Question

Let A be a square matrix defined by $A = \begin{pmatrix}-3&-1&-3\\ 5&2&5\\ -1&-1&-1\end{pmatrix}$

(a) Find the characteristic polynomial of A.

(b) Find the eigenvalues of A. Show that A is not diagonalizable over $\mathbb{R}$

(c) Show that A is diagonalizable over$\mathbb{C}$. Find the eigenspaces. Diagonalize A.

(d) Express $A^n$ in the form of $a_nA^2+b_ nA+c_nI_n$ where $(a_n), (b_n)$ and $(c_n)$ are real sequences to be specified.

$A=PDP^{-1},D=\begin{pmatrix}0&0&0\\ 0&-1-i&0\\ 0&0&-1+i\end{pmatrix},P=\begin{pmatrix}-1&-1-2i&-1+2i\\ 0&1+3i&1-3i\\ 1&1&1\end{pmatrix}$

Answer 1

Solution

(a) Find the characteristic polynomial of A.

$\implies P(\lambda)=|A-\lambda I|=\begin{vmatrix} -3-\lambda & -1 & -3\\ 5 & 3-\lambda & 5\\ 1- & -1 & -1 \end{vmatrix}$

$=(-3-\lambda)\begin{vmatrix} 2-\lambda & 5 \\ -1 & -1-\lambda \end{vmatrix}+\begin{vmatrix} 5 & 5 \\ -1 & -1-\lambda \end{vmatrix}-3\begin{vmatrix} 5 & 2-\lambda \\ -1 & -1 \end{vmatrix}$

Therefore. $P(\lambda)=-\lambda\bigg(\lambda^2+2\lambda+2\bigg)$

(b) Find the eigenvalues of A. Show that A is not diagonalizable over $\mathbb{R}$

since $P(\lambda)\hspace{2mm}is\hspace{2mm} not\hspace{2mm} splitted\hspace{2mm} over \hspace{2mm}\mathbb{R}$

Therefore. A is not diagonalizable over $\mathbb{R}$

c). Show that A is diagonalizable over $\mathbb{C}$. Find the eigenspaces. Diagonalize A.

we have  $P(\lambda)=-\lambda\bigg(\lambda^2+2\lambda+2\bigg)$

$\implies P(\lambda)=0\implies -\lambda\bigg(\lambda^2+2\lambda+2\bigg)=0$

$\iff \lambda_1=0,\lambda_2=-1+i,\lambda_3=-1-i$

Since over $\mathbb{C}$ : $P(\lambda)\hspace{2mm} is\hspace{2mm} splitted \implies P(\lambda)=-\lambda\bigg(\lambda+1+i\bigg)\bigg(\lambda+1-i\bigg)$

Then $am(\lambda)=gm(\lambda),\forall \lambda$  Thus. A is diagonalizable.

$\bullet\hspace{2mm}$Find diagonalize A

since. $A=PDP^{-1},D=\begin{pmatrix}0&0&0\\ 0&-1-i&0\\ 0&0&-1+i\end{pmatrix}$  

$For\hspace{2mm} \lambda_1=0\implies E_{\lambda_1}=\bigg\{x\in \mathbb{R} : Ax=0\bigg\}$

$\implies A=\begin{pmatrix}-3&-1&-3\\ 5&2&5\\ -1&-1&-1\end{pmatrix}\sim \begin{pmatrix}2&0&2\\ 3&0&3\\ 1&1&1\end{pmatrix}\sim \begin{pmatrix}1&1&1\\ 1&0&1\\ 0&0&0\end{pmatrix}$

Let $x_3=t\implies x_1=-t,x_2=0$

Thus.  $E_{\lambda_1}=span\left\{\begin{pmatrix}-1\\ 0\\ 1\end{pmatrix}\right\}$

$For\hspace{2mm} \lambda_2=-1-i$

$\implies A+(1+i)I=\begin{pmatrix}-2+i&-1&-3\\ \:5&3+i&5\\ -1\:&-1&i\end{pmatrix}\sim \begin{pmatrix}1&0&1+2i\\ 0&1&-1-3i\\ 0&0&0\end{pmatrix}$

Thus. $E_{\lambda_2}=span\left\{\begin{pmatrix}-1-2i\\ 1+3i\\ 1\end{pmatrix}\right\}$

$For\hspace{2mm} \lambda_3=-1+i$

$\implies A+(1-i)I=\begin{pmatrix}-2-i&-1&-3\\ 5&3-i&5\\ -1&-1&-i\end{pmatrix}\sim \begin{pmatrix}1&0&1-2i\\ 0&1&-1+3i\\ 0&0&0\end{pmatrix}$

Thus. $E_{\lambda_3}=span\left\{\begin{pmatrix}-1+2i\\ 1-3i\:\\1 \:\end{pmatrix}\right\}$

Therefore.  

$A=PDP^{-1},D=\begin{pmatrix}0&0&0\\ 0&-1-i&0\\ 0&0&-1+i\end{pmatrix},P=\begin{pmatrix}-1&-1-2i&-1+2i\\ 0&1+3i&1-3i\\ 1&1&1\end{pmatrix}$

(d) Express $A^n$ in the form of $a_nA^2+b_ nA+c_nI_n$where $(a_n), (b_n)$ and $(c_n)$ are real sequences to be specified.

we have $x^n=p(x)q(x)+a_nx^2+b_nx+c_n$

since. $P(0)=0,P(-1-i)=0\hspace{2mm}and\hspace{2mm}P(-1+i)=0$

$\implies \begin{cases} c_n=0 & \quad \\ (-1-i)^n=a_n(-1-i)^2+b_n(-1-i) \hspace{3mm}(1)& \quad \\ (-1+i)^n=a_n(-1+i)^2+b_n(-1+i) \hspace{3mm}(2) & \quad \end{cases}$

$(1)-(2) : (-1-i)^n-(-1+i)^n=b_n(-1-i+1-i)=b_n(-2i)$

$\implies b_n=\frac{1}{2}i\bigg[(-1-i)^n-(-1+i)^n\bigg]$

From (1) : $a_n=-2i(-1-i)^n-\bigg[(-1-i)^n-(-1+i)^n\bigg]$

Answer

Therefore.

a). $P(\lambda)=-\lambda\bigg(\lambda^2+2\lambda+2\bigg)$

b). A is not diagonalizable over $\mathbb{R}$

c). $A=PDP^{-1},D=\begin{pmatrix}0&0&0\\ 0&-1-i&0\\ 0&0&-1+i\end{pmatrix},P=\begin{pmatrix}-1&-1-2i&-1+2i\\ 0&1+3i&1-3i\\ 1&1&1\end{pmatrix}$

d). $A^n=a_nA^2+b_nA$