Question

1 2 1 Show that A=0 0 a diagonal D. 2 is diagonalizable by finding P and D such that P-1AP = D for 3 0

Answer 1

Solution: Given

A= 1 1 0 0 2 2 0 0 3

Now eigen values are given by

A - XI = 0

$\Rightarrow \begin{vmatrix} 1-\lambda & 1 & 0\\ 0 & 2-\lambda & 2\\ 0 & 0 &3-\lambda \end{vmatrix}=0$

(1 - 1) (2 - x)(3-X) = 0

>= 1,2,3

all eigen values are distinct and real so is diagonalizable.

Now, invertible matrix is given by the eigenvectors.

So let us find eigen vectors corresponding to each eigen value.

Now, eigen vector $v_1$ corresponsing to $\lambda=1$ is given by

$[A-I]v_1=0$

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

> y=0,2 = 0

By choosing $x=1$ , we get

$v_1=\begin{bmatrix} 1\\0 \\ 0 \end{bmatrix}$

Now, eigen vector $v_2$ corresponsing to $\lambda=2$ is given by

$[A-2I]v_2=0$

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

$\Rightarrow x=y,z=0$

By choosing $x=1$ , we get

$v_2=\begin{bmatrix} 1\\1 \\ 0 \end{bmatrix}$

Now, eigen vector $v_3$ corresponsing to $\lambda=3$ is given by

$[A-3I]v_3=0$

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

$\Rightarrow 2x-y=0,\ \ \ y-2z=0$

$\Rightarrow 2x=y,\ \ \ y=2z$

$\Rightarrow x=z,y=2z$

By choosing $z=1$ , we get

$v_3=\begin{bmatrix} 1\\2 \\ 1 \end{bmatrix}$

Thus,

$P=\begin{bmatrix} 1 & 1&1 \\ 0& 1& 2\\ 0& 0 & 1 \end{bmatrix}$

Now find inverse of .

Let

$P=IP$

$\Rightarrow \begin{bmatrix} 1 & 1&1 \\ 0& 1& 2\\ 0& 0 & 1 \end{bmatrix}= \begin{bmatrix} 1 & 0&0 \\ 0& 1& 0\\ 0& 0 & 1 \end{bmatrix}P$

applying $R_1\to R_1-R_3,\ R_2\to R_2-2R_1$

$\Rightarrow \begin{bmatrix} 1 & 1&0 \\ 0& 1& 0\\ 0& 0 & 1 \end{bmatrix}= \begin{bmatrix} 1 & 0&-1 \\ 0& 1& -2\\ 0& 0 & 1 \end{bmatrix}P$

applying $R_1\to R_1-R_2$

$\Rightarrow \begin{bmatrix} 1 & 0&0 \\ 0& 1& 0\\ 0& 0 & 1 \end{bmatrix}= \begin{bmatrix} 1 & -1&1 \\ 0& 1& -2\\ 0& 0 & 1 \end{bmatrix}P$

compare it with

$I=P^{-1}P$

we get

$P^{-1}= \begin{bmatrix} 1 & -1&1 \\ 0& 1& -2\\ 0& 0 & 1 \end{bmatrix}$

Hence,

$D=P^{-1}AP$

$\Rightarrow D=\begin{bmatrix} 1 & -1&1 \\ 0& 1& -2\\ 0& 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 & 0\\ 0 & 2 & 2\\ 0 & 0 &3 \end{bmatrix}\begin{bmatrix} 1 & 1&1 \\ 0& 1& 2\\ 0& 0 & 1 \end{bmatrix}$

$\Rightarrow D=\begin{bmatrix} 1 & -1&1 \\ 0& 1& -2\\ 0& 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3\\ 0 & 2 & 6\\ 0 & 0 &3 \end{bmatrix}$

$\Rightarrow D=\begin{bmatrix} 1 & 0&0 \\ 0& 2& 0\\ 0& 0 & 3 \end{bmatrix}$

Which is the required solution.

This complete the solution.