Question

Find all distinct eigenvalues of A. Then find the basic eigenvectors of A corresponding to each eigenvalue. For each eigenvalue, specify the number of basic eigenvectors corresponding to that eigenvalue, then enter the eigenvalue followed by the basic eigenvectors corresponding to that eigenvalue 2 12 6 A 0 -14 -8 0 24 14 Number of distinct eigenvalues: 1 Number of Vectors: 1 030

Answer 1

Solution: The characteristic equation is

$\small \lambda ^{3}-\left ( \mathrm{sum\: of\: diagonal\: elements }\right )\lambda ^{2}+\left ( \mathrm{sum\: of\: minor\: of\: diagonals }\right )\lambda -\left | A \right |=0$

$\small \therefore \lambda ^{3}-2\lambda ^{2}-4\lambda +8=0$

$\small \therefore \left ( \lambda -2 \right )\left ( \lambda ^{2}-4 \right )=0$

$\small \therefore \lambda =2\: ,\: 2\: ,\: -2$

For   $\small \lambda _{1}=2\: \: \: \: \: \: \left [ A-\lambda _{1}I \right ]X=0$

$\small \begin{bmatrix} 0 &12 &6 \\ 0 &-16 &-8 \\ 0& 24 & 12 \end{bmatrix}\begin{bmatrix} x\\ y\\ z\\ \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0\\ \end{bmatrix}$

By $\small R_{1}\times \frac{1}{6}\: \: ,\: \: R_{2}\times \frac{-1}{8}\: \: ,\: \: R_{3}\times \frac{1}{12}$

$\small \small \begin{bmatrix} 0 &2 &1 \\ 0 &2 &1 \\ 0& 2 & 1 \end{bmatrix}\begin{bmatrix} x\\ y\\ z\\ \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0\\ \end{bmatrix}$

By $\small R_{2}-R_{1}\: \: ,\: \: R_{3}-R_{1}$

0 210 0001 11,1

The rank of $\small A$   is $\small 1< 3$   the number of unknown .

$\small \therefore$ The system has nontrivial solution. The number of linearly independent solution is $\small 3-1=2$

$\small \therefore 0\cdot x+2y+z=0$

Put $\small x=s\: ,\: \: y=t$

$\small \therefore z=-2t$

$\small \therefore$ The solution space is

$\small X=\begin{bmatrix} s\\ t\\ -2t\\ \end{bmatrix}=s\begin{bmatrix} 1\\ 0\\ 0\\ \end{bmatrix}+t\begin{bmatrix} 0\\ 1\\ -2\\ \end{bmatrix}$

$\small \therefore X_{1}=\begin{bmatrix} 1\\ 0\\ 0\\ \end{bmatrix}$

$\small \therefore X_{2}=\begin{bmatrix} 0\\ 1\\ -2\\ \end{bmatrix}$

For   $\small \lambda _{2}=-2\: \: \: \: \: \: \left [ A-\lambda _{2}I \right ]X=0$

$\small \begin{bmatrix} 4 &12 &6 \\ 0 &-12 &-8 \\ 0& 24 & 16 \end{bmatrix}\begin{bmatrix} x\\ y\\ z\\ \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0\\ \end{bmatrix}$

By $\small R_{1}\times \frac{1}{2}\: \: ,\: \: R_{2}\times \frac{-1}{4}\: \: ,\: \: R_{3}\times \frac{1}{8}$

$\small \begin{bmatrix} 2 &6 &3 \\ 0 &3 &2 \\ 0& 3 & 2 \end{bmatrix}\begin{bmatrix} x\\ y\\ z\\ \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0\\ \end{bmatrix}$

By $\small R_{3}-R_{2}$

$\small \begin{bmatrix} 2 &6 &3 \\ 0 &3 &2 \\ 0& 0 & 0 \end{bmatrix}\begin{bmatrix} x\\ y\\ z\\ \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0\\ \end{bmatrix}$

The rank of $\small A$   is $\small 2< 3$   the number of unknown .

$\small \therefore$ The system has nontrivial solution. The number of linearly independent solution is $\small 3-2=1$

$\small 3y+2z=0$

$\small 2x+6y+3z=0$

Put   $\small z=3t$

$\small \therefore y=-2t$   and   $\small x=\frac{3}{2}t$

$\small \therefore$ The solution space is

$\small X_{3}=\begin{bmatrix} \frac{3}{2}\\ -2t\\ 3t\\ \end{bmatrix}=t\begin{bmatrix} \frac{3}{2}\\ -2\\ 3\\ \end{bmatrix}$

Number of distinct eigenvalues: $\small {\color{Red} 2}$

Number of vectors: $\small {\color{Magenta} 3}$

$\small {\color{DarkGreen} 2:\left \{ \begin{bmatrix} 1\\ 0\\ 0\\ \end{bmatrix}\: \: ,\: \: \begin{bmatrix} 0\\ 1\\ -2\\ \end{bmatrix} \right \}}$

$\small {\color{DarkRed} -2:\left \{ \begin{bmatrix} \frac{3}{2}\\ -2\\ 3\\ \end{bmatrix} \right \}}$