Question

Let A be a square matrix defined by \( A =\begin{pmatrix}-8&-3&-6\\ 4&0&4\\ 4&2&2\end{pmatrix} \)

(a) Find the characteristic polynomial of A.

(b) Find the eigenvalues and eigenspaces of A.

(c) Show that A is not diagonalizable, but it is triangularizable, then triangularize A.

(d) Write \( A^n \) in terms of \( I, A,A^2 \) and n.

 

Answer 1

Solution

(a) Find the characteristic polynomial of A.

we have \( A =\begin{pmatrix}-8&-3&-6\\ 4&0&4\\ 4&2&2\end{pmatrix} \)

\( \implies P(\lambda)=|A-\lambda I|=-\lambda^3-6\lambda^2-12\lambda-8=-\bigg(\lambda+2\bigg)^3 \)

So. \( P(\lambda)=-\bigg(\lambda+2\bigg)^3 \)

(b) Find the eigenvalues and eigenspaces of A.

\( \forall \lambda\in sp(A)\iff P(\lambda)=0\implies \lambda=2 \hspace{2mm}with \hspace{2mm}am(-2)=3 \)

\( \bullet \hspace{2mm}Find\hspace{2mm} eigenspace \)

\( A+2I=\begin{pmatrix}-6&-3&-6\\ 4&2&4\\ 4&2&4\end{pmatrix}\sim \begin{pmatrix}-6&-3&-6\\ 0&0&0\\ 0&0&0\end{pmatrix} \implies 2x_1+x_2+2x_3=0 \)

Thus, \( E_2=span\left\{\begin{pmatrix}1\\ -2\\ 0\end{pmatrix},\begin{pmatrix}0\\ -2\\ 1\end{pmatrix}\right\} \)

(c) Show that A is not diagonalizable, but it is triangularizable, then triangularize A.

since. \( P(\lambda) \) $ is splitted and \( dim(E_2)=2 \neq am(-2 \))

Then A is not diaginalizable but it's triangularizable.

\( \bullet \hspace{2mm}Triangularizable \hspace{2mm}it \)

That is \( A=PJP^{-1} \hspace{2mm} where\hspace{2mm} J=\begin{pmatrix}-2&1&0\\ 0&-2&0\\ 0&0&-2\end{pmatrix} \)

we have \( A+2I=\begin{pmatrix}-6&-3&-6\\ 4&2&4\\ 4&2&4\end{pmatrix} \) 

\( \implies \bigg(A+2I\bigg)^2=\begin{pmatrix}-6&-3&-6\\ 4&2&4\\ 4&2&4\end{pmatrix}\begin{pmatrix}-6&-3&-6\\ 4&2&4\\ 4&2&4\end{pmatrix}=0 \)

Then, \( G_{\lambda}=\left\{x\in \mathbb{R} : \bigg(A+2I\bigg)^2x=0\right\} \hspace{2mm} \)

\( \implies\hspace{2mm} G_{\lambda}=\left\{\begin{pmatrix}1\\ 0\\ 0\end{pmatrix},\begin{pmatrix}0\\ 1\\ 0\end{pmatrix},\begin{pmatrix}0\\ 0\\ 1\end{pmatrix}\right\} \)

\( V_2\in G_{\lambda}-R_{\lambda}\implies V_2=\begin{pmatrix}1\\ 0\\ 0\end{pmatrix}\hspace{2mm} \) then, \( V_1=\bigg(A+2I\bigg),V_2=\begin{pmatrix}-6&-3&-6\\ 4&2&4\\ 4&2&4\end{pmatrix}\begin{pmatrix}1\\ 0\\ 0\end{pmatrix}=\begin{pmatrix}-6\\ 4\\ 4\end{pmatrix} \)

Therefore , \( A=PDP^{-1}\hspace{2mm}J=\begin{pmatrix}-2&1&0\\ 0&-2&0\\ 0&0&-2\end{pmatrix},P=\begin{pmatrix}-6&1&1\\ 4&0&-2\\ 4&0&0\end{pmatrix} \)

(d) Write \( A^n \) in terms of \( I, A,A^2 \) and n.

we have \( P(\lambda)=-\bigg(\lambda+2\bigg)^3\hspace{2mm} or\hspace{2mm} P(x)=-\bigg(x+2\bigg)^3 \)

Let  \( f(x)=x^n \)

\( \implies x^n=P(x)Q(x)+Ax^2+Bx+C \hspace{2mm},A,B,C\in \mathbb{R} \)

\( nx^{n-1}=P'(x)Q(x)+Q'(x)P(x)+2Ax+B \)

\( n\bigg(n-1\bigg)x^{n-2}=P''(x)Q(x)+Q'(x)P'(x)+Q''(x)P(x)+P'(x)Q'(X)+2A \)

we have \( P(2)=P'(2)=P''(2)=0 \)

\( \iff \begin{cases} 2^n=4A+2B+C & \quad \\ n2^{n-1}=2^2A+B & \quad \\ n(n-1)2^{n-2}=2A & \quad \end{cases} \implies \begin{cases} A=n(n-1)2^{n-3} & \quad \\ B=n2^{n-1}-n(n-1)2^{n-1} & \quad \\ C=2^n-n(n-1)2^{n-1}-2B & \quad \end{cases} \)

\( \implies \begin{cases} A=n(n-1)2^{n-3} & \quad \\ B=2^{n-1}(2n-n^2)=n2^n-n^22^{n-1}=(2n-n^2)^{n-1} & \quad \\ C=2^n-n^22^{n-1}+n2^{n-1}-2n2^n+2n^22^{n-1}=2^{n-1}(2n^2+2-3n) & \quad \end{cases} \)

\( \iff \hspace{2mm}A^n=n(n-1)2^{n-3}A^2 +(2n-n^2)2^{n-1}A+(2n^2-3n+2)2^{n-1}I \)

Answer

Therefore.

a). \( P(\lambda)=-\bigg(\lambda+2\bigg)^3 \)

b). \( E_2=span\left\{\begin{pmatrix}1\\ -2\\ 0\end{pmatrix},\begin{pmatrix}0\\ -2\\ 1\end{pmatrix}\right\} \)

c). \( A=PDP^{-1}\hspace{2mm}J=\begin{pmatrix}-2&1&0\\ 0&-2&0\\ 0&0&-2\end{pmatrix},P=\begin{pmatrix}-6&1&1\\ 4&0&-2\\ 4&0&0\end{pmatrix} \)

d). \( A^n= n\left(n-1\right)2^{n-3}\begin{pmatrix}-8&-3&-6\\ \:4&0&4\\ \:4&2&2\end{pmatrix}^2+\left(2n-n^2\right)2^{n-1}A+\left(2n^2-3n+2\right)2^{n-1}\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix} \)