(1)

$\begin{aligned}&\begin{bmatrix}1-\lambda&-1&0\\-1&2-\lambda&-1\\0&-1&1-\lambda\end{bmatrix}\\ &\sim \begin{bmatrix}0&-1+2-3\lambda+\lambda^2&\lambda-1\\-1&2-\lambda&-1\\0&-1&1-\lambda\end{bmatrix} \\&\sim\begin{bmatrix}\lambda^2-3\lambda+1&\lambda-1\\-1&1-\lambda\end{bmatrix}\\&\sim(\lambda-1)\begin{bmatrix}\lambda^2-3\lambda+1&1\\-1&-1\end{bmatrix}\\&\sim(\lambda-1)(\lambda^2-3\lambda)=\lambda(\lambda-1)(\lambda-3)\end{aligned}$

$\begin{aligned}\lambda_1=0\\\begin{bmatrix}1&-1&0\\-1&2&-1\\0&-1&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}\\ \sim \begin{bmatrix}1&-1&0\\0&1&-1\\0&-1&1\end{bmatrix} \\ \sim \begin{bmatrix}1&0&-1\\0&1&-1\\0&0&0\end{bmatrix}\\v_1=\begin{bmatrix}1\\1\\1\end{bmatrix}\end{aligned}$

$\begin{aligned}\lambda_2=1\\\begin{bmatrix}0&-1&0\\-1&1&-1\\0&-1&0\end{bmatrix}\\ \sim \begin{bmatrix}0&-1&0\\-1&0&-1\\0&0&0\end{bmatrix}\\v_2=\begin{bmatrix}1\\0\\-1\end{bmatrix}\end{aligned}$

$\begin{aligned}\lambda_3=3\\\begin{bmatrix}-2&-1&0\\-1&-1&-1\\0&-1&-2\end{bmatrix}\\\sim\begin{bmatrix}-2&0&2\\-1&0&1\\0&-1&-2\end{bmatrix}\\\sim\begin{bmatrix}0&0&0\\-1&0&-1\\0&-1&-2\end{bmatrix}\\v_3=\begin{bmatrix}-1\\-2\\1\end{bmatrix}\end{aligned}$

$\begin{aligned}A \begin{bmatrix}v_1&v_2&v_3\end{bmatrix}=\begin{bmatrix}v_1&v_2&v_3\end{bmatrix} \begin{bmatrix}\lambda_1&0&0\\0&\lambda_2&0\\0&0&\lambda_3\end{bmatrix}\\V \equiv \begin{bmatrix}v_1&v_2&v_3\end{bmatrix}=\begin{bmatrix}1&1&-1\\1&0&-2\\1&-1&1\end{bmatrix}\\A=V \begin{bmatrix}\lambda_1&0&0\\0&\lambda_2&0\\0&0&\lambda_3\end{bmatrix} V^{-1}\end{aligned}$

note. if distinct eigenvalues, eigenvectors must be independent. Say, $\lambda_1$ is the eigenvalue of max absolute value

$\begin{aligned}v_1=\sum_{i>1} a_i v_i\\v_1=\sum_{i>1} a_i \frac{\lambda_i}{\lambda_1} v_i\end{aligned}$

Apply infinite times $\frac{1}{\lambda_1}A$, then $v_1=0$. contradiction to non-zero eigenvector.

(a)

$\begin{aligned}A^k=V \begin{bmatrix}\lambda_1^k&0&0\\0&\lambda_2^k&0\\0&0&\lambda_3^k\end{bmatrix} V^{-1}\\ poly(A)=V \begin{bmatrix}poly(\lambda_1)&0&0\\0&poly(\lambda_2)&0\\0&0&poly(\lambda_3)\end{bmatrix} V^{-1}\end{aligned}$

Therefore $A^3$ has also eigenvectors $v_1,v_2,v_3$ with eigenvalues $0,1,27$, $A+10I$ has eigenvectors $v_1,v_2,v_3$ with eigenvalues $10,11,13$

(b)

$A^T=(V^{-1})^T \begin{bmatrix}\lambda_1&0&0\\0&\lambda_2&0\\0&0&\lambda_3\end{bmatrix} V^T$ so eigenvectors are columns of $(V^{-1})^T$ with the same eigenvalues

(c)

yes

(d)

not invertible. if invertible, then $\frac{1}{\lambda}$ are the eigenvalues

(e)

if one of the eigenvalue is 0, not invertible

(2)

(a)

$A=\begin{bmatrix}1&0\\0&1\end{bmatrix}$

(b)

$A=\begin{bmatrix}1&0\\1&1\end{bmatrix}$

(c)

(1)'s A

(d)

$A=\begin{bmatrix}0&0\\1&0\end{bmatrix}$

No special relation

(3)

(a)

no. (2)'s (d) is an example

(b)

put two different invertible B and set $C=\begin{bmatrix}0&0\\0&1\end{bmatrix}$

(c)

$\begin{bmatrix}0&0\\1&0\end{bmatrix}$ and $\begin{bmatrix}0&0\\2&0\end{bmatrix}$ can not be similar

$B\begin{bmatrix}0&0\\1&0\end{bmatrix}=\begin{bmatrix}b_{12}&0\\b_{22}&0\end{bmatrix}$

$\begin{bmatrix}0&0\\2&0\end{bmatrix} B=\begin{bmatrix}0&0\\2b_{21}&2b_{22}\end{bmatrix}$

So $B=\begin{bmatrix}b_{11}&0\\0&0\end{bmatrix}$ not invertible

(d)

yes. by the note in (1), $A=U \Lambda U^{-1}, C=V \Lambda V^{-1}$ then $C=(V U^{-1})A (V U^{-1})^{-1}$

(4)

(a)

$A u_1= u_1 \cdot 0, A u_2=u_2 \cdot 5$

(b)

$U=\begin{bmatrix}u_1,&u_2\end{bmatrix}=\begin{bmatrix}2&1\\-1&2\end{bmatrix},\Lambda=\begin{bmatrix}0&0\\0&5\end{bmatrix},A=U \Lambda U^{-1}$

$u_1 \cdot 1 + u_2 \cdot 1 = U \begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}3\\0\end{bmatrix}$ $u_1 \cdot 0 + u_2 \cdot 5 = U \begin{bmatrix}0\\5\end{bmatrix}=\begin{bmatrix}5\\10\end{bmatrix}$

(e)

segment vector of $u_i$ direction times $\lambda_i$