Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $A=\begin{bmatrix}2&1&2\\4&2&4\\2&1&2\end{bmatrix}$

Its eigenvalues are $0,0,6$. I want to find its eigenvectors.

My solution:

when $\lambda_1=0$


$x_1=(-\frac{1}{2}s-t s t)=-\frac{1}{2}s(1, -2, 0)+t(-1, 0, 1)$ since 2nd and 3rd vairables are free variables. Thus $(1, -2, 0)$ and $(-1, 0, 1)$ are the eigenvectors.

BUT the answer is $(1, -2, 0)$ and $(0, -2, 1)$.

I can't understand why these are the true answers. What is wrong with my solution?

share|cite|improve this question
up vote 3 down vote accepted

Nothing, you just chose a different basis of the eigenspace corresponding to $0$. Note that $(0,-2,1)=(1,-2,0)+(-1,0,1)$.

share|cite|improve this answer
Oh, then both are right? Then it means that eigenvectors are not unique? – existence Nov 11 '12 at 12:19
Yes. Suppose $v_1$ and $v_2$ are eigenvectors to the eigenvalue $\lambda$, then $A(v_1+v_2)=Av_1+Av_2=\lambda v_1+\lambda v_2=\lambda(v_1+v_2)$ – Julian Kuelshammer Nov 11 '12 at 12:22
Now I understand! Thanks! – existence Nov 11 '12 at 12:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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