# problem in diagonalizing a matrix

• Nov 17th 2010, 06:08 AM
kumaran5555
problem in diagonalizing a matrix
$\displaystyle \begin{bmatrix}-1&-1&1\\0&-2&1\\0&0&-1\end{bmatrix}$

Eigenvalues of the matrix are -1,-1,-2.

-1 with algebric multiplicity 2.

Eigenvectors for -1 are $\displaystyle \begin{bmatrix}1\\0\\0\end{bmatrix}$ and $\displaystyle \begin{bmatrix}0\\1\\1\end{bmatrix}$

I am finding difficulty in getting eigenvectors for -2.

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

How to solve this.

I am getting z=0,y=0,and x=y-z.

Can anyone help me.
• Nov 17th 2010, 06:25 AM
FernandoRevilla
Quote:

Originally Posted by kumaran5555
I am getting z=0,y=0,and x=y-z.

Why $\displaystyle y=0$ ?. We have:

$\displaystyle (x,y,z)=(y,y,0)=y(1,1,0)$

Regards.
• Nov 17th 2010, 06:29 AM
kumaran5555
I am just lost !!
• Nov 17th 2010, 07:41 AM
HallsofIvy
You do NOT get y= 0.
$\displaystyle \begin{bmatrix}1&-1&1\\0&0&1\\0&0&1\end{bmatrix}\begin{bmatrix}x\\y\ \z\end{bmatrix}$
is equivalent to the equations x- y+ z= 0, z= 0, and z= 0. That is, the last two equations are both z= 0.

With that, x- y+ 0= 0 so y= x. Any eigenvector is of the form <x, y, z>= <x, x, 0>= x<1, 1, 0>.