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I have

$$A = \begin{bmatrix} 5 & -2 & 1 \\ -2 & 2 & -2 \\ -1 & -2 & 5 \end{bmatrix}$$

which has eigenvalues $\lambda_1 = \lambda_2 = 6$ and $\lambda_3 = 0$. I want to find the eigenvectors for these eigenvalues.

I've tried to turn it into equations and trying to solve them (for $\lambda_1 = \lambda_2 = 6$) $$ -x -2y - z = 0 \\ -2x - 4y - 2z = 0 \\ -x -2y - z = 0$$

But no matter how I try to approach this, I have no idea how to get my eigenvectors. I know from the answer that the eigenvector for $\lambda_1$ & $\lambda_2$ should be:

enter image description here

and eigenvector for $\lambda_3$$ should be:

enter image description here

But I have no idea how to arrive to that answer. Can someone explain (every step)?

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  • $\begingroup$ I don't understand everything, but you are saying both $\lambda2=6$ and $\lambda2=0$, which is contradictory. By the way, if you meant a subscript, type in underline in LaTeX mode $\lambda_2=6$ to get $\lambda_2=6$. $\endgroup$
    – Marc van Leeuwen
    May 30, 2016 at 7:33
  • $\begingroup$ Since its a 3×3 matrix so it should have only three eigen values $\endgroup$
    – Shona
    May 30, 2016 at 7:35
  • $\begingroup$ The eigenspace for $\lambda=6$ is $2$-dimensional, so you'll have to choose a basis with two elements for it. There is no guarantee that you will choose the same basis as provided in the answer; your vectors may look quite different even if they span the same subspace. $\endgroup$
    – Marc van Leeuwen
    May 30, 2016 at 7:42
  • $\begingroup$ Your first eqiation should be -x-2y+z =0 according to your question $\endgroup$
    – Shona
    May 30, 2016 at 7:42
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    $\begingroup$ For everyone's convenience, consider typing up your posts instead of using images. Formatting tips here. $\endgroup$
    – Em.
    May 30, 2016 at 8:04

1 Answer 1

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I am assuming you meant $$A = \begin{bmatrix} 5 & -2 & 1 \\ -2 & 2 & -2 \\ -1 & -2 & 5 \end{bmatrix}.$$

Let's find the eigenvector corresponding to $\lambda = 6$. So we wish to solve $A\vec{v} = 6\vec{v}$. Suppose $$\vec{v} = \begin{bmatrix} x \\ y \\ z\end{bmatrix}.$$

Then $$A\vec{v} = \begin{bmatrix} 5x -2y + z \\ -2x + 2y -2z \\ -x -2y + 5z\end{bmatrix} = \begin{bmatrix} 6x \\ 6y \\ 6z \end{bmatrix}$$.

Thus we get the three equations: $$ -x -2y + z = 0 \\ -2x -4y - 2z = 0 \\ -x - 2y - z = 0.$$

Can you find the general solution from here?

EDIT: Since in the question you were stuck at exactly this point, I'll complete this part for you.

Note that adding the first and third equation gives $x + 2y = 0$, so $x = -2y$. Since the second equation is identical to the third (upto a factor of $2$), it is redundant. So any solution of the form $(-2y, y, z)$ will solve this system. Thus, an arbitrary eigenvector will be of the form $$ \vec{v} = \begin{bmatrix} -2s \\ s \\ t \end{bmatrix} = s\begin{bmatrix} -2 \\ 1 \\ 0\end{bmatrix} + t \begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix}.$$ Does this help?

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