Finding eigenvectors

I-Think · #1 $Old$ November 11th, 2009, 07:16 PM

The matrix A, given by

$A=\left(\begin{array}{ccc}7&-4&6\\2&2&2\\-3&4&-2\end{array}\right)$

has eigenvalues 1,2,4. Find the set of corresponding vectors.

Every time I perform these operations I only get my answers as 0's. I'll geve an example.

$x(A-\delta{I})=0$ taking $\delta$ as 1

$\left(\begin{array}{ccc}6&-4&6\\2&1&2\\-3&4&-3\end{array}\right)$ $\left(\begin{array}{c}x\\y\\z\end{array}\right)=\left(\begin{array}{c}0\\0\\0\end{array}\right)$

Solving
$6x-4y+6z=0$
$2x+y+2z=0$
$-3x+4y-3z=0$

$-x+5y-z=0$
$x=5y-z$

$30y-6z-4y+6z=0$
$y=0$
And so on. Could someone please direct me as to what I am doing wrong?

Gamma · #2 $Old$ November 11th, 2009, 07:40 PM

Just by inspection, you can see that an eigenvector that will work is <1,0,-1>.

Jose27 · #3 $Old$ November 11th, 2009, 08:38 PM

You're solving the equations wrong (note that the coefficients of x and z will always vanish together if we try to sum up one equation to another) , the solutions for that eigenvalue should be x=-z