Proof. Suppose \(\dim V=n,\) \(\dim W=m\) and write \(\mathbf{b}=(v_1,\ldots,v_n)\) and \(\mathbf{c}=(w_1,\ldots,w_m).\)

We first show that \(\Theta\) is linear. Let \(s_1,s_2 \in \mathbb{K}\) and \(f_1,f_2 \in \mathrm{Hom}(V,W).\) By definition \[\Theta(s_1 f_1+s_2 f_2)=\mathbf{M}(s_1 f_1+s_2 f_2,\mathbf{b},\mathbf{c}),\] where we omit writing \(\cdot_{\mathrm{Hom}(V,W)}\) and where we write \(+\) instead of \(+_{\mathrm{Hom}(V,W)}.\) Linearity means that \[\Theta(s_1 f_1+s_2 f_2)=s_1 \mathbf{M}(f_1,\mathbf{b},\mathbf{c})+s_2 \mathbf{M}(f_2,\mathbf{b},\mathbf{c}).\] Hence we need to show that \[\mathbf{M}(s_1 f_1+s_2 f_2,\mathbf{b},\mathbf{c})=s_1 \mathbf{M}(f_1,\mathbf{b},\mathbf{c})+s_2\mathbf{M}(f_2,\mathbf{b},\mathbf{c}).\] Write \[\mathbf{M}(f_1,\mathbf{b},\mathbf{c})=(A_{ij})_{1\leqslant i\leqslant m,1\leqslant j\leqslant n}\qquad \text{and}\qquad \mathbf{M}(f_2,\mathbf{b},\mathbf{c})=(B_{ij})_{1\leqslant i\leqslant m,1\leqslant j\leqslant n}.\] Recall from Proposition 3.108 that this means that for all \(1\leqslant j\leqslant n,\) we have \[f_1(v_j)=\sum_{i=1}^m A_{ij}w_i\qquad \text{and}\qquad f_2(v_j)=\sum_{i=1}^m B_{ij}w_i.\] Therefore, for all \(1\leqslant j\leqslant n,\) we obtain \[(s_1 f_1+s_2 f_2)(v_j)=s_1 f_1(v_j)+s_2 f_2(v_j)=\sum_{i=1}^m (s_1 A_{ij}+s_2 B_{ij})w_i\] so that \[\mathbf{M}(s_1 f_1+s_2 f_2,\mathbf{b},\mathbf{c})=s_1\mathbf{M}(f_1,\mathbf{b},\mathbf{c})+s_2 \mathbf{M}(f_2,\mathbf{b},\mathbf{c})\] as claimed.

We next show that \(\Theta\) is surjective. Let \(\mathbf{A}=(A_{ij})_{1\leqslant i\leqslant m, 1\leqslant j \leqslant n} \in M_{m,n}(\mathbb{K})\) and define \(f : V \to W\) as follows. For \(v\in V\) there exist unique scalars \(s_1,\ldots,s_n\) such that \(v=\sum_{i=1}^n s_i v_i\) (since \(\mathbf{b}\) is an ordered basis of \(V\)). We define \[f(v)=\sum_{j=1}^n\sum_{i=1}^m A_{ij}s_j w_i.\] Then \(f\) satisfies \(f(v_j)=\sum_{i=1}^m A_{ij}w_i\) for all \(1\leqslant j\leqslant n.\) Hence \(\Theta(f)=\mathbf{M}(f,\mathbf{b},\mathbf{c})=\mathbf{A}\) and \(\Theta\) is surjective.

If mappings \(f,g\in \mathrm{Hom}(V,W)\) satisfy \(\Theta(f)=\mathbf{M}(f,\mathbf{b},\mathbf{c})=\Theta(g)=\mathbf{M}(g,\mathbf{b},\mathbf{c}),\) then they agree in particular on the ordered basis \(\mathbf{b}\) and hence agree by Lemma 3.93. It follows that \(\Theta\) is injective as well and hence bijective and thus an isomorphism. Since \(\Theta\) is an isomorphism we have \(\dim \mathrm{Hom}(V,W)=\dim M_{m,n}(\mathbb{K})=mn=\dim(V)\dim(W).\)

Proof. We need to show that for all \(s_1,s_2 \in \mathbb{K}\) and \(\omega_1,\omega_2 \in W^*,\) we have \[f^T(s_1\omega_1+s_2\omega_2)=s_1f^T(\omega_2)+s_2f^T(\omega_2).\] This is a condition that needs to hold for all \(v \in V\) and indeed, by definition, we have for all \(v \in V\) \[\begin{aligned} v \,\lrcorner\,f^T(s_1\omega_1+s_2\omega_2)&=f(v)\,\lrcorner\,(s_1\omega_1+s_2\omega_2)=s_1\omega_1(f(v))+s_2\omega_2(f(v))\\ &=s_1(v\,\lrcorner\,f^T(\omega_1))+s_2(v\,\lrcorner\,f^T(\omega_2)), \end{aligned}\] as claimed.

Proof. Let \(\mathbf{b}=(v_1,\ldots,v_n),\) \(\mathbf{c}=(w_1,\ldots,w_m)\) and \(\boldsymbol{\beta}=(\nu_1,\ldots,\nu_n),\) \(\boldsymbol{\gamma}=(\omega_1,\ldots,\omega_m).\) Then, by definition, we have for all \(1\leqslant j\leqslant m\) \[f^T(\omega_j)=\sum_{i=1}^n [\mathbf{M}(f^T,\boldsymbol{\gamma},\boldsymbol{\beta})]_{ij}\nu_i.\] Hence for all \(1\leqslant k\leqslant n,\) we obtain \[\begin{aligned} v_k \,\lrcorner\,f^T(\omega_j)&=v_k \,\lrcorner\,\sum_{i=1}^n [\mathbf{M}(f^T,\boldsymbol{\gamma},\boldsymbol{\beta})]_{ij}\nu_i=\sum_{i=1}^n[\mathbf{M}(f^T,\boldsymbol{\gamma},\boldsymbol{\beta})]_{ij}(v_k\,\lrcorner\,\nu_i)\\ &=\sum_{i=1}^n[\mathbf{M}(f^T,\boldsymbol{\gamma},\boldsymbol{\beta})]_{ij}\nu_i(v_k)=[\mathbf{M}(f^T,\boldsymbol{\gamma},\boldsymbol{\beta})]_{kj}, \end{aligned}\] where the last equality uses that \(\nu_i(v_k)=\delta_{ik}.\) By definition, we also have \[\begin{aligned} v_k\,\lrcorner\,f^T(\omega_j)&=f(v_k)\,\lrcorner\,\omega_j=\left(\sum_{i=1}^m [\mathbf{M}(f,\mathbf{b},\mathbf{c})]_{ik}w_i\right)\,\lrcorner\,\omega_j\\ &=\sum_{i=1}^m[\mathbf{M}(f,\mathbf{b},\mathbf{c})]_{ik}\omega_j(w_i)=[\mathbf{M}(f,\mathbf{b},\mathbf{c})]_{jk}=[\mathbf{M}(f,\mathbf{b},\mathbf{c})^T]_{kj}, \end{aligned}\] where the second last equality uses \(\omega_j(w_i)=\delta_{ji}.\)

Proof. The proof is an exercise.

Proof. Choose a complement \(U^{\prime}\) of \(U\) in \(V\) so that \(V=U\oplus U^{\prime}.\) Recall that such a complement exists by Corollary 6.11. Consequently, every vector \(v \in V\) can be written uniquely as \(v=u+u^{\prime}.\) We then define \(\Omega(v)=\omega(u).\)

Proof of Proposition 13.16. We use the rank-nullity Theorem 3.76. Recall that the identity mapping of \(U\) is the linear mapping from \(U\) to \(U\) which returns its input \(\mathrm{Id}_U(u)=u\) for all \(u \in U.\) Since \(U\subset V,\) we can also think of the identity mapping on \(U\) as a mapping into \(V,\) \(\mathrm{Id}_U : U \to V.\) Applying the rank-nullity theorem to the transpose \(\mathrm{Id}_U^T : V^* \to U^*,\) we obtain \[\dim V=\dim V^*=\dim \operatorname{Ker}(\mathrm{Id}_U^T)+\dim \operatorname{Im}(\mathrm{Id}_U^T),\] where the first equality uses Remark 13.4. By definition we have \[\operatorname{Ker}(\mathrm{Id}_U^T)=\left\{\nu \in V^*\,|\,\mathrm{Id}^T_U(\nu)=\nu\circ \mathrm{Id}_U=0_{U^*}\right\}.\] Again by definition \(0_{U^*}\) is the linear map \(o : U \to \mathbb{K}\) which satisfies \(o(u)=0\) for all \(u \in U.\) Therefore we have \[\operatorname{Ker}(\mathrm{Id}_U^T)=\left\{\nu \in V^*\,|\, \nu(u)=0\;\forall\; u \in U\right\}=U^0.\] We want to show next that \(\mathrm{Id}_U^T : V^* \to U^*\) is surjective. Let \(\omega \in U^*,\) by the previous lemma we have \(\Omega \in V^*\) so that \(\Omega(u)=\omega(u)\) for all \(u \in U.\) Now notice that for all \(u \in U\) we have \[u\,\lrcorner\,\mathrm{Id}^T_U(\Omega)=u\,\lrcorner\,(\Omega\circ \mathrm{Id}_U)=\Omega(u)=\omega(u)=u\,\lrcorner\,\omega\] and hence \(\mathrm{Id}^T_U\) is surjective. It follows that \(\dim \operatorname{Im}(\mathrm{Id}_U^T)=\dim U^*=\dim U.\) Putting all together, we obtain \[\dim V=\dim U+\dim U^0,\] as claimed.

Proof. (i) An element \(\omega \in W^*\) lies in the kernel of \(f^T : W^* \to V^*\) if and only if \[v\,\lrcorner\,f^T(\omega)=0=f(v) \,\lrcorner\,\omega\] for all \(v \in V.\) Equivalently, \(w \,\lrcorner\,\omega=0\) for all elements \(w\) in the image of \(f,\) that is, \(\omega \in (\operatorname{Im}f)^0.\)

(ii) We have \[\dim \operatorname{Ker}f^T=\dim (\operatorname{Im}f)^0=\dim W-\dim \operatorname{Im}f=\dim \operatorname{Ker}f+\dim W-\dim V.\] The first equality uses (i), the second equality uses Proposition 13.16 and the last equality uses the rank-nullity Theorem 3.76.

Proof. The linear map \(f : V \to W\) is surjective if and only if \(\operatorname{Im}(f)=W,\) equivalently \(\operatorname{Im}(f)^0=\{0_{W^*}\}=\operatorname{Ker}(f^T),\) where the second equality uses the previous proposition. By the characterisation of injectivity of a linear map, Lemma 3.31, we have \(\{0_{W^*}\}=\operatorname{Ker}(f^T)\) if and only if \(f^T\) is injective.

Proof. (i) We have \[\dim \operatorname{Im}(f^T)=\dim W^*-\dim \operatorname{Ker}(f^T)=\dim W^*-\dim \operatorname{Im}(f)^0=\dim \operatorname{Im}(f),\] where the first equality uses the rank-nullity Theorem 3.76, the second equality uses Proposition 13.18 and the third equality uses Proposition 13.16.

(ii) First suppose \(\nu \in \operatorname{Im}(f^T).\) Then there exists \(\omega \in W^*\) with \(f^T(\omega)=\nu.\) We want to argue that \(\nu \in (\operatorname{Ker}f)^0.\) By definition \[(\operatorname{Ker}f)^0=\left\{\nu \in V^* | \nu(v)=0\;\forall v \in \operatorname{Ker}f\right\}.\] Let \(v \in \operatorname{Ker}f,\) then \[\nu(v)=v\,\lrcorner\,\nu=v \,\lrcorner\,f^T(\omega)=f(v)\,\lrcorner\,\omega=0_W\,\lrcorner\,\omega=0.\] It follows that \(\operatorname{Im}(f^T)\subset (\operatorname{Ker}f)^0.\) We complete the proof by showing that \(\operatorname{Im}(f^T)\) and \((\operatorname{Ker}f)^0\) have the same dimension. We compute \[\dim \operatorname{Im}(f^T)=\dim \operatorname{Im}(f)=\dim V-\dim \operatorname{Ker}(f)=\dim \operatorname{Ker}(f)^0,\] where the first equality uses (i), the second equality uses the rank-nullity Theorem 3.76 and the last equality uses Proposition 13.16.

Proof. Recall that surjectivity of \(f^T\) means that \(\operatorname{Im}(f^T)=V^*.\) By the characterisation of injectivity, Lemma 3.31, \(f\) is injective if and only if \(\operatorname{Ker}f=\{0_V\},\) equivalently, \((\operatorname{Ker}f)^0=V^*=\operatorname{Im}(f^T),\) by the previous proposition.

Proof. The column rank of \(\mathbf{A}\) equals \(\dim \operatorname{Im}(f_\mathbf{A}).\) Now \[\dim \operatorname{Im}(f_\mathbf{A})=\dim \operatorname{Im}((f_\mathbf{A})^T)=\dim \operatorname{Im}(f_{\mathbf{A}^T}),\] where we first use Proposition 13.20 and then Proposition 13.10. Since the matrix transpose interchanges the role of rows and columns, \(\dim \operatorname{Im}(f_{\mathbf{A}^T})\) is equal to the number of linearly independent row vectors of \(\mathbf{A}.\)