The matrix $A = \begin{bmatrix}5&0&0\\-7&-2&0\\12&2&0\end{bmatrix}$ has eigenvalues $-2,0, $ and $5$. Find its associated eigenvectors for each eigenvalue
So I started with $\lambda_1 = -2$ to get $\begin{bmatrix}5-(-2)&0&0\\-7&-2-(-2)&0\\12&(2–2)&0\end{bmatrix}$ ~ $\begin{bmatrix} 7&0&0\\-7&0&0\\12&4&0\end{bmatrix}$
So I know I have
$7x = 0$ which doesn't tell me anything
so I see that $12x +4y = 0 \implies 12x = -4y \implies x = 1, y = -3$ so I thought that a eigenvector associated with the eigenvalue $\lambda_1 = -2$ would be $(1,-3,0)$ but for some reason I am incorrect. Am i in error in my method? Is there a difference between a corresponding eigenvector and a associated eigenvector? I was under the impression they were the same thing
UPDATE (Had subtracted lambda from the wrong row/column)
Now I get: $\begin{bmatrix} 7&0&0\\-7&0&0\\12&2&-2\end{bmatrix}$
so $7x = 0 \implies x = 0$ so then I have $2y = 2z \implies y = 1, z = 1$ so that the eigenvector associated with $-2$ is $(1,1,0)$ ? But that is incorrect also, where am I in error?
Best Answer
So, if $A = \begin{bmatrix}5&0&0\\-7&-2&0\\12&2&0\end{bmatrix}$ we want to find its associated eigenvectors for each eigenvalue.
Let's start with $\lambda_1 = -2$. We find that the matrix ($A-\lambda_1Id$) where $Id = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$ is
$A_1 = \begin{bmatrix}7&0&0\\-7&0&0\\12&2&2\end{bmatrix}$ (pay attention, it's 2 on $a_{33}$ not -2 because you are subtracting -2).
We now find that it must be $x=0$, $y=-z$. So the eigenvector corresponding to $\lambda_1$ is $(0,1,-1)$.
We do the same with the other eigenvalues, $\lambda_2=0$ and $\lambda_3=5$.
$A_2 = \begin{bmatrix}5&0&0\\-7&-2&0\\12&2&0\end{bmatrix}$
From which we see that $x=0, y=0, z=\alpha$ with $\alpha\in K$ where $K$ is the field you are considering. So the eigenvector corrisponding to $\lambda_2$ is $(0,0,\alpha)$.
$A_3 = \begin{bmatrix}0&0&0\\-7&-7&0\\12&2&-5\end{bmatrix}$
From which we see $x=-y, z=2x$ so the eigenvector related to $\lambda_3$ is $(1,-1,2)$.