1a (3pts) True: If \(\boldsymbol{u}\) is an eigenvector of both matrices \(A\) and \(B\), we have \(A\boldsymbol{u}=\lambda\boldsymbol{u}\) and \(B\boldsymbol{u}=\mu \boldsymbol{u}\): then \(AB \boldsymbol{u}= \mu A \boldsymbol{u}= \lambda\mu \boldsymbol{u}\). Vector \(\boldsymbol{u}\) is an eigenvector of \(AB\).
1b (3pts) False. For instance, the identity matrix \(I_n\) has only one eigenvalue.
1c (3pts) True. We have \(\det(\lambda I -A)= \det(\lambda I -A)^T= \det(\lambda I -A^T)\). \(A\) and \(A^T\) have the same characteristic polynomial, and thus have the same eigenvalues.
2a. We can write out matrix \(A=\text{mat}_{\cal B}L\) by finding \(a_{ij}=L(\boldsymbol{e}_j)\cdot\boldsymbol{e}_i\): \[ L(\boldsymbol{e}_1)= \omega_3\boldsymbol{e}_2-\omega_2\boldsymbol{e}_3,\quad L(\boldsymbol{e}_2)= -\omega_3 \boldsymbol{e}_1+\omega_1\boldsymbol{e}_3, \quad L(\boldsymbol{e}_3)=\omega_2\boldsymbol{e}_1-\omega_1\boldsymbol{e}_2 \] which yields \[ A= \begin{pmatrix} 0 & -\omega_3 & \omega_2\\ \omega_3 & 0 & -\omega_1\\ -\omega_2 &\omega_1 & 0 \end{pmatrix} \] Clearly, the matrix is skew-symmetric: \(a_{ji}=-a_{ij}\) and in particular \(a_{ii}=0\).
Note: operator \(L\) is skew-symmetric: \(L(\boldsymbol{u})\cdot\boldsymbol{v}= -\boldsymbol{u}\cdot L(\boldsymbol{v})\). This stems from the fact that \(L(\boldsymbol{u})=\boldsymbol{\omega}\times \boldsymbol{u}\).
We easily verify that \(L(\boldsymbol{\omega})=\boldsymbol{0}\) (or \(A\boldsymbol{\omega}=\boldsymbol{0}\)): this shows that \(\boldsymbol{\omega}\in\ker(A)\).
2b. Starting with \(\boldsymbol{e}_1\), we find vectors \(A\boldsymbol{e}_1\) and \(A^2 \boldsymbol{e}_1\): \[ A\boldsymbol{e}_1 = \begin{pmatrix} 0 \\ \omega_3 \\-\omega_2 \end{pmatrix}=\omega_3\boldsymbol{e}_2-\omega_2\boldsymbol{e}_3 \Longrightarrow A^2\boldsymbol{e}_1 = \omega_3 A\boldsymbol{e}_2-\omega_2 A\boldsymbol{e}_3 = \begin{pmatrix} -\omega_2^2-\omega_3^2 \\ \omega_1\omega_2 \\ \omega_1\omega_3 \end{pmatrix} =-\boldsymbol{e}_1 +\omega_1\boldsymbol{\omega} \] where we have used \(-\omega_2^2-\omega_3^2=\omega^2_1 -1\). Finally, \[ A^3\boldsymbol{e}_1= -A\boldsymbol{e}_1 +\omega_1 A\boldsymbol{\omega}= -A\boldsymbol{e}_1 \] using \(A\boldsymbol{\omega}=\boldsymbol{0}\). We conclude that \((A^3+A)\boldsymbol{e}_1= \boldsymbol{0}\). Therefore \(A^3+A=O\): this implies that \(p(A)= A^3+A\) is the minimum polynomial of \(A\).
We know that the zeros of the minimal polynomial of \(A\) are the eigenvalues of \(A\): if \(\lambda\) is an eigenvalue of \(A\), then there exists a non-zero vector \(\boldsymbol{v}\) such that \(A\boldsymbol{v}= \lambda\boldsymbol{v}\), leading to \[ A^3\boldsymbol{v}=\lambda^3 \boldsymbol{v}= -A\boldsymbol{v}= -\lambda\boldsymbol{v} \Longrightarrow (\lambda^3 +\lambda)\boldsymbol{v}=\boldsymbol{0} \Longrightarrow \lambda(\lambda^2+1)=0\Longrightarrow \lambda= 0,\pm i. \] We conclude that matrix \(A\) admits 3 distinct eigenvalues \(\lambda= 0,-i,i\): it is necessarily diagonalizable.
2c. Since \(\boldsymbol{\omega}\in\ker(A)\), we have \(A\boldsymbol{\omega}=\boldsymbol{0}\): \(\boldsymbol{\omega}\) is an eigenvector for \(\lambda=0\) and \(E(0,A)=\text{span}(\boldsymbol{\omega})\).
2d. To find \(\exp(At)\) we use the identity \(I+ At+A^2\frac{t^2}{2!}+\cdots + A^n \frac{t^n}{n!}+\cdots\). We then use the fact that \(A^3=-A\), \(A^4=-A^2\), \(A^5= A\), \(A^6=A^2\), etc: \[ A^{2p}= (-1)^{p+1} A^2,\qquad A^{2p-1}= (-1)^{p-1} A \quad (p\geq 1) \] This gives \[ \exp(At)= I + A\Big(t- \frac{t^3}{3!}+ \cdots+ (-1)^{p} \frac{t^{2p+1}}{(2p+1)!}+ \cdots\Big) +A^2 \Big( \frac{t^2}{2!} - \frac{t^4}{4!}+\cdots+ (-1)^{p+1} \frac{t^{2p}}{(2p)!}+\cdots\Big) \] which is recognized to be \[ \exp(At)= I +\sin(t) A+ \big(1-\cos(t)\big)A^2 \]
3a. We find matrix \(A\) in the standard basis of \(\mathbb{R}^3\): \[ A= \begin{pmatrix} 3 & -3 & 4\\ 0 & -1 & 4\\ 0 & -1 & 3 \end{pmatrix} \] whose characteristic polynomial is \[ p_A(\lambda)=\det(\lambda I-A)= (\lambda-3) \begin{vmatrix} \lambda+1 & -4 \\ 1 & \lambda-3 \end{vmatrix} =(\lambda-3)\big[(\lambda+1)(\lambda-3)+4\big] =(\lambda-3)(\lambda^2-2\lambda+1) =(\lambda-3)(\lambda-1)^2 \] We conclude that the eigenvalues of matrix \(A\) are \(\lambda=3\) (simple) and \(\lambda=1\) (double).
3b. Matrix \(J\) is the Jordan form of \(A\) (which is not diagonalizable). To find basis \((\boldsymbol{w}_1,\boldsymbol{w}_2,\boldsymbol{w}_3)\), we identify
vector \(\boldsymbol{w}_1\) as an eigenvector of \(A\) for eigenvalue \(\lambda=1\): we solve \((A-I)\boldsymbol{w}_1= \boldsymbol{0}\) \[ \begin{cases} 2x-3y+4z=0 \\ -2y+4z=0 \\ -y+2z=0 \end{cases} \Longrightarrow \begin{cases} x=z \\ y=2z \end{cases} \Longrightarrow (x,y,z)= z(1,2,1) \] We choose \(\boldsymbol{w}_1= (1,2,1)^T\) (we confirm that \(A\) is not diagonalizable since \(\dim E(1,A)\neq 2\)).
vector \(\boldsymbol{w}_2\) as a generalized eigenvector of \(A\) for eigenvalue \(\lambda=1\): we solve \((A- I)\boldsymbol{w}_2= \boldsymbol{w}_1\) \[ \begin{cases} 2x-3y+4z=1 \\ -2y+4z=2 \\ -y+2z=1 \end{cases} \Longrightarrow (x,y,z)= (-1,-1,0)+z(1,2,1) \] We choose \(\boldsymbol{w}_2= (-1,-1,0)^T\).
vector \(\boldsymbol{w}_3\) as an eigenvector of \(A\) for eigenvalue \(\lambda=3\): we solve \((A-3I)\boldsymbol{w}_3= \boldsymbol{0}\) \[ \begin{cases} -3y+4z=0 \\ -4y+4z=0 \\ -y=0 \end{cases} \Longrightarrow (x,y,z)= x(1,0,0) \] We choose \(\boldsymbol{w}_3= (1,0,0)^T\).
3c. We can write \(J= D+N\) with \[ D= \text{diag}(1,1,3), \qquad N= \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \] Matrix \(N\) is nilpotent since \(N^2 =O\). We have \(J^n= (D+N)^n\) and, since \(D\) and \(N\) commute, we have \[ J^n= D^n + n D^{n-1}N + \frac{n(n-1)}{2}D^{n-2}N^2 + \cdots = D^n + n D^{n-1}N \] where all other terms are zero since \(N^p=O\) for \(p\geq 2\). We find \[ J^n= \text{diag}(1,1,3^n)+ n \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 3^{n-1} \end{pmatrix} \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & n & 0\\ 0 & 1 & 0 \\ 0 & 0 & 3^n \end{pmatrix} \]
3d. Using \(P= (\boldsymbol{w}_1|\boldsymbol{w}_2|\boldsymbol{w}_3)\), we can write \[ A= P J P^{-1}= P D P^{-1}+ PNP^{-1}= B+ M \] where \(B= PDP^{-1}\) is diagonalizable, and \(M= PNP^{-1}\) is nilpotent since \(M^2 = PN^2 P^{-1}= O\).