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?

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    $\begingroup$ Why did you subtract $-2$ from the $(3,2)$ entry rather than the $(3,3)$ entry? $\endgroup$ – carmichael561 Jul 14 '16 at 3:51
  • $\begingroup$ @carmichael561 OP has been updated $\endgroup$ – Yusha Jul 14 '16 at 3:56
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    $\begingroup$ The last row should be $12,2,2$, right? $\endgroup$ – carmichael561 Jul 14 '16 at 3:58
  • $\begingroup$ Augh, wow. @carmichael561 $\endgroup$ – Yusha Jul 14 '16 at 3:59
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    $\begingroup$ I think $(0,1,-1)$ should work. If you apply $A$ to this vector you get $(0,-2,2)$, right? $\endgroup$ – carmichael561 Jul 14 '16 at 4:02

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 $(1,-1,2)$.


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