Question

Find the Eigenvalues, Eigenvectors, If possible find an invertible matrix P, such that P-AP is in diagonalized form. -3 1 A = 4 3 () 0 0 -2 A = 1 2 1 0 3

Answer 1

Solution:

-1 -3 A = 4 3

The characteristic equation is

(sum of diagonal element) A + |A = 0

2-2A9= 0

A122i V

Thus, the eigenvalues are   .

12 2i, 1-2/2i

for
A, = 1 +2v2i V
,   
A 1I X 0

1 1-2/2i -3 3 1 2 2i 4

-2 2/21 -3 2 2 2i

. 4ar(2-2v2i y 02x (- V2i)y 0 -1+v2 2 y

$\small \therefore v_{1}=\begin{pmatrix} -1+\sqrt{2}i\\ 2\\ \end{pmatrix}$

for $\small \lambda_{2} =1- 2\sqrt{2}i$ ,   

A 2I X =0

1 1 2/2i -3 3 12 2i, 4

$\small \begin{pmatrix} -2+2\sqrt{2}i &-3 \\ 4 & 2+2\sqrt{2}i \end{pmatrix}\begin{pmatrix} x\\ y\\ \end{pmatrix}=\begin{pmatrix} 0\\ 0\\ \end{pmatrix}$

$\small \therefore 4x+\left ( 2+2\sqrt{2}i \right )y=0\Rightarrow 2x+\left ( 1+\sqrt{2} i\right )y=0\Rightarrow x=\frac{-1-\sqrt{2}i}{2}y$

$\small \therefore v_{2}=\begin{pmatrix} -1-\sqrt{2}i\\ 2\\ \end{pmatrix}$

Thus, the eigenvectors are  

$\small \begin{pmatrix} -1+\sqrt{2}i\\ 2\\ \end{pmatrix}\: \: \: \: \:$, $\small \: \: \begin{pmatrix} -1-\sqrt{2}i\\ 2\\ \end{pmatrix}$

Since the eigenvalues are distinct , $\small A$ is diagonalisable.

Now,  

$\small P=\begin{pmatrix} -1+\sqrt{2} i& -1-\sqrt{2}i\\ 2& 2 \end{pmatrix}$

$\small P^{-1}AP=\begin{pmatrix} -1+\sqrt{2} i& -1-\sqrt{2}i\\ 2& 2 \end{pmatrix}^{-1}\begin{pmatrix} -1 & -3\\ 4 & 3 \end{pmatrix}\begin{pmatrix} -1+\sqrt{2} i& -1-\sqrt{2}i\\ 2& 2 \end{pmatrix}$

  $\small =\begin{pmatrix} 1+2\sqrt{2}i & 0\\ 0 & 1-2\sqrt{2}i \end{pmatrix}$

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The characteristic polynomial is

$\small \lambda ^{2}-\left (\mathrm{ sum\: of\: \: diagonal\: element} \right )\lambda ^{2}+\left ( \mathrm{sum\: of\: minor\: of\: diagonal\: element} \right )\lambda -\left | A \right |$

$\small \therefore \lambda ^{3}-5\lambda ^{2}+8\lambda -4$

1,2,2

for  $\small \lambda _{1}=2$ ,

A 1I X 0

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

$\small \therefore x=-z$

$\small \therefore v_{1}=z\begin{pmatrix} -1\\ 0\\ 1\\ \end{pmatrix}+y\begin{pmatrix} 0\\ 1\\ 0\\ \end{pmatrix}$

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

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

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

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

$\small \therefore x=-2z,\: \: \: y=z$

-2 1 . v2 1

Thus, the eigenvalues and eigenvectors are

$\small 1:\left \{ \begin{pmatrix} -2\\ 1\\ 1\\ \end{pmatrix} \right \}$

$\small 2:\left \{ \begin{pmatrix} -1\\ 0\\ 1\\ \end{pmatrix} ,\: \: \begin{pmatrix} 0\\ 1\\ 0\\ \end{pmatrix}\right \}$

Since the algebraic multiplicity of each eigenvalue is same as the geometric multiplicity , $\small A$ is diagonalisable.

$\small P=\begin{pmatrix} -2 &-1 &0 \\ 1 & 0& 1\\ 1 & 1&0 \end{pmatrix}$

Also,

$\small P^{-1}AP=\begin{pmatrix} -2 &-1 &0 \\ 1 & 0& 1\\ 1 & 1&0 \end{pmatrix}^{-1}\begin{pmatrix} 0 &0 &-2 \\ 1& 2 &1 \\ 1& 0 & 3 \end{pmatrix}\begin{pmatrix} -2 &-1 &0 \\ 1 & 0& 1\\ 1 & 1&0 \end{pmatrix}=\begin{pmatrix} 1 & 0 &0 \\ 0& 2 & 0\\ 0 & 0 & 2 \end{pmatrix}$