{\vb x_j}{\sum_ic_i\vb x_i}=\braket{\vb x_j}{\vb0}=0$$ for $j\in\{1,\ldots,n\}$. Thus $(\vb x_1,\ldots,\vb x_n)$ is independent. $\blacksquare$

Proof. Define a function $F:N\times X^n\to N\times X^n$ by:

• if $i\lt n$, then $F(i,(\vb x_1,\ldots,\vb x_n))=F\p{i+1,\p{\vb x_1,\ldots,\vb y,\ldots,\vb x_n}}$, where $\vb y$ is the $(i+1)$-th element of the tuple and $$\vb y=\vb x_{i+1}-\sum_{j=1}^{i}\frac{\braket{\vb x_{i+1}}{\vb x_j}}{\braket{\vb x_j}{\vb x_j}}\vb x_j$$
• if $i\ge n$, then $F(i,(\vb x_1,\ldots,\vb x_n))=F\p{i+1,(\vb x_1,\ldots,\vb x_n)}$.

Now let $B=(\vb b_1,\ldots,\vb b_n)$ be a basis of $X$. By recursion theorem, we can find a function $u:N\to N\times X^n$ such that $u(0)=(0,B)$ and $u(i+1)=F(u(i))$. Denote the second element of $u(i)$ by $B_i$, we will show by induction that, for $i\in\{1,\ldots,n\}$,

• the $(i+1)$-th to the $n$-th elements of $B_i$ are $\vb b_{i+1},\ldots,\vb b_n$,
• the first $i$ elements of $B_i$ are pairwise orthogonal and non-zero, and
• the $j$-th element of $B_i$ is in the span of $(\vb b_1,\ldots,\vb b_j)$ for $j\in\{1,\ldots,i\}$.

Consider $i=1$ the base case. Since $B_1=B_0=B$, all conditions are trivially satisfied. Now let $i\in\{1,\ldots,n-1\}$. Suppose the inductive hypothesis holds. Denote $B_i$ by $(\vb a_1,\ldots,\vb a_n)$, then the $(i+1)$-th element of $B_{i+1}$ is $\vb y=\vb a_{i+1}-\sum_{j=1}^{i}\frac{\braket{\vb a_{i+1}}{\vb a_j}}{\braket{\vb a_j}{\vb a_j}}\vb a_j$. Since the $(i+2)$-th to the $n$-th elements of $B_{i+1}$ are the $(i+2)$-th to the $n$-th elements of $B_{i}$, they are $\vb b_{i+2},\ldots,\vb b_n$. Let $k\in\{1,\ldots,i\}$, then $$\braket{\vb a_k}{\vb y}=\braket{\vb a_k}{\vb a_{i+1}-\sum_{j=1}^{i}\frac{\braket{\vb a_{i+1}}{\vb a_j}}{\braket{\vb a_j}{\vb a_j}}\vb a_j} =\braket{\vb a_k}{\vb a_{i+1}}-\braket{\vb a_k}{\frac{\braket{\vb a_{i+1}}{\vb a_k}}{\braket{\vb a_k}{\vb a_k}}\vb a_k}=0$$ Hence the first $i+1$ elements of $B_{i+1}$ are pairwise orthogonal. Since for each $j\in\{1,\ldots,i\}$, $\vb a_j$ is in the span of $(\vb b_1,\ldots,\vb b_j)$ and $\vb a_{i+1}=\vb b_{i+1}$:

• if $\vb y=\vb0$, then $\vb b_{i+1}$ is in the span of $(\vb b_1,\ldots,\vb b_j)$, a contradiction, implying $\vb y$ (and thus the first $i+1$ elements of $B_{i+1}$) is non-zero; and
• clearly $\vb y$ is in the span of $(\vb b_1,\ldots,\vb b_{i+1})$.

And the induction is complete. We conclude that the elements of $B_n$ are pairwise orthogonal and non-zero. Divide each vector in $B_n$ by its norm and we have an orthonormal $n$-tuple of vectors, which is independent, and therefore a basis of $X$. $\blacksquare$

Proof. Most of the above are obvious. We will only prove the less obvious one.

$A(BC)=(AB)C$: Suppose $A$ is $m\times n$, $B$ is $n\times p$, $C$ is $p\times q$. Then both $A(BC)$ and $(AB)C$ are $m\times q$. Suppose $i\in\{1,\ldots,m\}$, $j\in\{1,\ldots,n\}$, $k\in\{1,\ldots,p\}$, $l\in\{1,\ldots,q\}$. Then $(A(BC))_{il}=\sum_jA_{ij}(BC)_{jl}=\sum_jA_{ij}\sum_kB_{jk}C_{kl}=\sum_j\sum_kA_{ij}B_{jk}C_{kl} =\sum_k\sum_jA_{ij}B_{jk}C_{kl}=\sum_k(\sum_jA_{ij}B_{jk})C_{kl}=\sum_k(AB)_{ik}C_{kl}=((AB)C)_{il}$. Hence $A(BC)=(AB)C$. $\blacksquare$

Proof. Most of the above are obvious. We will only prove the less obvious one.

$(AB)^T=B^TA^T$: Suppose $A$ is $m\times n$, $B$ is $n\times p$. Then both $(AB)^T$ and $B^TA^T$ are $p\times m$. Suppose $i\in\{1,\ldots,m\}$, $j\in\{1,\ldots,n\}$, $k\in\{1,\ldots,p\}$. Then $(AB)^T_{ki}=(AB)_{ik}=\sum_jA_{ij}B_{jk}=\sum_jB^T_{kj}A^T_{ji}=(B^TA^T)_{ki}$. Hence $(AB)^T=B^TA^T$. $\blacksquare$

Proof. For injectivity, suppose $f_1,f_2\in L(X,Y)$ and $\phi(f_1)=\phi(f_2)$, then $\phi(f_1)_{ji}=\phi(f_2)_{ji}$, so $\psi_Y(f_1(\vb x_i))_j=\psi_Y(f_2(\vb x_i))_j$, thus $\psi_Y(f_1(\vb x_i))=\psi_Y(f_2(\vb x_i))$, hence $f_1(\vb x_i)=f_2(\vb x_i)$. So for all $\vb x\in X$, $f_1(\vb x)=f_1(\sum_i\psi_X(\vb x)_i\vb x_i)=\sum_i\psi_X(\vb x)_if_1(\vb x_i) =\sum_i\psi_X(\vb x)_if_2(\vb x_i)=f_2(\sum_i\psi_X(\vb x)_i\vb x_i)=f_2(\vb x)$, which means $f_1=f_2$.

For surjectivity, given $A\in F^{m\times n}$, we can define a function $f^A:X\to Y$ such that $f^A(\vb x)=\psi_Y^{-1}(A\psi_X(\vb x))$. Now given $\vb x,\vb y\in F^n$ and $c\in F$, we have $$f^A(c\vb x)=\psi_Y^{-1}(A\psi_X(c\vb x))=\psi_Y^{-1}(A(c\psi_X(\vb x)))=\psi_Y^{-1}(cA\psi_X(\vb x))=c\psi_Y^{-1}(A\psi_X(\vb x))=cf^A(\vb x)$$ and $$f^A(\vb x+\vb y)=\psi_Y^{-1}(A\psi_X(\vb x+\vb y))=\psi_Y^{-1}(A(\psi_X(\vb x)+\psi_X(\vb y))) =\psi_Y^{-1}(A\psi_X(\vb x)+A\psi_X(\vb y))=\psi_Y^{-1}(A\psi_X(\vb x))+\psi_Y^{-1}(A\psi_X(\vb y))=f^A(\vb x)+f^A(\vb y)$$ implying $f^A$ is linear, meaning $f^A\in L(X,Y)$. Note that $\phi(f^A)_{ji}=\psi_Y(f^A(\vb x_i))_j=\psi_Y(\psi_Y^{-1}(A\psi_X(\vb x_i)))_j=(A\psi_X(\vb x_i))_j=(A\vb e_i)_j=A_{ji}$, hence $\phi(f^A)=A$. We have shown that there exists $f\in L(X,Y)$ such that $\phi(f)=A$, namely $f^A$.

Since $\phi$ is bijective, $\phi^{-1}$ exists. Given $A\in F^{m\times n}$, we have $f^A\in L(X,Y)$ with $\phi(f^A)=A$, hence $\phi^{-1}(A)=f^A$. $\blacksquare$

Proof. $\phi(f)$ is $m\times n$, $\psi_X(\vb x)$ is implicitly regarded as $n\times 1$, $\psi_Y(\vb y)$ is implicitly regarded as $m\times 1$, thus the dimensions make sense. Now we have $$(\phi(f)\psi_X(\vb x))_j=\sum_i\phi(f)_{ji}\psi_X(\vb x)_i=\sum_i\psi_Y(f(\vb x_i))_j\psi_X(\vb x)_i =(\sum_i\psi_X(\vb x)_i\psi_Y(f(\vb x_i)))_j=(\psi_Y(\sum_i\psi_X(\vb x)_if(\vb x_i)))_j=(\psi_Y(f(\sum_i\psi_X(\vb x)_i\vb x_i)))_j =(\psi_Y(f(\vb x)))_j=(\psi_Y(\vb y))_j$$, hence $\phi(f)\psi_X(\vb x)=\psi_Y(\vb y)$.

$A$ is $m\times n$, $\vb v$ is implicitly regarded as $n\times 1$, $\vb u$ is implicitly regarded as $m\times 1$, thus the dimensions make sense. Now we have $$\phi^{-1}(A)(\psi_X^{-1}(\vb v))=\psi_Y^{-1}(A\psi_X(\psi_X^{-1}(\vb v)))=\psi_Y^{-1}(A\vb v)=\psi_Y^{-1}(\vb u)$$ $\blacksquare$

Proof. Clearly, $cf,f\pm g\in L(X,Y)$ and $h\circ f\in L(X,Z)$, so $\phi(cf),\phi(f\pm g)\in F^{m\times n}$ and $\phi(h\circ f)\in F^{l\times n}$. Note that $cA,A\pm B\in F^{m\times n}$ and $CA\in F^{l\times n}$. Since we also have $$\phi(cf)_{ji}=\psi_Y((cf)(\vb x_i))_j=\psi_Y(cf(\vb x_i))_j=(c\psi_Y(f(\vb x_i)))_j=c\psi_Y(f(\vb x_i))_j=c\phi(f)_{ji}=(cA)_{ji}$$ $$\phi(f\pm g)_{ji}=\psi_Y((f\pm g)(\vb x_i))_j=\psi_Y(f(\vb x_i)\pm g(\vb x_i))_j=(\psi_Y(f(\vb x_i))\pm\psi_Y(g(\vb x_i)))_j =\psi_Y(f(\vb x_i))_j\pm\psi_Y(g(\vb x_i))_j=\phi(f)_{ji}\pm\phi(g)_{ji}=(A\pm B)_{ji}$$ $$\phi(h\circ f)_{ki}=\psi_Z((h\circ f)(\vb x_i))_k=\psi_Z(h(\psi_Y^{-1}(A\psi_X(\vb x_i))))_k =\psi_Z(\psi_Z^{-1}(C\psi_Y(\psi_Y^{-1}(A\psi_X(\vb x_i)))))_k=C(A\vb e_i)_k=(CA_{*,i})_k=C_{k,*}A_{*,i}=(CA)_{ki}$$ we conclude that $\phi(cf)=cA$, $\phi(f\pm g)=A\pm B$, and $\phi(h\circ f)=CA$. $\blacksquare$

Proof. Let $AB=I$. Suppose for contradiction that $B$ is not surjective. Let $S$ denote the range of $B$, then $S$ is a subspace of $F^n$ but $S\neq F^n$. Thus $S$ has dimensionality less than $n$. So there exist $\vb v_1,\ldots,\vb v_{n-1}\in S$ such that $(\vb v_1,\ldots,\vb v_{n-1})$ spans $S$. Then the range of $AB$ is spanned by $(A\vb v_1,\ldots,A\vb v_{n-1})$. So the range of $AB$ has dimensionality less than $n$, but that implies the range of $I$ has dimensionality less than $n$, a contradiction. Hence, $B$ is surjective.

Since $B=BI=B(AB)=(BA)B$, we have $(BA)B-B=0$, implying $(BA-I)B=0$. For all $\vb v\in F^n$, since $B$ is surjective, there exists $\vb u\in F^n$ such that $B\vb u=\vb v$, so $(BA-I)\vb v=(BA-I)B\vb u=0\vb u=\vb 0$. That means $(BA-I)_{*,i}=(BA-I)\vb e_i=\vb 0$ for all $i$, hence $BA-I=0$, so $BA=I$. $\blacksquare$

Proof. Suppose $AB=I=AC$, then $BA=I=CA$, so $B=BI=B(AC)=(BA)C=IC=C$. $\blacksquare$

Proof. Suppose $A$ is invertible, then there exists $B\in F^{n\times n}$ such that $BA=I$. By the same reasoning as a proof above, $A$ is surjective. Now suppose $A\vb v=A\vb u$, then $\vb v=I\vb v=(BA)\vb v=B(A\vb v)=B(A\vb u)=(BA)\vb u=I\vb u=\vb u$. So $A$ is injective. Hence $A$ is bijective.

Suppose $A$ is bijective, then it has an inverse function that is also linear, and hence has a standard matrix representation $B$. Note that for all $\vb v\in F^n$, $B(A\vb v)=\vb v$. So $(BA)_{*,i}=(BA)\vb e_i=\vb e_i$ for all $i$. Hence $BA=I$, implying $A$ is invertible. $\blacksquare$

Proof. Suppose $A$ is injective. If $\sum_ic_iA\vb e_i=\vb 0$, then $A(\sum_ic_i\vb e_i)=\vb 0$, and by injectivity of $A$, $\sum_ic_i\vb e_i=0$, then by independence of standard basis, $c_i=0$ for all $i$. Hence, $(A\vb e_1,\ldots,A\vb e_n)$ is independent, so it spans $F^n$, implying $A$ is surjective.

Suppose $A$ is surjective, then $(A\vb e_1,\ldots,A\vb e_n)$ spans $F^n$, so it is independent. If $A\vb x=\vb0$, then for some $c_1,\ldots,c_n$, we have $\vb x=\sum_ic_i\vb e_i$, then we have $A(\sum_ic_i\vb e_i)=\vb 0$, so $\sum_ic_iA\vb e_i=\vb 0$, implying $c_i=0$ for all $i$ by independence, hence $\vb x=\vb 0$. Therefore, if $A\vb x=A\vb y$, then $A(\vb x-\vb y)=0$, implying $\vb x-\vb y=0$, or $\vb x=\vb y$, hence $A$ is injective. $\blacksquare$

Proof. Suppose $m\lt n$. Then $(A\vb e_1,\ldots,A\vb e_n)$ are not independent, implying $A\vb v=\vb0$ for some non-zero $\vb v$. Thus $A$ is not injective.

Suppose $m\gt n$. Then $(A\vb e_1,\ldots,A\vb e_n)$ does not span $F^m$. Thus some $\vb v\in F^m$ is not in the range of $A$, implying $A$ is not surjective. $\blacksquare$

Proof. $$(cA)(c^{-1}A^{-1})=c^{-1}cAA^{-1}=I$$ $$(AB)(B^{-1}A^{-1})=A(BB^{-1})A^{-1}=I$$ $$A^T(A^{-1})^T=(A^{-1}A)^T=I$$

Proof. Let $f\in L(X,Y)$, where $\dim X=n$ and $\dim X=m$, be bijective. Fix any coordinate maps $\psi_X,\psi_Y$ of $X,Y$ respectively. Define $\phi:L(X,Y)\to F^{m\times n}$ based on $\psi_X,\psi_Y$.

Suppose $\vb v,\vb w\in F^n$ and $\phi(f)\vb v=\phi(f)\vb w$. Then $f(\psi_X^{-1}(\vb v))=f(\psi_X^{-1}(\vb w))$, implying $\psi_X^{-1}(\vb v)=\psi_X^{-1}(\vb w)$, which implies that $\vb v=\vb w$. We have shown that $\phi(f)$ is injective.

Let $\vb u\in F^m$. Then $\psi_Y^{-1}(\vb u)\in Y$, thus there exists $\vb x\in X$ such that $f(\vb x)=\psi_Y^{-1}(\vb u)$, implying $\phi(f)\psi_X(\vb x)=\vb u$. We have shown that $\phi(f)$ is surjective.

Since $\phi(f)\in F^{m\times n}$ and $\phi(f)$ is bijective, $m=n$. Hence $\phi(f)$ is invertible. $\blacksquare$

Proof. Let $B_1=(\vb v_1,\ldots,\vb v_n)$. Given $\vb v\in V$, there exist $c_1,\ldots,c_n$ such that $\vb v=\sum_ic_i\vb v_i$. Thus $$\psi_{B_2}(B_1)\varphi_{B_1}(\vb v) =\psi_{B_2}(B_1)\varphi_{B_1}\p{\sum_ic_i\vb v_i} =\sum_ic_i\psi_{B_2}(B_1)\varphi_{B_1}(\vb v_i) =\sum_ic_i\psi_{B_2}(B_1)_{*,i} =\sum_ic_i\varphi_{B_2}(\vb v_i) =\varphi_{B_2}\p{\sum_ic_i\vb v_i} =\varphi_{B_2}(\vb v)$$ $\blacksquare$

Proof. Let $A$ be the transition matrix from $B_1$ to $B_2$. Then let $B$ be the transition matrix from $B_2$ to $B_1$. Let $B_1=(\vb v_1,\ldots,\vb v_n)$ and let $\vb x\in F^n$. Then $\varphi_{B_1}(\sum_ix_i\vb v_i)=\vb x$. Thus $$BA\vb x=B\p{A\varphi_{B_1}\p{\sum_ix_i\vb v_i}}=B\varphi_{B_2}\p{\sum_ix_i\vb v_i}=\varphi_{B_1}\p{\sum_ix_i\vb v_i}=\vb x$$ implying $BA=I$. Hence $A$ is invertible, and $A^{-1}=B$. $\blacksquare$

Proof. Let $f_B$ denote the matrix representation of $f$ with respect to basis $B$. Let $\vb v_B$ denote the coordinates of $\vb v\in V$ with respect to basis $B$. Then for all $\vb v\in V$, $$f_{B_2}P\vb v_{B_1}=f_{B_2}\vb v_{B_2}=f(\vb v)_{B_2}=Pf(\vb v)_{B_1}=PA\vb v_{B_1}$$ implying $(P^{-1}f_{B_2}P)\vb v_{B_1}=A\vb v_{B_1}$, and thus $P^{-1}f_{B_2}P=A$, therefore $f_{B_2}=PAP^{-1}$. $\blacksquare$

Proof. Trivially, $0\in\{\norm{A\vb x}:\Vert\vb x\Vert\le1\}$, so it is non-empty.

Let $\vb x\in R^n$ with $\Vert x\Vert\le1$. Note that there exists $c_1,\ldots,c_n\in R$ such that $\vb x=\sum_ic_i\vb e_i$. Since $\Vert x\Vert\le1$, $\abs{c_i}\le1$ for all $i$. Hence $$\norm{A\vb x}=\norm{\sum_ic_iA\vb e_i}\le\sum_i\abs{c_i}\norm{A\vb e_i}\le\sum_i\norm{A\vb e_i}$$ implying that $\{\norm{A\vb x}:\Vert\vb x\Vert\le1\}$ is bounded. $\blacksquare$

Proof.

• Note that $\norm{A0}=0$, thus $\norm{A}\ge0$.
• If $A=0$, then trivially, $\norm{A}=0$. If $A\neq0$, then for some index $(i,j)$, $A_{ij}\neq0$. Then $A\vb e_j\neq0$, thus $\norm{A}\ge\norm{A\vb e_j}\gt0$.
• Given $\vb x\in R^n$, $\norm{(cA)\vb x}=\norm{c(A\vb x)}=\abs{c}\norm{A\vb x}$. Thus $\{\norm{(cA)\vb x}:\Vert\vb x\Vert\le1\}=\{\abs{c}r:r\in\{\norm{A\vb x}:\Vert\vb x\Vert\le1\}\}$. And we conclude that $\norm{cA}=\abs{c}\norm{A}$.
• Clearly, $\norm{A\vb 0}=0=\norm{A}\Vert\vb 0\Vert$. Now suppose $\vb x\neq0$. Then $\frac{\norm{A\vb x}}{\Vert\vb x\Vert}=\norm{\frac{A\vb x}{\Vert\vb x\Vert}}=\norm{A\frac{\vb x}{\Vert\vb x\Vert}}$. Since $\norm{\frac{\vb x}{\Vert\vb x\Vert}}=1$, we have $\frac{\norm{A\vb x}}{\Vert\vb x\Vert}\le\norm{A}$, hence $\norm{A\vb x}\le\norm{A}\Vert\vb x\Vert$.
• Given $\Vert\vb x\Vert\le1$, $$\norm{(A\pm B)\vb x}=\norm{A\vb x\pm B\vb x}\le\norm{A\vb x}+\norm{B\vb x}\le(\norm{A}+\norm{B})\Vert\vb x\Vert\le\norm{A}+\norm{B}$$ Hence $\norm{A\pm B}\le\norm{A}+\norm{B}$.
• Given $\Vert\vb x\Vert\le1$, $$\norm{(AB)\vb x}=\norm{A(B\vb x)}\le\norm{A}\norm{B\vb x}\le\norm{A}\norm{B}\Vert\vb x\Vert\le\norm{A}\norm{B}$$ Hence $\norm{AB}\le\norm{A}\norm{B}$.

$\blacksquare$

Proof. Note that, since $0\in R^{n\times n}$ is not invertible, $A^{-1}\neq0$, so $\norm{A^{-1}}\gt0$. Let $B\in R^{n\times n}$ such that $d(A,B)\lt\frac{1}{\norm{A^{-1}}}$. Let $\alpha$ denote $\frac{1}{\norm{A^{-1}}}$ and $\beta$ denote $d(A,B)$, or $\norm{A-B}$. For all $\vb x\in R^n$, we have $$\alpha\Vert\vb x\Vert=\alpha\norm{A^{-1}A\vb x}\le\alpha\norm{A^{-1}}\norm{A\vb x} =\norm{A\vb x}\le\norm{(A-B)\vb x}+\norm{B\vb x}\le\beta\Vert\vb x\Vert+\norm{B\vb x}$$ so $$\p{\alpha-\beta}\Vert\vb x\Vert\le\norm{B\vb x}$$ Since $\alpha-\beta\gt0$, if $\vb x\neq\vb0$ then $B\vb x\neq\vb0$. Given $B\vb x=B\vb y$, we have $B(\vb x-\vb y)=\vb 0$, so $\vb x-\vb y=0$, or $\vb x=\vb y$. Hence $B$ is injective, and thus surjective, implying it is bijective, and therefore invertible. $\blacksquare$

Proof. Let $A\in\Omega$, then for all $B\in R^{n\times n}$ such that $d(A,B)\lt\frac{1}{\norm{A^{-1}}}$, $B\in\Omega$. Hence $\Omega$ is an open subset of $R^{n\times n}$.

Let $A\in\Omega$. For every $\varepsilon\gt0$, let $\delta=\inf(\alpha/2,\alpha^2\varepsilon/2)$, then for all $B\in R^{n\times n}$ such that $d(A,B)\lt\delta$, we have $\beta\lt\alpha$, so $B\in\Omega$ and we still have $\p{\alpha-\beta}\Vert\vb x\Vert\le\norm{B\vb x}$ for all $\vb x\in R^n$. Let $\vb y\in R^n$ such that $\Vert\vb y\Vert\le1$, we have $$\norm{B^{-1}\vb y}\le\frac{\norm{B(B^{-1}\vb y)}}{\alpha-\beta}=\frac{\Vert\vb y\Vert}{\alpha-\beta}\le\frac{1}{\alpha-\beta}$$ thus $\norm{B^{-1}}\le\frac{1}{\alpha-\beta}$. Now we have $$d(\varphi(A),\varphi(B))=\norm{B^{-1}-A^{-1}}=\norm{B^{-1}(A-B)A^{-1}}\le\norm{B^{-1}}\norm{A-B}\norm{A^{-1}} \le\frac{\beta}{\alpha(\alpha-\beta)}\lt\frac{2\beta}{\alpha^2}\lt\varepsilon$$ We have shown that $\varphi$ is continuous. $\blacksquare$

Proof. Note that $\sigma\mapsto\sigma^{-1}$ is a bijection from $S_n$ to $S_n$, hence $$\det(A) =\sum_{\sigma\in S_n}\sgn(\sigma)\prod_{i=1}^nA_{i,\sigma_i} =\sum_{\sigma\in S_n}\sgn(\sigma)\prod_{i=1}^nA^T_{\sigma_i,i} =\sum_{\sigma\in S_n}\sgn(\sigma)\prod_{i=1}^nA^T_{i,\sigma^{-1}_i} =\sum_{\sigma\in S_n}\sgn(\sigma^{-1})\prod_{i=1}^nA^T_{i,\sigma_i} =\sum_{\sigma\in S_n}\sgn(\sigma)\prod_{i=1}^nA^T_{i,\sigma_i} =\det(A^T)$$ $\blacksquare$

Proof. $$\det(A[*,j:A_{*,j}+\vb b]) =\det(A^T[j,*:A_{*,j}+\vb b]) =\sum_{\sigma\in S_n}\sgn(\sigma)\p{\prod_{i\in\{1,\ldots,n\}}^{i\neq j}A^T_{i,\sigma_i}}(A_{\sigma_j,j}+b_{\sigma_j}) =\sum_{\sigma\in S_n}\sgn(\sigma)\p{\prod_{i=1}^nA^T_{i,\sigma_i}+\prod_{i=1}^nA^T[j,*:\vb b]_{i,\sigma_i}} =\det(A^T)+\det(A^T[j,*:\vb b]) =\det(A)+\det(A[*,j:\vb b])$$ $$\det(A[*,j:cA_{*,j}]) =\sum_{\sigma\in S_n}\sgn(\sigma)c\prod_{i=1}^nA_{i,\sigma_i} =c\det(A)$$ $$\det(A[\tau]) =\sum_{\sigma\in S_n}\sgn(\sigma)\prod_{i=1}^nA[\tau]_{i,\sigma_i} =\sum_{\sigma\in S_n}\sgn(\sigma)\prod_{i=1}^nA_{i,\tau\sigma_i} =\sum_{\sigma\in S_n}\sgn(\tau^{-1}\sigma)\prod_{i=1}^nA_{i,\tau\tau^{-1}\sigma_i} =\sum_{\sigma\in S_n}\sgn(\tau^{-1})\sgn(\sigma)\prod_{i=1}^nA_{i,\sigma_i} =-\sum_{\sigma\in S_n}\sgn(\sigma)\prod_{i=1}^nA_{i,\sigma_i} =-\det(A)$$ $\blacksquare$

Proof. Suppose $A$ has repeated columns $A_{*,i}=A_{*,j}$ where $i\neq j$. Let $\tau\in S_n$ be the transposition swapping $i$ and $j$. Then $\det(A)=\det(A[\tau])=-\det(A)$, thus $\det(A)=0$. Suppose $A$ has repeated rows, then $\det(A)=\det(A^T)=0$. $\blacksquare$

Proof. Note that given $\tau\in S_n$, $\sigma\mapsto\sigma\tau$ is a bijection from $S_n$ to $S_n$. Also, given a map $j:n\to n$, either $j$ is an injection, in which case $j\in S_n$, or for some $a,b\in\{1,\ldots,n\}$ such that $a\neq b$, $j_a=j_b$, in which case $\sum_{\sigma\in S_n}\sgn(\sigma)\prod_{i=1}^nB_{j_i,\sigma_i}=0$. Therefore, $$\det(AB) =\sum_{\sigma\in S_n}\sgn(\sigma)\prod_{i=1}^n\sum_{j=1}^nA_{i,j}B_{j,\sigma_i} =\sum_{\sigma\in S_n}\sgn(\sigma)\sum_{j:n\to n}\prod_{i=1}^nA_{i,j_i}B_{j_i,\sigma_i} =\sum_{j:n\to n}\prod_{i=1}^nA_{i,j_i}\sum_{\sigma\in S_n}\sgn(\sigma)\prod_{k=1}^nB_{j_k,\sigma_k} $$ $$ =\sum_{\tau\in S_n}\prod_{i=1}^nA_{i,\tau_i}\sum_{\sigma\in S_n}\sgn(\sigma)\prod_{k=1}^nB_{\tau_k,\sigma_k} =\sum_{\tau\in S_n}\prod_{i=1}^nA_{i,\tau_i}\sum_{\sigma\in S_n}\sgn(\sigma\tau)\prod_{k=1}^nB_{k,\sigma_k} =\sum_{\tau\in S_n}\prod_{i=1}^nA_{i,\tau_i}\sgn(\tau)\det(B) =\det(A)\det(B) $$ $\blacksquare$

Proof. Suppose $i=j$, then $$(A\text{adj}(A))_{ij} =\sum_{k=1}^nA_{jk}\text{adj}(A)_{kj} =\sum_{k=1}^nA_{jk}\sum_{\sigma\in S_n[j,k]}\sgn(\sigma)\prod_{l=\{1,\ldots,n\}}^{l\neq j}A_{l,\sigma_l} =\sum_{k=1}^n\sum_{\sigma\in S_n[j,k]}\sgn(\sigma)\prod_{l=1}^nA_{l,\sigma_l} =\sum_{\sigma\in S_n}\sgn(\sigma)\prod_{l=1}^nA_{l,\sigma_l} =\det(A)$$ Suppose $i\neq j$, let $A'$ be obtained from $A$ by replacing the $j$-th row with the $i$-th row, then $$(A\text{adj}(A))_{ij} =\sum_{k=1}^nA_{ik}\text{adj}(A)_{kj} =\sum_{k=1}^nA'_{jk}\sum_{\sigma\in S_n[j,k]}\sgn(\sigma)\prod_{l=\{1,\ldots,n\}}^{l\neq j}A'_{l,\sigma_l} =\det(A') =0$$ Hence $A\text{adj}(A)=\det(A)I$. $\blacksquare$

Proof. Suppose $A$ is invertible, then $AA^{-1}=I$, hence $\det(A)\det(A^{-1})=\det(I)=1$, implying $\det(A)\neq0$.

Suppose $\det(A)\neq0$, then $A\p{\frac{1}{\det(A)}\text{adj}(A)}=I$, implying $A$ is invertible. $\blacksquare$

Proof. $$\text{adj}(A)=A^{-1}A\text{adj}(A)=A^{-1}(\det(A)I)=\det(A)A^{-1}$$ Since $\det(A)\neq0$, we have $A^{-1}=\frac{1}{\det(A)}\text{adj}(A)$ $\blacksquare$

Proof. We will propose an algorithm with $n(m+1)$ steps that takes the current number of steps and current form of the matrix, and performs the next step. Coupled with recursion theorem, this algorithm can be used to find $A'$.

Let $j\in\{0,\ldots,n-1\}$.

• On step $j(m+1)+1$, let the current form of the matrix be $B$, and let the matrix formed by the first $j$ columns of $B$ be $B'$. If there are no zero rows in $B'$ (when $j=0$, all rows of $B'$ are empty and considered zero rows), perform the identity operation; otherwise let the index of the first zero row of $B'$ be $i$. If $B_{i,j+1}$ is zero and for some zero row $i'$ of $B'$, $B_{i',j+1}$ is non-zero, swap row $i$ and row $i'$; otherwise perform the identity operation.
• On step $j(m+1)+2$, find $i$ or perform the identity operation as above. If $B_{i,j+1}$ is non-zero, multiply row $i$ by $1/B_{i,j+1}$; otherwise perform the identity operation.
• On steps $j(m+1)+3$ through $(j+1)(m+1)$, find $i$ or perform the identity operation as above. For each row $i'$ other than $i$, add $-B_{i',j+1}$ times row $i$ to row $i'$.

It's easy to see by induction that after step $n(m+1)$, the matrix is in echelon form.

Now suppose we have $B=S_k\ldots S_1A$ and $C=S'_l\ldots S'_1A$ where $S_1,\ldots,S_k,S'_1,\ldots,S'_l$ are elementary row operations and $B$ and $C$ are in echelon form. Note that if we denote the matrix formed by the first $i$ columns of $A$ by $A_i$ for $i\in\{1,\ldots,n\}$, and same for $B$ and $C$, then $B_i=S_k\ldots S_1A_i$ and $C_i=S'_l\ldots S'_1A_i$, and $B_i$ and $C_i$ are in echelon form. Also note that if $S$ and $T$ differ by one elementary row operation, then a column $i$ of $S$ being a linear combination of the columns of $S$ implies the column $i$ of $T$ is a linear combination of the columns of $T$ with the same scalar coefficients. Thus given $j\in\{1,\ldots,i\}$, the column $j$ of $B_i$ is a linear combination of the columns of $B_i$ with coefficients $c_1,\ldots,c_i$ if and only if the column $j$ of $C_i$ is a linear combination of the columns of $C_i$ with coefficients $c_1,\ldots,c_i$. We will now use induction to show that $B_i=C_i$:

• If the first column of $A$ is a zero column, then $B_1$ and $C_1$ are both zero matrixes. If the first column of $A$ is not a zero column, then $B_1$ and $C_1$ are both non-zero, and any non-empty entry is a pivot. Note that the pivot must be the first entry of the column, and all other entries must be $0$. Since this holds for both $B_1$ and $C_1$, they are equal.
• Now suppose $B_i=C_i$ where $i\in\{1,\ldots,n-1\}$. Note that $B_i$ and $C_i$ are $B_{i+1}$ and $C_{i+1}$ without the last column. Hence the first $i$ columns of $B_{i+1}$ and $C_{i+1}$ are equal. If the last column of $B_{i+1}$ or $C_{i+1}$ contains a pivot, then it is on the first zero row of $B_i$ or $C_i$, and all other entries of the column are zero, implying the last column cannot be a linear combination of previous columns. If the last column of $B_{i+1}$ or $C_{i+1}$ does not contain a pivot, then the column must have zero on all zero rows of $B_i$ or $C_i$, implying the last column must be a linear combination of previous columns that contain pivots. Therefore, the last column of $B_{i+1}$ and $C_{i+1}$ must both contain a pivot or both do not. If they both contain a pivot, then they are clearly equal. If they both do not contain a pivot, note that they are the same linear combination of the same pivot columns and are thus equal.

Therefore, $B=B_n=C_n=C$. And we have shown the uniqueness of $A'$. $\blacksquare$

Proof. If the echelon form of $A$ is the identity matrix, then $A$ can be decomposed into elementary row operations, which is invertible.

If $A$ is invertible, then it has non-zero determinant. If the echelon form of $A$ is not the identity matrix, then it must have at least one zero row, and thus a zero determinant. Since a matrix with zero determinant still has zero determinant after an elementary row operation, $A$ itself must also have zero determinant, a contradiction. Thus the echelon form of $A$ is the identity matrix. $\blacksquare$

Proof. $V$ has a basis $A$. Let $B$ be a basis of $V$ and let $P$ be the transition matrix from $A$ to $B$, then $$\det(f_B)=\det(Pf_AP^{-1})=\det(P)\det(f_A)\det(P^{-1})=\det(f_A)\det(I)=\det(f_A)$$ Uniqueness is trivial. $\blacksquare$

Proof. $$\tr(cA)=\sum_ica_{ii}=c\sum_ia_{ii}=c\tr(A)$$ $$\tr(A+B)=\sum_i(a_{ii}+b_{ii})=\sum_ia_{ii}+\sum_ib_{ii}=\tr(A)+\tr(B)$$ $$\tr(CD)=\sum_i\sum_jC_{ij}D_{ji}=\sum_j\sum_iD_{ji}C_{ij}=\tr(DC)$$ $\blacksquare$

Proof. Since coordinate maps are linear, $\mu^i\in V^*$ for all $i$.

Suppose $\sum_j c_j\mu^j=0$. Then $$c_i=\sum_j c_j\phi(e_i)^j=\sum_j c_j\mu^j(e_i)=0(e_i)=0$$ for all $i$. We have shown that $(\mu^1,\ldots,\mu^n)$ is independent.

Suppose we have $f\in V^*$. Define $g=\sum_j f(e_j)\mu^j$, then $g\in V^*$, and for all $v\in V$, $$g(v)=\sum_i\phi(v)^ig(e_i)=\sum_i\phi(v)^i\sum_j f(e_j)\mu^j(e_i)=\sum_i\phi(v)^i\sum_j f(e_j)\phi(e_i)^j=\sum_i\phi(v)^if(e_i)=f(v)$$ Hence $(\mu^1,\ldots,\mu^n)$ spans $V^*$.

Therefore, $(\mu^1,\ldots,\mu^n)$ is a basis of $V^*$. $\blacksquare$

Proof. Let $B=(e_1,\ldots,e_n)$ be a basis of $V$, and $(\mu^1,\ldots,\mu^n)$ be the dual basis of $V^*$ for $B$. Then $\mu^j(v)=0$ for all $j$. Thus, let $\phi$ be the coordinate map with respect to $B$, then $\phi(v)^j=0$ for all $j$, implying $v=\sum_i0e_i=0$. $\blacksquare$

Proof. Let $v\in V$, then $w\mapsto w(v)$ defines a function $u_v:V^*\to F$. Let $w,w'\in V^*$ and $c\in F$, then $$u_v(w+w')=(w+w')(v)=w(v)+w'(v)=u_v(w)+u_v(w')$$ $$u_v(cw)=(cw)(v)=cw(v)=cu_v(w)$$ Thus $u_v\in V^{**}$. Then $v\mapsto u_v$ defines a function $\Phi:V\to V^{**}$. Note that for all $v\in V$, for all $v^*\in V^*$, $\Phi(v)(v^*)=v^*(v)$.

Let $v,v'\in V$ and $c\in F$, then for all $w\in V^*$, $$\Phi(v+v')(w)=u_{v+v'}(w)=w(v+v')=w(v)+w(v')=u_v(w)+u_{v'}(w)=\Phi(v)(w)+\Phi(v')(w)=(\Phi(v)+\Phi(v'))(w)$$ $$\Phi(cv)(w)=u_{cv}(w)=w(cv)=cw(v)=cu_v(w)=c\Phi(v)(w)=(c\Phi(v))(w)$$ Thus $\Phi$ is linear.

Suppose $v,v'\in V$ and $\Phi(v)=\Phi(v')$. Then for all $w\in V^*$, $w(v-v')=w(v)-w(v')=u_v(w)-u_{v'}(w)=\Phi(v)(w)-\Phi(v')(w)=0$. Thus $v-v'=0$, implying $v=v'$. Hence $\Phi$ is injective. Since $\dim V=\dim V^{**}$, $\Phi$ is bijective.

Since $\Phi$ is linear and bijective, it is an isomorphism. Uniqueness is trivial. $\blacksquare$

Proof. Let $(e_1,\ldots,e_n)$ be a basis of $V$, let $(\mu^1,\ldots,\mu^n)$ be its dual basis, and let $(\omega_1,\ldots,\omega_n)$ be the dual basis of the dual basis. Let $j\in\{1,\ldots,n\}$. Let $v^*\in V^*$, then $v^*=\sum_ic_i\mu^i$ for some $c_1,\ldots,c_n\in F$. Then $$\Phi(e_j)(v^*)=v^*(e_j)=\sum_ic_i\mu^i(e_j)=c_j=\sum_ic_i\omega_j(\mu^i)=\omega_j\p{\sum_ic_i\mu^i}=\omega_j(v^*)$$ Thus $\Phi(e_j)=\omega_j$. $\blacksquare$

Proof. Let $(e_1,\ldots,e_n),(e'_1,\ldots,e'_n)\in\mathcal B$ and $\mathcal F(e_1,\ldots,e_n)=\mathcal F(e'_1,\ldots,e'_n)$. Let $(\mu^1,\ldots,\mu^n)$ denote their common dual basis. Then $$\phi(e_i)^j=\mu^j(e_i)=\mu^j(e'_i)=\phi(e'_i)^j$$ where $\phi$ is the coordinate map defined with $(e_1,\ldots,e_n)$. Thus $(e_1,\ldots,e_n)=(e'_1,\ldots,e'_n)$. We have shown that $\mathcal F$ is injective.

Let $\Phi:V\to V^{**}$ be the canonical isomorphism. Define the function $\mathcal F^*:\mathcal B^*\to\mathcal B$ that maps a basis to its dual basis, then transforms it into a basis of $V$ by applying $\Phi^{-1}$ on each entry. Let $B\in\mathcal B$, let $B^*$ denote its dual, and let $B^{**}$ denote its double dual, then $\mathcal F^*(\mathcal F(B))$ is a basis of $V$ whose double dual is also $B^{**}$. By injectivity of dual maps, $\mathcal F^*(\mathcal F(B))=B$. Now let $B^*\in\mathcal B^*$ and let $B^{**}$ denote its dual. Then $\mathcal F^*(B^*)$ is a basis of $V$ whose double dual is $B^{**}$. Again, by injectivity of dual maps, $B^*$ is the dual of $\mathcal F^*(B^*)$, thus $\mathcal F(\mathcal F^*(B^*))=B^*$. We have shown that $\mathcal F$ is bijective, and $\mathcal F^{-1}=\mathcal F^*$. $\blacksquare$

Proof. $$ w(v) =\sum_jw_j\mu^j\p{\sum_iv^ie_i} =\sum_jw_jv^j =\sum_j\phi^*(w)_j\phi(v)^j =\phi^*(w)\phi(v) $$ $\blacksquare$

Proof. Let $B_1=(\mu^1,\ldots,\mu^j)$. Given $w\in V^*$, there exist $c_1,\ldots,c_n$ such that $w=\sum_ic_i\mu^i$. Thus $$\varphi_{B_1}(w)\psi_{B_2}(B_1) =\varphi_{B_1}\p{\sum_ic_i\mu^i}\psi_{B_2}(B_1) =\sum_ic_i\varphi_{B_1}(\mu^i)\psi_{B_2}(B_1) =\sum_ic_i\psi_{B_2}(B_1)_{i,*} =\sum_ic_i\varphi_{B_2}(\mu^i) =\varphi_{B_2}\p{\sum_ic_i\mu^i} =\varphi_{B_2}(w)$$ $\blacksquare$

Proof. Let $\phi,\phi'$ be the coordinate maps corresponding to $e,e'$, and $\psi,\psi'$ to $\mu,\mu'$. Then $$ (BA)_{ji} =B_{j,*}A_{*,i} =\psi(\mu^j)BA\phi(e_i) =\psi'(\mu^j)\phi'(e_i) =\mu^j(e_i) =I_{ji} $$ Thus $AB=BA=I$. $\blacksquare$

Proof. Trivial. $\blacksquare$

Proof. Let $T\in L(V_1,\ldots,V_k,F)$. Then for all $(v_1,\ldots,v_k)\in V_1\times\ldots\times V_k$, let $v_j^i$ denote the $i$-th coordinate of $v_j$ for each $j$, we have $$ \sum_{i\in I}T(e_{i_1}^1,\ldots,e_{i_k}^k)\bigotimes_j\mu_j^{i_j}(v_1,\ldots,v_k) =\sum_{i\in I}T(e_{i_1}^1,\ldots,e_{i_k}^k)\prod_j\mu_j^{i_j}(v_j) =\sum_{i\in I}T(e_{i_1}^1,\ldots,e_{i_k}^k)\prod_jv_j^{i_j} =T\p{\sum_iv_1^ie_i^1,\ldots,\sum_iv_k^ie_i^k} =T(v_1,\ldots,v_k) $$ Thus $(\bigotimes_j\mu_j^{i_j})_{i\in I}$ spans $L(V_1,\ldots,V_k,F)$.

Let $(c_i)_{i\in I}$ be elements of $F$ such that $\sum_{i\in I}c_i\bigotimes_j\mu_j^{i_j}=0$. Then for all $i'\in I$, $$ c_{i'} =\sum_{i\in I}c_i\prod_j\mu_j^{i_j}(e_{i'_j}^j) =\sum_{i\in I}c_i\bigotimes_j\mu_j^{i_j}(e_{i'_1}^1,\ldots,e_{i'_k}^k) =0 $$ Thus $(\bigotimes_j\mu_j^{i_j})_{i\in I}$ is linearly independent. $\blacksquare$

Proof. Trivial. $\blacksquare$

Proof. Given $f\in\mathcal F(S)$, $\sum_{\{u\in S|f(u)\neq0\}}f(u)A(u)\in X$. This defines a function $\overline A:\mathcal F(S)\to X$, which clearly satisfies the requirements. Suppose we have another linear function $\overline A':\mathcal F(S)\to X$ that satisfies the requirements, then for all $f\in\mathcal F(S)$, $$\overline A(f) =\sum_{\{u\in S|f(u)\neq0\}}f(u)A(u) =\sum_{\{u\in S|f(u)\neq0\}}f(u)\overline A'(u_{\mathcal F}) =\overline A'\p{\sum_{\{u\in S|f(u)\neq0\}}f(u)u_{\mathcal F}} =\overline A'(f)$$ Thus $\overline A=\overline A'$. $\blacksquare$

Proof. We will denote the map $u\mapsto u_{\mathcal F}$ as $\varphi$ and the map $v\mapsto v+\mathcal R$ as $\psi$, then $\pi=\psi\circ\varphi$. By the above lemma, $\overline T:\mathcal F(V_1\times\ldots\times V_k)\to X$ is a linear function such that $T(u)=\overline T(u_{\mathcal F})$ for all $u\in V_1\times\ldots\times V_k$, implying $T=\overline T\circ\varphi$. Let $u\in V_1\otimes\ldots\otimes V_k$, then for some $v_u\in\mathcal F(V_1\times\ldots\times V_k)$, $u=v_u+\mathcal R$, and we have $\overline T(v_u)\in X$. This defines a function $\tilde T:V_1\otimes\ldots\otimes V_k\to X$. Let $u\in\mathcal F(V_1\times\ldots\times V_k)$, then $\tilde T(u+\mathcal R)=\overline T(v)$, where $v\in\mathcal F(V_1\times\ldots\times V_k)$ and $u+\mathcal R=v+\mathcal R$, implying $v-u\in\mathcal R$. Let $(v_1,\ldots,v_k)\in V_1\times\ldots\times V_k,i\in\{1,\ldots,k\},c\in F$, then $$\overline T((v_1,\ldots,cv_i,\ldots,v_k)_{\mathcal F})=T(v_1,\ldots,cv_i,\ldots,v_k)=cT(v_1,\ldots,v_k) =c\overline T((v_1,\ldots,v_k)_{\mathcal F})=\overline T(c(v_1,\ldots,v_k)_{\mathcal F})$$ Let $(v_1,\ldots,v_k)\in V_1\times\ldots\times V_k,i\in\{1,\ldots,k\},v_i'\in V_i$, then $$\overline T((v_1,\ldots,v_i+v_i',\ldots,v_k)_{\mathcal F})=T(v_1,\ldots,v_i+v_i',\ldots,v_k)=T(v_1,\ldots,v_k)+T(v_1,\ldots,v_i',\ldots,v_k) =\overline T((v_1,\ldots,v_k)_{\mathcal F})+\overline T((v_1,\ldots,v_i',\ldots,v_k)_{\mathcal F}) =\overline T((v_1,\ldots,v_k)_{\mathcal F}+(v_1,\ldots,v_i',\ldots,v_k)_{\mathcal F})$$ We have shown that for all $w\in\mathcal R$, $\overline T(w)=0$. Thus $\tilde T(u+\mathcal R)=\overline T(v)=\overline T(u+(v-u))=\overline T(u)$. We have shown that $\tilde T\circ\psi=\overline T$, thus $$\tilde T\circ\pi=\tilde T\circ\psi\circ\varphi=\overline T\circ\varphi=T$$ Let $u,w\in V_1\otimes\ldots\otimes V_k$, then $v_{u+w}+\mathcal R=u+w=(v_u+\mathcal R)+(v_w+\mathcal R)=(v_u+v_w)+\mathcal R$. Thus $$\tilde T(u+w)=\overline T(v_{u+w})=\overline T(v_u+v_w)=\overline T(v_u)+\overline T(v_w)=\tilde T(u)+\tilde T(w)$$ Let $u\in V_1\otimes\ldots\otimes V_k$ and $c\in F$, then $v_{cu}+\mathcal R=cu=c(v_u+\mathcal R)=(cv_u)+\mathcal R$. Thus $$\tilde T(cu)=\overline T(v_{cu})=\overline T(cv_u)=c\overline T(v_u)=c\tilde T(u)$$ We have shown that $\tilde T$ is linear.

Suppose we have $\tilde T'$ that also satisfies the conditions. Let $u\in V_1\otimes\ldots\otimes V_k$, then there exists $v\in\mathcal F(V_1\times\ldots\times V_k)$ such that $u=\psi(v)$, and there exist $x_1,\ldots,x_l\in V_1\times\ldots\times V_k$ and $c^1,\ldots,c^l\in F$ for some $l\in N$, such that $v=\sum_ic^i\varphi(x_i)$. Thus $$\tilde T(u)=\tilde T\p{\psi\p{\sum_ic^i\varphi(x_i)}}=\tilde T\p{\sum_ic^i\pi(x_i)}=\sum_ic^iT(x_i)$$ Similarly, we have $\tilde T'(u)=\sum_ic^iT(x_i)=\tilde T(u)$. We have shown that $\tilde T'=\tilde T$. $\blacksquare$

Proof. For each $j$, let $(\mu_j^1,\ldots,\mu_j^{n_j})$ denote the dual basis of $(e_1^j,\ldots,e_{n_j}^j)$.

Let $T\in V_1\otimes\ldots\otimes V_k$. Then for some $u_1,\ldots,u_l\in V_1\times\ldots\times V_k$ where $l\in N$, $T$ is a linear combination of $\pi(u_1),\ldots,\pi(u_l)$, and each of which is a linear combination of $\p{\bigotimes_{j=1}^ke_{i_j}^j}_{i\in I}$, implying $T$ is a linear combination of $\p{\bigotimes_{j=1}^ke_{i_j}^j}_{i\in I}$. We have shown that $\p{\bigotimes_{j=1}^ke_{i_j}^j}_{i\in I}$ spans $V_1\otimes\ldots\otimes V_k$.

Let $(c^i)_{i\in I}$ be elements of $F$ such that $\sum_{i\in I}c^i\bigotimes_je_{i_j}^j=0$. Let $m\in I$, given $(v_1,\ldots,v_k)\in V_1\times\ldots\times V_k$, we have $\prod_j\mu_j^{m_j}(v_j)\in F$, which defines a function $\tau^m:V_1\times\ldots\times V_k\to F$ that is multilinear. By the above lemma, we have linear $\tilde{\tau^m}:V_1\otimes\ldots\otimes V_k\to F$ such that $\tilde{\tau^m}\circ\pi=\tau^m$. Then $$c^m=\sum_{i\in I}c^i\tau^m(e_{i_1}^1,\ldots,e_{i_k}^k)=\sum_{i\in I}c^i\tilde{\tau^m}(e_{i_1}^1\otimes\ldots\otimes e_{i_k}^k) =\tilde{\tau^m}\p{\sum_{i\in I}c^i\bigotimes_je_{i_j}^j}=0$$ We have shown that $\p{\bigotimes_{j=1}^ke_{i_j}^j}_{i\in I}$ is linearly independent. $\blacksquare$

Proof. Let $n,m$ denote $\dim V,\dim W$. Let $(a_1,\ldots,a_n),(b_1,\ldots,b_m)$ be bases of $V,W$ respectively. Note that the map $(v,w)\mapsto w\otimes v$ from $V\times W$ to $W\otimes V$, denoted $A$, is multilinear. Thus there exists linear $\tilde A:V\otimes W\to W\otimes V$ such that $\tilde A\circ\pi=A$. Then for all $v\in V$ and $w\in W$, $$\tilde A(v\otimes w)=w\otimes v$$ Let $T\in W\otimes V$, then $T=\sum_{j,i}T^{ji}b_j\otimes a_i$, implying $$ \tilde A\p{\sum_{j,i}T^{ji}a_i\otimes b_j} =\sum_{j,i}T^{ji}\tilde A(a_i\otimes b_j) =\sum_{j,i}T^{ji}b_j\otimes a_i =T $$ Thus $\tilde A$ is surjective. Since $V\otimes W$ and $W\otimes V$ share the same dimensionality, $\tilde A$ is bijective. Hence $\tilde A$ is an isomorphism. Suppose $\tilde A'$ is another such isomorphism, then $\tilde A$ and $\tilde A'$ agree on $a_i\otimes b_j$ for each $(i,j)$, implying they are equal. $\blacksquare$

Proof. Note that the map $(v_1,\ldots,v_{n_k})\mapsto\bigotimes_{j=1}^k\bigotimes_{i=n_{j-1}+1}^{n_j}v_i$, denoted $A$, is multilinear. Thus there exists linear $\tilde A:\bigotimes_{i=1}^{n_k}V_i\to\bigotimes_{j=1}^k\bigotimes_{i=n_{j-1}+1}^{n_j}V_i$ such that $\tilde A\circ\pi=A$. Then for all $(v_1,\ldots,v_{n_k})\in V_1\times\ldots\times V_{n_k}$, $$\tilde A\p{\bigotimes_{i=1}^{n_k}v_i}=\bigotimes_{j=1}^k\bigotimes_{i=n_{j-1}+1}^{n_j}v_i$$ Note that $\{\bigotimes_{j=1}^ku_j:(u_1,\ldots,u_k)\in\bigtimes_{j=1}^k\bigotimes_{i=n_{j-1}+1}^{n_j}V_i\}$ spans $\bigotimes_{j=1}^k\bigotimes_{i=n_{j-1}+1}^{n_j}V_i$. And for each $(u_1,\ldots,u_k)\in\bigtimes_{j=1}^k\bigotimes_{i=n_{j-1}+1}^{n_j}V_i$, $\bigotimes_{j=1}^ku_j$ is spanned by $\{\bigotimes_{j=1}^k\bigotimes_{i=n_{j-1}+1}^{n_j}v_i:(v_1,\ldots,v_{n_k})\in\bigtimes_{i=1}^{n_k}V_i\}$. Thus $\bigotimes_{j=1}^k\bigotimes_{i=n_{j-1}+1}^{n_j}V_i$ is spanned by $\{\tilde A(\bigotimes_{i=1}^{n_k}v_i):(v_1,\ldots,v_{n_k})\in\bigtimes_{i=1}^{n_k}V_i\}$. Hence $\tilde A$ is surjective. Since $\bigotimes_{i=1}^{n_k}V_i$ and $\bigotimes_{j=1}^k\bigotimes_{i=n_{j-1}+1}^{n_j}V_i$ share the same dimensionality, $\tilde A$ is bijective. We have shown that $\tilde A$ is an isomorphism. Suppose $\tilde A'$ is another such isomorphism, then $\tilde A$ and $\tilde A'$ agree on a basis of $\bigotimes_{i=1}^{n_k}V_i$, implying they are equal. $\blacksquare$

Proof. Let $(v_1,\ldots,v_k)\in V_1\times\ldots\times V_k$, then $\Phi_1(v_1)\otimes\ldots\otimes\Phi_k(v_k)\in W_1\otimes\ldots\otimes W_k$. This defines a function $A:V_1\times\ldots\times V_k\to W_1\otimes\ldots\otimes W_k$, which is multilinear. Thus we have linear $\tilde A:V_1\otimes\ldots\otimes V_k\cong W_1\otimes\ldots\otimes W_k$ such that $\tilde A\circ\pi=A$. Then for all $(v_1,\ldots,v_k)\in V_1\times\ldots\times V_k$, $$\tilde A(v_1\otimes\ldots\otimes v_k)=\Phi_1(v_1)\otimes\ldots\otimes\Phi_k(v_k)$$ Note that for each $j$, $\Phi_j$ sends a basis of $V_j$ to a basis of $W_j$. This means that $\tilde A$ sends a basis of $V_1\otimes\ldots\otimes V_k$ to a basis of $W_1\otimes\ldots\otimes W_k$, implying it is surjective. Since $V_1\otimes\ldots\otimes V_k$ and $W_1\otimes\ldots\otimes W_k$ share the same dimensionality, $\tilde A$ is bijective, and thus is an isomorphism. Suppose $\tilde A'$ is another such isomorphism, then $\tilde A$ and $\tilde A'$ agree on a basis of $V_1\otimes\ldots\otimes V_k$, implying they are equal. $\blacksquare$

Proof. Let $(w^1,\ldots,w^k)\in V_1^*\times\ldots\times V_k^*$, then $w^1\otimes\ldots\otimes w^k\in L(V_1,\ldots,V_k;F)$. This defines a function $A:V_1^*\times\ldots\times V_k^*\to L(V_1,\ldots,V_k;F)$, which is multilinear. Thus we have linear $\tilde A:V_1^*\otimes\ldots\otimes V_k^*\to L(V_1,\ldots,V_k;F)$ such that $\tilde A\circ\pi=A$. Then for all $(w^1,\ldots,w^k)\in V_1^*\times\ldots\times V_k^*$, the abstract tensor product $w^1\otimes\ldots\otimes w^k$ corresponds to the tensor product $w^1\otimes\ldots\otimes w^k$ by $\tilde A$. This means that $\tilde A$ sends a basis of $V_1^*\otimes\ldots\otimes V_k^*$ to a basis of $L(V_1,\ldots,V_k;F)$, implying it is surjective. Since $V_1^*\otimes\ldots\otimes V_k^*$ and $L(V_1,\ldots,V_k;F)$ share the same dimensionality, $\tilde A$ is bijective, and thus is an isomorphism. Suppose $\tilde A'$ is another such isomorphism, then $\tilde A$ and $\tilde A'$ agree on a basis of $V_1^*\otimes\ldots\otimes V_k^*$, implying they are equal. $\blacksquare$

Proof. $$ T =\sum_{I,J}T^I_Je_{I_1}\otimes\ldots\otimes e_{I_k}\otimes e^{J_1}\otimes\ldots\otimes e^{J_l} =\sum_{I,J}T^I_J\bigotimes_i\p{\sum_{i'}{A^{i'}}_{I_i}e'_{i'}}\otimes\bigotimes_j\p{\sum_{j'}{B^{J_j}}_{j'}e'^{j'}} =\sum_{I,J,I',J'}T^I_J\bigotimes_i\p{{A^{I'_i}}_{I_i}e'_{I'_i}}\otimes\bigotimes_j\p{{B^{J_j}}_{J'_j}e'^{J'_j}} =\sum_{I',J'}\p{\sum_{I,J}T^I_J\prod_i{A^{I'_i}}_{I_i}\prod_j{B^{J_j}}_{J'_j}}e'_{I'_1}\otimes\ldots\otimes e'_{I'_k}\otimes e'^{J'_1}\otimes\ldots\otimes e'^{J'_l} $$ $\blacksquare$

Proof. We will show for the special case where contraction is done on $T\in T^k_l(V)$ and there is only one pair of indexes next to each other to be summed over. The general case follows similarly. We will use the same notations as above, and use $S,S'$ to denote the resulting tensors of contraction obtained with $e,e'$ respectively. Then $$ \sum_b{T'^{I'b}}_{bJ'} =\sum_b\sum_{I,a,a',J}{T^{Ia}}_{a'J}\prod_i{A^{I'_i}}_{I_i}\p{{A^b}_a{B^{a'}}_{b}}\prod_j{B^{J_j}}_{J'_j} =\sum_{I,a,a',J}\p{\sum_b\p{{B^{a'}}_{b}{A^b}_a}}{T^{Ia}}_{a'J}\prod_i{A^{I'_i}}_{I_i}\prod_j{B^{J_j}}_{J'_j} =\sum_{I,a,a',J}\delta^{a'}_a{T^{Ia}}_{a'J}\prod_i{A^{I'_i}}_{I_i}\prod_j{B^{J_j}}_{J'_j} =\sum_{I,J}\sum_a{T^{Ia}}_{aJ}\prod_i{A^{I'_i}}_{I_i}\prod_j{B^{J_j}}_{J'_j} $$ Thus $$ S =\sum_{I,J}\p{\sum_a{T^{Ia}}_{aJ}}e_{I_1}\otimes\ldots\otimes e_{I_{k-1}}\otimes e^{J_1}\otimes\ldots\otimes e^{J_{l-1}} =\sum_{I',J'}\p{\sum_{I,J}\sum_a{T^{Ia}}_{aJ}\prod_i{A^{I'_i}}_{I_i}\prod_j{B^{J_j}}_{J'_j}}e'_{I'_1}\otimes\ldots\otimes e'_{I'_{k-1}}\otimes e'^{J'_1}\otimes\ldots\otimes e'^{J'_{l-1}} =\sum_{I',J'}\p{\sum_b{T'^{I'b}}_{bJ'}}e'_{I'_1}\otimes\ldots\otimes e'_{I'_{k-1}}\otimes e'^{J'_1}\otimes\ldots\otimes e'^{J'_{l-1}} =S' $$ $\blacksquare$

Proof. Clearly, given $T\in T^k(V^*)$, $(v_1,\ldots,v_k)\mapsto\frac{1}{k!}\sum_{\sigma\in S_k}\sgn(\sigma)T(v_{\sigma_1},\ldots,v_{\sigma_k})$ defines a function $A_T:V^k\to F$ such that $A_T\in T^k(V^*)$. Let $(v_1,\ldots,v_k)\in V^k$ and $\tau\in S_k$ be a transposition, then the function $\sigma\mapsto\tau\sigma$ is clearly a bijection from $S_k$ to $S_k$. Hence $$A_T(v_1,\ldots,v_k) =\frac{1}{k!}\sum_{\sigma\in S_k}\sgn(\sigma)T(v_{\sigma_1},\ldots,v_{\sigma_k}) =\frac{1}{k!}\sum_{\sigma\in S_k}\sgn(\tau\sigma)T(v_{\tau\sigma_1},\ldots,v_{\tau\sigma_k}) =-\frac{1}{k!}\sum_{\sigma\in S_k}\sgn(\sigma)T(v_{\tau\sigma_1},\ldots,v_{\tau\sigma_k}) =-A_T(v_{\tau_1},\ldots,v_{\tau_k})$$ We have shown that $A_T\in\Lambda^k(V^*)$. Thus $T\mapsto A_T$ defines a function $\Alt:T^k(V^*)\to\Lambda^k(V^*)$, and for all $T\in T^k(V^*)$, for all $(v_1,\ldots,v_k)\in V^k$, $$\Alt(T)(v_1,\ldots,v_k)=\frac{1}{k!}\sum_{\sigma\in S_k}\sgn(\sigma)T(v_{\sigma_1},\ldots,v_{\sigma_k})$$ $\blacksquare$

Proof. Let $\sigma\in S_k$. If $\sgn(\sigma)=1$, then $\sigma$ is a composition of an even number of elementary transpositions, hence $\omega(v_{\sigma_1},\ldots,v_{\sigma_k})=\omega(v_1,\ldots,v_k)$ for all $(v_1,\ldots,v_k)\in V^k$. If $\sgn(\sigma)=-1$, then $\sigma$ is a composition of an odd number of elementary transpositions, hence $\omega(v_{\sigma_1},\ldots,v_{\sigma_k})=-\omega(v_1,\ldots,v_k)$ for all $(v_1,\ldots,v_k)\in V^k$. In both cases, $\sgn(\sigma)\omega(v_{\sigma_1},\ldots,v_{\sigma_k})=\omega(v_1,\ldots,v_k)$. Hence $\Alt(\omega)(v_1,\ldots,v_k)=\omega(v_1,\ldots,v_k)$. $\blacksquare$

Proof. If $I$ is not injective, then given $v_1,\ldots,v_k\in V$, $\p{\mu^{I_i}(v_j)}_{ij}$ has repeated rows, thus $e^I=0$. Suppose $I$ and $J$ are injective and their ranges are equal. Suppose $\sigma$ is a permutation of $k$ and $\tau$ is an elementary transposition of $k$, and $e^{I\sigma}=e^I\sgn(\sigma)$, then given $v_1,\ldots,v_k\in V$, $e^{I\sigma\tau}(v_1,\ldots,v_k)=\det\p{\p{\mu^{{I\sigma\tau}_i}(v_j)}_{ij}}=-\det\p{\p{\mu^{{I\sigma}_i}(v_j)}_{ij}}=-e^{I\sigma}(v_1,\ldots,v_k)$. Thus $e^{I\sigma\tau}=-e^{I\sigma}=-e^I\sgn(\sigma)=e^I\sgn(\sigma\tau)$. By induction, we have $e^J=e^{II^{-1}J}=e^I\sgn(I^{-1}J)$. $\blacksquare$

Proof. If $I$ or $J$ is not injective, then $(\delta^{I_i}_{J_j})_{ij}$ has repeated rows or columns, thus $\delta^I_J=0$. If the range of $I$ does not equal the range of $J$, then $(\delta^{I_i}_{J_j})_{ij}$ has a zero row or a zero column, thus $\delta^I_J=0$. Suppose otherwise, then $(\delta^{I_i}_{I_j})_{ij}$ is the identity matrix, thus $\delta^I_I=1=\sgn(e)$ where $e$ is the identity permutation of $k$. Suppose $\sigma$ is a permutation of $k$ and $\tau$ is an elementary transposition of $k$, and $\delta^I_{I\sigma}=\sgn(\sigma)$, then $\delta^I_{I\sigma\tau}=\det\p{(\delta^{I_i}_{I\sigma\tau_j})_{ij}}=-\det\p{(\delta^{I_i}_{I\sigma_j})_{ij}}=-\delta^I_{I\sigma}=-\sgn(\sigma)=\sgn(\sigma\tau)$. By induction, $\delta^I_J=\delta^I_{II^{-1}J}=\sgn(I^{-1}J)$. $\blacksquare$

Proof. Let $\mathcal I$ denote the set of $k$-tuples of $n$. Let $\omega\in\Lambda^k(V^*)$. Let $I\in\mathcal I$.

• If $I$ is not injective, then $$ \sum_{J\in\mathcal J}\omega(e_{J_1},\ldots,e_{J_k})e^J(e_{I_1},\ldots,e_{I_k}) =\sum_{J\in\mathcal J}\omega(e_{J_1},\ldots,e_{J_k})\delta^J_I =0 =\omega(e_{I_1},\ldots,e_{I_k}) $$
• If $I$ is injective, then $$ \sum_{J\in\mathcal J}\omega(e_{J_1},\ldots,e_{J_k})e^J(e_{I_1},\ldots,e_{I_k}) =\sum_{J\in\mathcal J}\omega(e_{J_1},\ldots,e_{J_k})\delta^J_I =\omega(e_{J'_1},\ldots,e_{J'_k})\delta^{J'}_I =\omega(e_{I_1},\ldots,e_{I_k})\sgn(I^{-1}J')\sgn({J'}^{-1}I) =\omega(e_{I_1},\ldots,e_{I_k}) $$ where $J'\in\mathcal J$ uniquely shares the same range as $I$.

Thus given any $(v_1,\ldots,v_k)\in V^k$, $$ \sum_{J\in\mathcal J}\omega(e_{J_1},\ldots,e_{J_k})e^J(v_1,\ldots,v_k) =\sum_{J\in\mathcal J}\omega(e_{J_1},\ldots,e_{J_k})\sum_{I\in\mathcal I}e^J(e_{I_1},\ldots,e_{I_k})c^I =\sum_{I\in\mathcal I}\sum_{J\in\mathcal J}\omega(e_{J_1},\ldots,e_{J_k})e^J(e_{I_1},\ldots,e_{I_k})c^I =\sum_{I\in\mathcal I}\omega(e_{I_1},\ldots,e_{I_k})c^I =\omega(v_1,\ldots,v_k) $$ where $c^I$ is the product of the coordinates $v_1^{I_1},\ldots,v_k^{I_k}$. Hence $$\sum_{J\in\mathcal J}\omega(e_{J_1},\ldots,e_{J_k})e^J=\omega$$ We have shown that $(e^J)_{J\in\mathcal J}$ spans $\Lambda^k(V^*)$.

Suppose for some scalars $(c_J)_{J\in\mathcal J}$, $\sum_{J\in\mathcal J}c_Je^J=0$. Then for each $I\in\mathcal J$, $$ c_I =\sum_{J\in\mathcal J}c_J\delta^J_I =\sum_{J\in\mathcal J}c_Je^J(e_{I_1},\ldots,e_{I_k}) =0(e_{I_1},\ldots,e_{I_k}) =0 $$ We have shown that $(e^J)_{J\in\mathcal J}$ is independent. $\blacksquare$

Proof. Let $K$ be a $(k+l)$-tuple of $n$. Then $e^{IJ}(e_{K_1},\ldots,e_{K_{k+l}})=\delta^{IJ}_K$ and $$ e^I\wedge e^J(e_{K_1},\ldots,e_{K_{k+l}}) =\frac{(k+l)!}{k!l!}\Alt(e^I\otimes e^J)(e_{K_1},\ldots,e_{K_{k+l}}) =\frac{(k+l)!}{k!l!}\frac{1}{(k+l)!}\sum_{\sigma\in S_{k+l}}\sgn(\sigma)(e^I\otimes e^J)(e_{K\sigma_1},\ldots,e_{K\sigma_{k+l}}) =\frac{1}{k!l!}\sum_{\sigma\in S_{k+l}}\sgn(\sigma)e^I(e_{K\sigma_1},\ldots,e_{K\sigma_k})e^J(e_{K\sigma_{k+1}},\ldots,e_{K\sigma_{k+l}}) =\frac{1}{k!l!}\sum_{\sigma\in S_{k+l}}\sgn(\sigma)\delta^I_{K\sigma[1:k]}\delta^J_{K\sigma[k+1:k+l]} $$
Suppose $K$ is not injective, then both $e^I\wedge e^J(e_{K_1},\ldots,e_{K_{k+l}})$ and $e^{IJ}(e_{K_1},\ldots,e_{K_{k+l}})$ are clearly $0$. Now suppose $K$ is injective and $IJ$ is not injective, then for some distinct indexes $a,b$, $(IJ)_a=(IJ)_b$, and $e^{IJ}(e_{K_1},\ldots,e_{K_{k+l}})=0$.

• If $a,b\in\{1,\ldots,k\}$, then each $\delta^I_{K\sigma[1:k]}$ is $0$, thus $e^I\wedge e^J(e_{K_1},\ldots,e_{K_{k+l}})=0$.
• If $a,b\in\{k+1,\ldots,k+l\}$, then each $\delta^J_{K\sigma[k+1:k+l]}$ is $0$, thus $e^I\wedge e^J(e_{K_1},\ldots,e_{K_{k+l}})=0$.
• If one of $a$ and $b$ is in $\{1,\ldots,k\}$ and the other in $\{k+1,\ldots,k+l\}$, then for each $\sigma\in S_{k+l}$, either $\delta^I_{K\sigma[1:k]}=0$ or $\delta^J_{K\sigma[k+1:k+l]}=0$, thus $e^I\wedge e^J(e_{K_1},\ldots,e_{K_{k+l}})=0$.

Now suppose both $K$ and $IJ$ are injective. If the ranges of $K$ and $IJ$ do not match, then for each $\sigma\in S_{k+l}$, the ranges of $K\sigma$ and $IJ$ also do not match. Thus both $e^I\wedge e^J(e_{K_1},\ldots,e_{K_{k+l}})$ and $e^{IJ}(e_{K_1},\ldots,e_{K_{k+l}})$ are $0$. Now suppose the range of $K$ equals the range of $IJ$, then $K^{-1}(IJ)$ is a $(k+l)$-permutation. Let $\mathcal S$ denote the subset of $S_{k+l}$ such that $\sigma\in\mathcal S$ if and only if the restriction of $\sigma$ on $\{1,\ldots,k\}$ is a $k$-permutation. Then $$ e^I\wedge e^J(e_{K_1},\ldots,e_{K_{k+l}}) =\frac{1}{k!l!}\sum_{\sigma\in S_{k+l}}\sgn(K^{-1}(IJ)\sigma)\delta^I_{(IJ)\sigma[1:k]}\delta^J_{(IJ)\sigma[k+1:k+l]} =\frac{1}{k!l!}\sum_{\sigma\in\mathcal S}\sgn(K^{-1}(IJ)\sigma)\delta^I_{(IJ)\sigma[1:k]}\delta^J_{(IJ)\sigma[k+1:k+l]} =\frac{1}{k!l!}\sum_{\tau\in S_k}\sum_{\eta\in S_l}\sgn(K^{-1}(IJ))\sgn(\tau)\sgn(\eta)\delta^I_{I\tau}\delta^J_{J\eta} $$ $$ =\sgn(K^{-1}(IJ))\p{\frac{1}{k!}\sum_{\tau\in S_k}\sgn(\tau)\delta^I_{I\tau}}\p{\frac{1}{l!}\sum_{\eta\in S_l}\sgn(\eta)\delta^J_{J\eta}} =\sgn(K^{-1}(IJ))\Alt(e^I)(e_{I_1},\ldots,e_{I_k})\Alt(e^J)(e_{J_1},\ldots,e_{J_l}) =\sgn(K^{-1}(IJ))e^I(e_{I_1},\ldots,e_{I_k})e^J(e_{J_1},\ldots,e_{J_l}) =\sgn((IJ)^{-1}K) =e^{IJ}(e_{K_1},\ldots,e_{K_{k+l}}) $$
We have shown that for all $(k+l)$-tuple $K$ of $n$, $e^I\wedge e^J(e_{K_1},\ldots,e_{K_{k+l}})=e^{IJ}(e_{K_1},\ldots,e_{K_{k+l}})$. Given any $(v_1,\ldots,v_{k+l})\in V^{k+l}$, $$ e^I\wedge e^J(v_1,\ldots,v_{k+l}) =\sum_{K\in\mathcal K}e^I\wedge e^J(e_{K_1},\ldots,e_{K_{k+l}})c^K =\sum_{K\in\mathcal K}e^{IJ}(e_{K_1},\ldots,e_{K_{k+l}})c^K =e^{IJ}(v_1,\ldots,v_{k+l}) $$ where $\mathcal K$ is the set of $(k+l)$-tuples of $n$, and each $c^K$ is a scalar. Therefore, $$e^I\wedge e^J=e^{IJ}$$ $\blacksquare$

Proof. $$ (\omega_1+\omega_2)\wedge\eta =\frac{(k+l)!}{k!l!}\Alt((\omega_1+\omega_2)\otimes\eta) =\frac{(k+l)!}{k!l!}\Alt(\omega_1\otimes\eta+\omega_2\otimes\eta) =\frac{(k+l)!}{k!l!}\Alt(\omega_1\otimes\eta)+\frac{(k+l)!}{k!l!}\Alt(\omega_2\otimes\eta) =\omega_1\wedge\eta+\omega_2\wedge\eta $$ $$ \omega\wedge(\eta_1+\eta_2) =\frac{(k+l)!}{k!l!}\Alt(\omega\otimes(\eta_1+\eta_2)) =\frac{(k+l)!}{k!l!}\Alt(\omega\otimes\eta_1+\omega\otimes\eta_2) =\frac{(k+l)!}{k!l!}\Alt(\omega\otimes\eta_1)+\frac{(k+l)!}{k!l!}\Alt(\omega\otimes\eta_2) =\omega\wedge\eta_1+\omega\wedge\eta_2 $$ $$ (c\omega)\wedge\eta =\frac{(k+l)!}{k!l!}\Alt((c\omega)\otimes\eta) =\frac{(k+l)!}{k!l!}\Alt(c(\omega\otimes\eta)) =c\frac{(k+l)!}{k!l!}\Alt(\omega\otimes\eta) =c(\omega\wedge\eta) $$ $$ \omega\wedge(c\eta) =\frac{(k+l)!}{k!l!}\Alt(\omega\otimes(c\eta)) =\frac{(k+l)!}{k!l!}\Alt(c(\omega\otimes\eta)) =c\frac{(k+l)!}{k!l!}\Alt(\omega\otimes\eta) =c(\omega\wedge\eta) $$ Let $(e_1,\ldots,e_n)$ be a basis of $V$. $$ (\omega\wedge\eta)\wedge\theta =\p{\sum_I\omega_Ie^I\bigwedge\sum_J\eta_Je^J}\bigwedge\sum_K\theta_Ke^K =\sum_I\sum_J\sum_K\omega_I\eta_J\theta_K((e^I\wedge e^J)\wedge e^K) =\sum_I\sum_J\sum_K\omega_I\eta_J\theta_Ke^{IJK} =\sum_I\sum_J\sum_K\omega_I\eta_J\theta_K(e^I\wedge(e^J\wedge e^K)) =\sum_I\omega_Ie^I\bigwedge\p{\sum_J\eta_Je^J\bigwedge\sum_K\theta_Ke^K} =\omega\wedge(\eta\wedge\theta) $$ $$ \omega\wedge\eta =\sum_I\omega_Ie^I\bigwedge\sum_J\eta_Je^J =\sum_I\sum_J\omega_I\eta_Je^{IJ} =\sum_J\sum_I\eta_J\omega_Ie^{JI}(-1)^{kl} =(-1)^{kl}\p{\sum_J\eta_Je^J\bigwedge\sum_I\omega_Ie^I} =(-1)^{kl}\eta\wedge\omega $$ $\blacksquare$

Proof. Let $(e_1,\ldots,e_n)$ be a basis of $V$ and let $(\mu^1,\ldots,\mu^n)$ be the corresponding dual basis. $$ (w^1\wedge\ldots\wedge w^k)(v_1,\ldots,v_k) =\sum_Ic_I(\mu^{I_1}\wedge\ldots\wedge\mu^{I_k})(v_1,\ldots,v_k) =\sum_Ic_Ie^I(v_1,\ldots,v_k) =\sum_Ic_I\det\p{\p{\mu^{I_i}(v_j)}_{ij}} =\det\p{\p{w^i(v_j)}_{ij}} $$ $\blacksquare$