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). \( A^n=a_nA^2+b_nA \)