problem in diagonalizing a matrix

**kumaran5555** · November 17th 2010, 06:08 AM

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

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

-1 with algebric multiplicity 2.

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

I am finding difficulty in getting eigenvectors for -2.

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

How to solve this.

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

Can anyone help me.

**FernandoRevilla** · November 17th 2010, 06:25 AM

Originally Posted by kumaran5555

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

Why $y=0$ ?. We have:

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

Regards.

**kumaran5555** · November 17th 2010, 06:29 AM

I am just lost !!

**HallsofIvy** · November 17th 2010, 07:41 AM

You do NOT get y= 0.
$\begin{bmatrix}1&-1&1\\0&0&1\\0&0&1\end{bmatrix}\begin{bmatrix}x\\y\ \z\end{bmatrix}$
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>.

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