1
$\begingroup$

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)?

|cite|improve this question
$\endgroup$
  • $\begingroup$ 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 '16 at 7:33
  • $\begingroup$ Since its a 3×3 matrix so it should have only three eigen values $\endgroup$ – Shona May 30 '16 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 '16 at 7:42
  • $\begingroup$ Your first eqiation should be -x-2y+z =0 according to your question $\endgroup$ – Shona May 30 '16 at 7:42
  • 2
    $\begingroup$ For everyone's convenience, consider typing up your posts instead of using images. Formatting tips here. $\endgroup$ – Em. May 30 '16 at 8:04
3
$\begingroup$

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?

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.