2023 Fall Final Exam Solution Set

Jan 12, 2024

Choice Answer ${15}^{\prime }$

• 1 $D$
• 2 $C$
• 3 $B$
• 4 $B$
• 5 $A$

二 ${25}^{\prime }$

(1)

$u^{\mathrm{\top }}u+1,1;1,n-1$

(2)

$\sqrt{5},\sqrt{7}$

(3)

$\left[\begin{array}{ccccc}0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 1& 2024& 1012\times 2023\\ 0& 0& 0& 1& 2024\\ 0& 0& 0& 0& 1\end{array}\right]$

(4)

$\frac{1}{2\sqrt{2}}[(1+\sqrt{2}{)}^{100}-(1-\sqrt{2}{)}^{100}]$

(5)

1,0

Question 3: ${12}^{\prime }$

(a)

There are two eigenvalues of $A:{\lambda }_1=0$ \＆${\lambda }_2=1$

$A$ basis for the eigen spacer $N(A)$ corresponding to ${\lambda }_1=0$ is:

$$\left[\begin{array}{c}-1\\ -1\\ 1\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}3\\ -2\\ 0\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}7\\ -3\\ 0\\ 0\\ 1\end{array}\right]$$

A basis for the eigenspace, $N(A-I)$, Corresponding to ${\lambda }_2=1$ is:

$$\left[\begin{array}{c}1\\ 1\\ -1\\ 1\\ 0\end{array}\right],\left[\begin{array}{l}3\\ 0\\ 0\\ 0\\ 1\end{array}\right]$$

Therefore,

$$S=\left[\begin{array}{ccccc}-1& 3& 7& 1& 3\\ -1& -2& -3& 1& 0\\ 1& 0& 0& -1& 0\\ 0& 1& 0& 1& 0\\ 0& 0& 1& 0& 1\end{array}\right]$$

(b)

Since $S^{-1}AS=\mathrm{\Lambda }$=$\left[\begin{array}{lllll}0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 1\end{array}\right]$,

$$\begin{array}{r}A=S\mathrm{\Lambda }{S}^{-1}\Rightarrow \\ A^k& =S{\mathrm{\Lambda }}^k{S}^{-1}\\ & =S\mathrm{\Lambda }{S}^{-1}=A\end{array}$$

Question 4: $8^{\prime }$

Orthonormalize the columns of $A$ by Gram-Schmidt to obtain

$$Q=\left[\begin{array}{ccc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{6}}& -\frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{6}}& \frac{1}{\sqrt{3}}\\ 0& \frac{2}{\sqrt{6}}& \frac{1}{\sqrt{3}}\end{array}\right]$$

Question 5: ${20}^{\prime }$

5-a

$$A^H={\left[\begin{array}{ccc}0& -i& 0\\ i& 1& i\\ 0& -i& 0\end{array}\right]}^H=\left[\begin{array}{ccc}0& -i& 0\\ i& 1& i\\ 0& -i& 0\end{array}\right]=A$$

$\Rightarrow A$ is Hermitian

5-b

$$\begin{array}{rl}& |A-\lambda I|=\left|\begin{array}{ccc}0-\lambda & -i& 0\\ i& 1-\lambda & i\\ 0& -i& 0-\lambda \end{array}\right|=(-\lambda )(-1{)}^{1+1}[(1-\lambda )(-\lambda )+i^2]\\ & +(-i)(-1{)}^{1+2}[(-\lambda )i]\\ & \Rightarrow -\lambda ({\lambda }^2-\lambda -1)+\lambda =-\lambda ({\lambda }^2-\lambda -2)=0\\ & \Rightarrow \lambda =0,-1,2\end{array}$$$$\begin{array}{ll}\lambda =0:(A-\lambda I)x=0& \Rightarrow x=\left[\begin{array}{c}1\\ 0\\ -1\end{array}\right]\\ \lambda =-1:(A-\lambda I)x=0& \Rightarrow x=\left[\begin{array}{c}1\\ -i\\ 1\end{array}\right]\\ \lambda =2:(A-\lambda I)x=0\Rightarrow x=\left[\begin{array}{c}1\\ 2i\\ 1\end{array}\right]\end{array}$$

5-c

$$U=\left[\begin{array}{ccc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{3}}& \frac{1}{\sqrt{6}}\\ 0& -\frac{i}{\sqrt{3}}& 2i/\sqrt{6}\\ -\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{3}}& \frac{1}{\sqrt{6}}\end{array}\right]$$

Question 6: ${10}^{\prime }$

6-a

$$A=\left[\begin{array}{ccc}\lambda -3& 2& 0\\ 2& \lambda & 0\\ 0& 0& \lambda -1\end{array}\right]$$$$\begin{array}{rl}& (\lambda -4)(\lambda +1)>0\\ & (\lambda -1)(\lambda -4)(\lambda +1)<0\end{array}$$

6-b

(B) $A$ is negativedefinite if and only if

$$\begin{array}{rl}& \text{ 1) }\det A_1=\lambda -3<0\\ & \text{ 2) }\det A_2={\lambda }^2-3\lambda -4>0\\ & \text{ 3) }\det A_3=(\lambda -1)({\lambda }^2-3\lambda -4)<0\\ & \begin{array}{c}\lambda -4>0\text{ or }\lambda <-1\\ \lambda -3<0\\ \lambda -1<0\end{array}\}\Rightarrow \lambda <-1\end{array}$$

Question 7

7- (a)

Since $A$ has $n$ distinct eigenvalues, $A$ is diagonalizable, ie.there exists an invertible $P$, such that

$$P^{-1}AP=\left[\begin{array}{llll}{\lambda }_1& & & \\ & {\lambda }_2& & \\ & & \ddots & \\ & & & {\lambda }_n\end{array}\right]$$$$AB=BA\Rightarrow P^{-1}AP{P}^{-1}BP=P^{-1}ABP=P^{-1}BAP$$$$=P^{-1}BP{P}^{-1}AP$$

Therefore, we can assume $A$ is a diagonal matrix, that is,

$$A=\left[\begin{array}{llll}{\lambda }_1& & & \\ & {\lambda }_2& & \\ & & \ddots & \\ & & & {\lambda }_n\end{array}\right]$$$$AB=\left[\begin{array}{cccc}{\lambda }_1{b}_{11}& {\lambda }_1{b}_{12}& ...& {\lambda }_1{b}_{1n}\\ {\lambda }_2{b}_{12}& {\lambda }_2{b}_{22}& ...& {\lambda }_2{b}_{2n}\\ ...& ...& ...& ...\\ {\lambda }_n{b}_{1n}& {\lambda }_n{b}_{n2}& ...& {\lambda }_n{b}_{nn}\end{array}\right]=\left[\begin{array}{cccc}{\lambda }_1{b}_{11}& {\lambda }_2{b}_{12}& ...& {\lambda }_n{b}_{1n}\\ {\lambda }_1{b}_{21}& {\lambda }_2{b}_{22}& ...& {\lambda }_n{b}_{2n}\\ ...& ...& ...& ...\\ {\lambda }_1{b}_{n1}& {\lambda }_2{b}_{n2}& ...& {\lambda }_n{b}_{nn}\end{array}\right]$$

Comparing the entries, we see that ${\lambda }_i{b}_{ij}={\lambda }_j{b}_{ij},{\lambda }_i\ne {\lambda }_j$, $(i\ne j$ hence $b_{ij}=0(i\ne j)$ .It follows that $B$ is a diagonal matrix

7-b

${}^{(c)}A$ and $B$ can be diagonalized by some invertible matrix $P$, icc

$$P^{-1}AP=\left[\begin{array}{cccc}{\lambda }_1& & & \\ & {\lambda }_2& & \\ & & \ddots & \\ & & & {\lambda }_n\end{array}\right],P^{-1}BP=\left[\begin{array}{cccc}{\mu }_1& & & \\ & {\mu }_2& & \\ & & \ddots & \\ & & & {\mu }_n\end{array}\right]$$

Where ${\lambda }_i,{\mu }_i$ are the eigenvalues of $A$ and $B$ Correspondingly

Choose $n$ polynomials as follows,

$$f_i(x)=\frac{(x-{\lambda }_1)\cdots (x-{\lambda }_{i-1})(x-{\lambda }_{i+1})\cdots (x-{\lambda }_n)}{({\lambda }_i-{\lambda }_1)\cdots ({\lambda }_i-{\lambda }_{i-1})({\lambda }_i-{\lambda }_{i+1})\cdots ({\lambda }_i-{\lambda }_n)},i=1,2,\cdots ,n$$

Let $f(x)={\mu }_1{f}_1(x)+{\mu }_2{f}_2(x)+\cdots +{\mu }_n{f}_n(x)$ .Then it can be verified that $f({\lambda }_i)={\mu }_i,i=1,2,\cdots ,n$

$$P^{-1}BP=\left[\begin{array}{llll}f({\lambda }_1)& & & \\ & f({\lambda }_2)& & \\ & & \ddots & \\ & & & f({\lambda }_n)\end{array}\right]=f(P^{-1}AP)=P^{-1}f(A)P$$

thus $B=f(A)$