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I was tasked with finding the eigenvalues and eigenvectors for the matrix:

$$A=\begin{pmatrix} -2&2&-3\\ 2&1&-6\\ -1&-2&0\\ \end{pmatrix}$$

I had found the eigen values as $\lambda=5,-3,-3$. Now, I proceeded to find eigenvectors. For $\lambda=5$, I got the characteristic equation as:

$$A-5I=\begin{pmatrix} -7&2&-3\\ 2&-4&-6\\ -1&-2&-5\\ \end{pmatrix}$$

I used cross multiplication to get the eigen vector.

However for $\lambda=-3$, I'm facing a problem. The characteristic matrix is:

$$A+3I=\begin{pmatrix} 1&2&-3\\ 2&4&-6\\ -1&-2&3\\ \end{pmatrix}$$

Here, all the equations are proportional to each other, and cross multiplication only gave me a trivial eigenvectors. How can I solve it in this case?

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If we look at the matrix $A+3I$, we can see that the second and third row are multiples of the first row.

An eigenvector corresponds to eigenvalue $-3$ satisfies $$(A+3I)v=0$$

You just have to solve for all the solution to

$$v_1+2v_2-3v_3=0$$

Let $v_2=s$ and $v_3=t$, $v_1=-2s+3v_3$.

$$\begin{bmatrix} v_1 \\ v_2 \\ v_3\end{bmatrix}= s\begin{bmatrix} -2 \\ 1 \\ 0\end{bmatrix}+ t\begin{bmatrix} 3 \\ 0 \\ 1\end{bmatrix}$$

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  • $\begingroup$ I don't understand how you got the result from "Let $v_2=s$ and $v_3=t$, $v_1=-2s+3t$". Can you elaborate there? $\endgroup$
    – Pritt Balagopal
    Nov 11, 2017 at 7:49
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    $\begingroup$ The first row of the vector read $v_1 = -2s+3t$, the second line reads $v_2 = s$, the third line is $v_3=t$. I just write them in vector form. $\endgroup$
    – Siong Thye Goh
    Nov 11, 2017 at 7:50
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    $\begingroup$ In other words, $\begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix} = \begin{bmatrix}-2v_2+3v_3\\v_2\\v_3\end{bmatrix} = \begin{bmatrix}-2v_2\\v_2\\0\end{bmatrix} + \begin{bmatrix}3v_3\\0\\v_3\end{bmatrix} = v_2 \begin{bmatrix}-2\\1\\0\end{bmatrix} + v_3 \begin{bmatrix}3\\0\\1\end{bmatrix}$. $\endgroup$
    – Misha Lavrov
    Nov 11, 2017 at 8:04
  • $\begingroup$ @MishaLavrov Thanks $\endgroup$
    – Pritt Balagopal
    Nov 11, 2017 at 8:07
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    $\begingroup$ Sure, in fact there are infinitely many eigenvectors but we have found them all. To let $v_1=0$, let $s=1.5$ and $t=1$. The set of all the eigenvectors correponding to eigenvalue $-3$ are the solution set to the linear system $(A+3I)v=0$. $\endgroup$
    – Siong Thye Goh
    Nov 11, 2017 at 8:11

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