Question

Let A be a square matrix defined by \( A = \begin{pmatrix}6&2&3\\ -3&-1&-1\\ -5&-2&-2\end{pmatrix} \)L be a map from\( \hspace{2mm} \mathbb{R}^3\hspace{2mm} \)into\( \hspace{2mm}\mathbb{R}^3\hspace{2mm} \)by\( \hspace{2mm} L(v) = Av. \)

(a) Show that L is a linear operator on \( \hspace{2mm}\mathbb{R}^3. \)

(b) Find the characteristic polynomial of L with respect to standard basis for \( \mathbb{R}^3 \) Derive the determinant of L then deduce that L is invertible.

(c) Find the eigenvalues and eigenspaces of L.

(d) Show that L is not diagonalizable, but it is triangularizable, then triangularize L.

(e) Write \( L^n \) in term of n, where \( L^n = L(L(...(L)..)) \), the n compositions of L.

Answer 1

Solution

(a) Show that L is a linear operator on \( \hspace{2mm}\mathbb{R^3}. \)

Let \( u,v\in \mathbb{R}^3 \hspace{2mm},\alpha \in \mathbb{R}\implies L(u+\alpha v)=L(u)+\alpha L(v). \)

Threfore  L is linear.

(b) Find the characteristic polynomial of L with respect to standard basis for \( \mathbb{R}^3 \). Derive the determinant of L then deduce that L is invertible.

we have \( A = \begin{pmatrix}6&2&3\\ -3&-1&-1\\ -5&-2&-2\end{pmatrix} \)

\( \implies P(\lambda)=|A-\lambda I|=\begin{vmatrix} 6-\lambda& 2 & 3 \\ -3 & -1-\lambda & -1 \\ -5 & -2 & -2-\lambda \end{vmatrix} \)

                    \( =-\lambda^3+3\lambda^2-3\lambda+1=-\bigg(\lambda-1\bigg)^3 \)

Therefore . \( P(\lambda)=-\bigg(\lambda-1\bigg)^3 \) and \( det\bigg([L]_B\bigg)=det(A)=P(0)=1 \)

Then  \( [L]_B \)  is invertible.

(c) Find the eigenvalues and eigenspaces of L.

if \( \lambda \in SP(A) \implies \lambda =1 \) so , \( SP(A)=\left\{1\right\} \)

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

\( \iff [L]_B-\lambda I=A-\lambda I=A-I=\begin{pmatrix}5&2&3\\ -3&-2&-1\\ -5&-2&-3\end{pmatrix} \sim \begin{pmatrix}5&2&3\\ 2&0&2\\ 0&0&0\end{pmatrix} \)

Let \( x_3=t \hspace{2mm},x_1=-t\implies 2x_2=-5x_1-3x_3=2t\implies x_2=t \)

Therefore . \( E_\lambda=span\left\{\begin{pmatrix}-1\\ 1\\ 1\end{pmatrix}\right\} \)

(d) Show that L is not diagonalizable, but it is triangularizable, then triangularize L.

since. \( Am(\lambda)\neq gm(\lambda) \) so L is not diagonalizable.  But L is triangularizable .

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

we have \( A-\lambda I=\begin{pmatrix}5&2&3\\ -3&-2&-1\\ -5&-2&-3\end{pmatrix} \hspace{2mm},\lambda=1 \)

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

Then \( E_{\lambda}^2=span\left\{\begin{pmatrix}0\\ 1\\ 0\end{pmatrix},\begin{pmatrix}1\\ 0\\ -1\end{pmatrix}\right\} \hspace{2mm}and \hspace{2mm}\bigg(A-\lambda I\bigg)^3=0 \)

\( \implies G_\lambda =E_{\lambda}^3=span\left\{\begin{pmatrix}1\\ 0\\ 0\end{pmatrix},\begin{pmatrix}0\\ 1\\ 0\end{pmatrix},\begin{pmatrix}0\\ 0\\ 1\end{pmatrix}\right\} \)

\( V_3\in E_{\lambda}^3-E_{\lambda}^2\implies V_3=\begin{pmatrix}1\\ 0\\ 0\end{pmatrix} \)

\( \implies V_2=\bigg(A-I\bigg)V_3=\begin{pmatrix}5&2&3\\ -3&-2&-1\\ -5&-2&-3\end{pmatrix}\begin{pmatrix}1\\ 0\\ 0\end{pmatrix}=\begin{pmatrix}5\\ -3\\ -5\end{pmatrix} \)

\( \implies V_1=\bigg(A-I\bigg)V_2=\begin{pmatrix}5&2&3\\ -3&-2&-1\\ -5&-2&-3\end{pmatrix}\begin{pmatrix}5\\ -3\\ -5\end{pmatrix}=\begin{pmatrix}4\\ -4\\ -4\end{pmatrix} \)

Therefore, \( A=PJP^{-1}=[L]_B \hspace{2mm},J=\begin{pmatrix}1&1&0\\ 0&1&1\\ 0&0&1\end{pmatrix},P=\begin{pmatrix}4&5&1\\ -4&-3&0\\ -4&-5&0\end{pmatrix} \)

(e) Write \( L^n \) in term of n, where \( L^n = L(L(...(L)..)) \), the n compositions of L.

we have \( L^n=A^nr=PI^nP^{-1}V \)

where  \( L^n=\bigg(I+M\bigg)^n \)Then \( M=\begin{pmatrix}0&1&0\\ 0&0&1\\ 0&0&0\end{pmatrix}\hspace{2mm} \)and\( \hspace{2mm} M^3=0 \)

so,  \( J^n=I+nM+\frac{n(n-1)}{2}M^2=\begin{pmatrix}1&n&\frac{n\left(n-1\right)}{2}\\ 0&1&n\\ 0&0&1\end{pmatrix} \)

Answer

Therefore. 

a). L is Linear

b). \( P(\lambda)=-\bigg(\lambda-1\bigg)^3 \) and \( det\bigg([L]_B\bigg)=det(A)=P(0)=1 \)

c). \( E_\lambda=span\left\{\begin{pmatrix}-1\\ 1\\ 1\end{pmatrix}\right\} \)

d). \( A=PJP^{-1}=[L]_B \hspace{2mm},J=\begin{pmatrix}1&1&0\\ 0&1&1\\ 0&0&1\end{pmatrix},P=\begin{pmatrix}4&5&1\\ -4&-3&0\\ -4&-5&0\end{pmatrix} \)

e). \( L^n=A^n \times V=\bigg(PJ^nP^{-1}\bigg)v\hspace{2mm}, J^n=\begin{pmatrix}1&n&\frac{n\left(n-1\right)}{2}\\ 0&1&n\\ 0&0&1\end{pmatrix} \)