1. Solution: If $T$ is invertible, then there exists $S\in\ca L(V)$ such that $TS=ST=I$. Then it follows from 10.4 that\[\ca M(S,(v_1,\cdots,v_n))\ca M(T,(v_1,\cdots,v_n))=\ca M(ST,(v_1,\cdots,v_n))=I\]\[\ca M(T,(v_1,\cdots,v_n))\ca M(S,(v_1,\cdots,v_n))=\ca M(TS,(v_1,\cdots,v_n))=I.\]Hence $\ca M(T,(v_1,\cdots,v_n))$ is invertible.
If $\ca M(T,(v_1,\cdots,v_n))$ is invertible, then there exists $B\in \mb F^{n,n}$ such that \[B\ca M(T,(v_1,\cdots,v_n))=\ca M(T,(v_1,\cdots,v_n))B.\]Note that by 3.60, we have an isomorphism between $\ca L(V)$ and $\mb F^{(n,n)}$ by taking $W=V$ and $w_i=v_i$ for all $i$, we can choose $S\in\ca L(V)$ such that $B=\ca M(S,(v_1,\cdots,v_n))$. Again by 10.4, we have\[\ca M(ST,(v_1,\cdots,v_n))=B\ca M(T,(v_1,\cdots,v_n))=I,\]\[\ca M(TS,(v_1,\cdots,v_n))=\ca M(T,(v_1,\cdots,v_n))B=I.\]Therefore $ST=I$ and $TS=I$, hence $T$ is invertible.
See also Linear Algebra Done Right Solution Manual Chapter 10 Problem 1.
2. Solution: Let $V$ be $\mb F^{n}$, where $n$ is the number of rows of $A$. Fix a basis $v_1,\cdots,v_n$ of $V$. By 3.60, we choose $T,S\in\ca L(V)$ such that\[\ca M(T,(v_1,\cdots,v_n))=A\quad\text{and}\quad\ca M(S,(v_1,\cdots,v_n))=B.\]Since $AB=I$, it follows from 10.4 that\[\ca M(TS,(v_1,\cdots,v_n))=\ca M(T,(v_1,\cdots,v_n))\ca M(S,(v_1,\cdots,v_n))=AB=I,\]hence $TS=I$. By Problem 10 of Exercise 3D, we have $ST=I$. Again by 10.4, we have\[BA=\ca M(S,(v_1,\cdots,v_n))\ca M(T,(v_1,\cdots,v_n))=\ca M(ST,(v_1,\cdots,v_n))=I.\]See also Linear Algebra Done Right Solution Manual Chapter 10 Problem 2.
3. Solution: See Linear Algebra Done Right Solution Manual Chapter 10 Problem 3.
4. Solution: See Linear Algebra Done Right Solution Manual Chapter 10 Problem 4.
5. Solution: Let $V=\C^n$, where $n$ is the number of the rows of $B$. Choose a basis $v_1,\cdots,v_n$ of $V$. Let $T\in\ca L(V)$ so that\[\ca M(T,(v_1,\cdots,v_n))=B.\]By 8.29, there exists another basis $w_1,\cdots,w_n$ of $V$ such that $\ca M(T,(w_1,\cdots,w_n))$ is an upper-triangular matrix.
Let $A=\ca M(I,(w_1,\cdots,w_n),(v_1,\cdots,v_n))$, then $A$ is intertible. It follows from 10.7 that\[\ca M(T,(w_1,\cdots,w_n)=A^{-1}\ca M(T,(v_1,\cdots,v_n))A=A^{-1}BA.\]Since $\ca M(T,(w_1,\cdots,w_n))$ is an upper-triangular matrix, so is $A^{-1}BA$.
See also Linear Algebra Done Right Solution Manual Chapter 10 Problem 5.
6. Solution: Let $V=\R^2$, then $e_1=(1,0)$, $e_2=(0,1)$ is a basis of $V$. Let $T$ be the unique operator in $\ca L(V)$ such that $Te_1=e_2$ and $Te_2=-e_1$. Then \[T^2e_1=Te_2=-e_1\quad \text{and}\quad T^2e_2=-Te_1=-e_2.\]Hence $\ca M(T^2,(e_1,e_2))=-I$. In particular, $\m{trace}~T^2=-2 < 0$.
See also Linear Algebra Done Right Solution Manual Chapter 10 Problem 6.
7. Solution: See Linear Algebra Done Right Solution Manual Chapter 10 Problem 7.
8. Solution: See Linear Algebra Done Right Solution Manual Chapter 10 Problem 8.
9. Solution: See Linear Algebra Done Right Solution Manual Chapter 10 Problem 9.
10. Solution: Choose an orthonormal basis $e_1,\cdots,e_n$ of $V$. By 7.10, we have
\[\ca M(T^*,(e_1,\cdots,e_n))=\overline{\ca M(T,(e_1,\cdots,e_n))}^T.\]By the definition of the trace of a matrix, we have $\m{trace}~A^T=\m{trace}~A$ for any square matrix $A$. Therefore, by 10.16, we have\begin{align*}\m{trace}~T^*=&\m{trace}(\ca M(T^*,(e_1,\cdots,e_n)))\\=&\m{trace}(\overline{\ca M(T,(e_1,\cdots,e_n))}^T)\\=&\m{trace}(\overline{\ca M(T,(e_1,\cdots,e_n))})\\ =&\overline{\m{trace}(\ca M(T,(e_1,\cdots,e_n)))}\\ =&\overline{\m{trace}~T}.\end{align*}
Here we use the fact that $\m{trace}~\overline A=\overline{\m{trace}~A}$ for any square matrix $A$. Why? Prove it.
11. Solution: By 7.35, we know that $T$ is self-adjoint and all the eigenvalues of $T$ are nonnegative. By the Spectral Theorem, $T$ has a diagonal matrix with respect to some orthonormal basis of $V$. Fix this orthonormal basis of $V$. Then $\m{trace}~T=\m{trace}(\ca M(T))=0$.
Note that $\ca M(T)$ is a diagonal matrix with the eigenvalues of $T$ on the diagonal entries and the fact that all the eigenvalues of $T$ are nonnegative, it follows from $\m{trace}(\ca M(T))=0$ that all the eigenvalues of $T$ are zero. Hence $\ca M(T)=0$, which implies $T=0$.
12. Suppose $V$ is an inner product space and $P,Q\in\ca L(V)$ are orthogonal projections. Prove that $\m{trace}(PQ)\geq 0$.
13. Solution: Since $\m{trace}(T)$ is the sum of all eigenvalues and the sum of the diagonal entries of $\ca M(T)$. Hence the third eigenvalue of $T$ is\[51+(-40)+1-(-48)-24=36.\]See also Linear Algebra Done Right Solution Manual Chapter 10 Problem 12.
14. Solution: Choose a basis of $V$. We have $\m{trace}(cT)=\m{trace}(\ca M(cT))$ and $\m{trace}(T)=\m{trace}{\ca M(T)}$ by 10.16. Note that by 3.38, we have $\ca M(cT)=c\ca M(T)$. Therefore, we have\[\m{trace}(\ca M(cT))=\m{trace}(c\ca M(T))=c\m{trace}(\ca M(T)).\]Hence\[\m{trace}(cT)=\m{trace}(\ca M(cT))=c\m{trace}(\ca M(T))=c\m{trace}(T).\]
15. Solution: Choose a basis of $V$. By 10.4, 10.14 and 10.16, we have\begin{align*}&\m{trace}(ST)\\ \text{by 10.16}\quad=&\m{trace}(\ca M(ST))\\ \text{by 10.4}\quad=&\m{trace}(\ca M(S)\ca M(T))\\ \text{by 10.14}\quad=&\m{trace}(\ca M(T)\ca M(S))\\ \text{by 10.4}\quad=&\m{trace}(\ca M(ST))\\ \text{by 10.16}\quad=&\m{trace}(TS).\end{align*}
16. Solution: Let $V=\mb F^2$. Fix a basis $v_1,v_2$ of $V$. Let $S=T$, then $\m{trace}(ST)=\m{trace}(I)=2$ and\[\m{trace}(S)=\m{trace}(T)=2.\]However\[\m{trace}(ST)=2\ne 4=\m{trace}(S)\m{trace}(T).\]See also Linear Algebra Done Right Solution Manual Chapter 10 Problem 14.
17. Solution: We assume that $\dim V=n$. If $T\ne 0$, then $\m{Ker}(T)\ne V$. Let $v_{m+1},\cdots,v_n$ be a basis of $\m{Ker}(T)$, then $m\geq 1$. Extend $v_{m+1},\cdots,v_n$ to a basis $v_1,\dots,v_n$ of $V$. By the proof of 3.22, we have $Tv_1,\cdots,Tv_m$ is linearly independent in $V$. Hence we can extend $Tv_1,\cdots,Tv_m$ to a basis $Tv_1,\cdots,Tv_m,w_{m+1},\cdots,w_m$. Hence there exists $S\in \ca L(V)$ such that $S(Tv_1)=v_1$ (since $m\geq 1$, this is possible), $S(Tv_i)=0$ for all $i=2,\cdots,m$ and $Sw_j=0$ for all $j=m+1,\cdots,n$. Then we have \[STv_1=v_1,\quad STv_i=0,\quad STv_j=S0=0,\quad i=2,\cdots,m,\quad j=m+1,\dots,n.\]Thus $1,0$ are all the eigenvalues of $ST$. Moreover, the eigenvalue $1$ has multiplicity one while the eigenvalue $0$ has multiplicities $n-1$. Since trace is the sum of eigenvalues with mltiplicity, we have $\m{trace}(ST)=1$ and we get a contradiction. Therefore $T=0$.
See also Linear Algebra Done Right Solution Manual Chapter 10 Problem 15.
18. Solution: See Linear Algebra Done Right Solution Manual Chapter 10 Problem 16.
19. Solution: See Linear Algebra Done Right Solution Manual Chapter 10 Problem 18.
20. Solution: See Linear Algebra Done Right Solution Manual Chapter 10 Problem 17.
21. Solution: See Linear Algebra Done Right Solution Manual Chapter 10 Problem 19.
