Proof. The sum \(\sum_{i=1}^n U_i\) is non-empty, since it contains the zero vector \(0_V.\) Let \(v\) and \(v^{\prime} \in \sum_{i=1}^n U_i\) so that \[v=v_1+v_2+\cdots +v_n \qquad \text{and} \qquad v^{\prime}=v^{\prime}_1+v^{\prime}_2+\cdots+v^{\prime}_n\] for vectors \(v_i,v^{\prime}_i \in U_i,\) \(i=1,\ldots,n.\) Each \(U_i\) is a vector subspace of \(V.\) Therefore, for all scalars \(s,t\in \mathbb{K},\) the vector \(sv_i+tv^{\prime}_i\) is an element of \(U_i,\) \(i=1,\ldots,n.\) Thus \[sv+tv^{\prime}=sv_1+tv^{\prime}_1+\cdots+sv_n+tv^{\prime}_n\] is an element of \(U_1+\cdots +U_n.\) By Definition 3.21, it follows that \(U_1+\cdots +U_n\) is a vector subspace of \(V.\)

Proof. Part of an exercise.

Proof. Suppose \((v_1,\ldots,v_m)\) is an ordered basis of \(U.\) By Theorem 3.64, there exists a basis \(\{v_1,\ldots,v_m,v_{m+1},\ldots,v_n\}\) of \(V.\) Defining \(U^{\prime}=\operatorname{span}\{v_{m+1},\ldots,v_n\},\) Proposition 6.8 implies the claim.

Proof. Let \(r=\dim(U_1\cap U_2)\) and let \(\{u_1,\ldots,u_r\}\) be a basis of \(U_1\cap U_2.\) These vectors are linearly independent and elements of \(U_1,\) hence by Theorem 3.64, there exist vectors \(v_1,\ldots,v_{m-r}\) so that \(\mathcal{S}_1=\{u_1,\ldots,u_r,v_1,\ldots,v_{m-r}\}\) is a basis of \(U_1.\) Likewise there exist vectors \(w_1,\ldots,w_{n-r}\) such that \(\mathcal{S}_2=\{u_1,\ldots,u_r,w_1,\ldots,w_{n-r}\}\) is a basis of \(U_2.\) Here \(m=\dim U_1\) and \(n=\dim U_2.\)

Now consider the set \(\mathcal{S}=\{u_1,\ldots,u_r,v_1,\ldots,v_{m-r},w_1,\ldots,w_{n-r}\}\) consisting of \(r+m-r+n-r=n+m-r\) vectors. If this set is a basis of \(U_1+U_2,\) then the claim follows, since then \(\dim(U_1+U_2)=n+m-r=\dim(U_1)+\dim(U_2)-\dim(U_1\cap U_2).\)

We first show that \(\mathcal{S}\) generates \(U_1+U_2.\) Let \(y \in U_1+U_2\) so that \(y=x_1+x_2\) for vectors \(x_1 \in U_1\) and \(x_2 \in U_2.\) Since \(\mathcal{S}_1\) is a basis of \(U_1,\) we can write \(x_1\) as a linear combination of elements of \(\mathcal{S}_1.\) Likewise we can write \(x_2\) as a linear combination of elements of \(\mathcal{S}_2.\) It follows that \(\mathcal{S}\) generates \(U_1+U_2.\)

We need to show that \(\mathcal{S}\) is linearly independent. So suppose we have scalars \(s_1,\ldots,s_r,\) \(t_1,\ldots,t_{m-r},\) and \(r_{1},\ldots,r_{n-r},\) so that \[\underbrace{s_1u_1+\cdots+s_r u_r}_{=u} +\underbrace{t_1v_1+\cdots+t_{m-r}v_{m-r}}_{=v}+\underbrace{r_1w_1+\cdots+r_{n-r}w_{n-r}}_{=w}=0_V.\] Equivalently, \(w=-u-v\) so that \(w \in U_1.\) Since \(w\) is a linear combination of elements of \(\mathcal{S}_2,\) we also have \(w \in U_2.\) Therefore, \(w \in U_1\cap U_2\) and there exist scalars \(\hat{s}_1,\ldots,\hat{s}_r\) such that \[w=\underbrace{\hat{s}_1u_1+\cdots+\hat{s}_r u_r}_{=\hat{u}}\] This is equivalent to \(w-\hat{u}=0_V,\) or written out \[r_1w_1+\cdots+r_{n-r}w_{n-r}-\hat{s}_1u_1-\cdots-\hat{s}_r u_r=0_V.\] Since the vectors \(\{u_1,\ldots,u_r,w_1,\ldots,w_{n-r}\}\) are linearly independent, we conclude that \(r_1=\cdots=r_{n-r}=\hat{s}_1=\cdots=\hat{s}_r=0.\) It follows that \(w=0_V\) and hence \(u+v=0_V.\) Again, since \(\{u_1,\ldots,u_r,v_1,\ldots,v_{m-r}\}\) are linearly independent, we conclude that \(s_1=\cdots=s_r=t_1=\cdots=t_{m-r}=0\) and we are done.

Proof. (i) We take \(\mathbf{C}=\mathbf{1}_{n}.\)

(ii) Suppose \(\mathbf{A}\) is similar to \(\mathbf{B}\) so that \(\mathbf{B}=\mathbf{C}\mathbf{A}\mathbf{C}^{-1}\) for some invertible matrix \(\mathbf{C}\in M_{n,n}(\mathbb{K}).\) Multiplying with \(\mathbf{C}^{-1}\) from the left and \(\mathbf{C}\) from the right, we get \[\mathbf{C}^{-1}\mathbf{B}\mathbf{C}=\mathbf{C}^{-1}\mathbf{C}\mathbf{A}\mathbf{C}^{-1}\mathbf{C}=\mathbf{A},\] so that the similarity follows for the choice \(\hat{\mathbf{C}}=\mathbf{C}^{-1}.\)

(iii) We have \(\mathbf{B}=\mathbf{C}\mathbf{A}\mathbf{C}^{-1}\) and \(\mathbf{X}=\mathbf{D}\mathbf{B}\mathbf{D}^{-1}\) for invertible matrices \(\mathbf{C},\mathbf{D}.\) Then we get \[\mathbf{X}=\mathbf{D}\mathbf{C}\mathbf{A}\mathbf{C}^{-1}\mathbf{D}^{-1},\] so that the similarity follows for the choice \(\hat{\mathbf{C}}=\mathbf{D}\mathbf{C}.\)

Proof. Let \(\mathbf{A}=(A_{ij})_{1\leqslant i,j\leqslant n}\) and \(\mathbf{B}=(B_{ij})_{1\leqslant i,j\leqslant n}.\) Then \[[\mathbf{A}\mathbf{B}]_{ij}=\sum_{k=1}^n A_{ik}B_{kj} \qquad \text{and}\qquad [\mathbf{B}\mathbf{A}]_{kj}=\sum_{i=1}^n B_{ki}A_{ij},\] so that \[\operatorname{Tr}(\mathbf{A}\mathbf{B})=\sum_{i=1}^n\sum_{k=1}^n A_{ik}B_{ki}=\sum_{k=1}^n\sum_{i=1}^n B_{ki}A_{ik}=\operatorname{Tr}(\mathbf{B}\mathbf{A}).\]

Proof. (i) Fix an ordered basis \(\mathbf{b}\) of \(V.\) Then, using Corollary 3.115 and Proposition 6.19, we obtain \[\begin{aligned} \operatorname{Tr}(f\circ g)&=\operatorname{Tr}\left(\mathbf{M}(f\circ g,\mathbf{b},\mathbf{b})\right)=\operatorname{Tr}\left(\mathbf{M}(f,\mathbf{b},\mathbf{b})\mathbf{M}(g,\mathbf{b},\mathbf{b})\right)\\ &=\operatorname{Tr}\left(\mathbf{M}(g,\mathbf{b},\mathbf{b})\mathbf{M}(f,\mathbf{b},\mathbf{b})\right)=\operatorname{Tr}\left(\mathbf{M}(g\circ f,\mathbf{b},\mathbf{b})\right)=\operatorname{Tr}(g\circ f). \end{aligned}\] The proof of (ii) is analogous, but we use Proposition 5.21 instead of Proposition 6.19.

Proof. The equivalence of the first three statements follows from Corollary 3.77. We fix an ordered basis \(\mathbf{b}\) of \(V.\) Suppose \(g\) is bijective with inverse \(g^{-1} : V \to V.\) Then we have \[\det (g\circ g^{-1})=\det(g)\det\left(g^{-1}\right)=\det\left(\mathrm{Id}_V\right)=\det\left(\mathbf{M}(\mathrm{Id}_V,\mathbf{b},\mathbf{b})\right)=\det\left(\mathbf{1}_{\dim V}\right)=1.\] It follows that \(\det(g)\neq 0\) and moreover that \[\det\left(g^{-1}\right)=\frac{1}{\det g}.\] Conversely, suppose that \(\det g\neq 0.\) Then \(\det \mathbf{M}(g,\mathbf{b},\mathbf{b}) \neq 0\) so that \(\mathbf{M}(g,\mathbf{b},\mathbf{b})\) is invertible by Corollary 5.22 and Proposition 3.116 implies that \(g\) is bijective.