# Finding eigenvector only knowing others eigenvectors.

The matrix $$A \in M_3(\mathbb{R})$$ satisfy $$A^t=A$$ and $$(1,2,1), (-1,1,0)$$ are eigenvectors of $$A$$. Which vector is also an eigenvector of $$A$$? Alternatives: $$(0,0,1)$$; $$(1,1,-3)$$; $$(1,1,3)$$; There is no other eigenvector.

The problem with this exercise is that I don't know the matrix $$A$$, and I don't have any eigenvalue to start with. I can get a matrix with less variables using $$A = A^t$$, but there's still 6 variables. Any tips or guidance is appreciated.

• There is some sloppiness in the language here. Of course there are more eigenvectors. In fact, there is an infinite amount of them. For instance, for every nonzero $\lambda$, $\lambda (1, 2, 1)$ is an eigenvector. – Andreas Rejbrand Feb 17 at 12:16
• The question only makes sense if the eigenvalues are distinct. Counter-example: if $A = 0$ every vector is an eigenvector. – alephzero Feb 17 at 13:49

Since $$A$$ is symmetric, the eigenvectors (for distinct eigenvalues) are orthogonal.

So, find which of the vectors is orthogonal to the first two.

(1,1,-3) is.

• I think I need to rethink it. I'm leaning towards $(1,1,-3)$ because it's orthogonal to both. Will edit. – Chris Custer Feb 17 at 7:01
• The other two could be eigenvectors. If so, the matrix would have to be a multiple of the identity, but I don’t see anything in the problem statement that would preclude this. – amd Feb 17 at 7:08
• @amd true. Good catch. – Chris Custer Feb 17 at 7:14
• @amd the orthogonal one must be an eigenvector though, if I'm not mistaken. – Chris Custer Feb 17 at 7:23
• @ChrisCuster Yeah the problem could probably have been better stated as "which vector is also necessarily an eigenvector" – angryavian Feb 17 at 7:45

Hint: the condition $$A^t = A$$ allows you to use the spectral theorem.

Hint: Specifically, the spectral theorem implies there is an orthonormal basis of eigenvectors of $$A$$.

• That is correct, but unless the eigenvalues are distinct the theorem does not say that no other vectors except for the basis vectors are eigenvectors. – alephzero Feb 17 at 15:21
• @PaulSinclair No, I think "no other eigenvector" is a possible "choice" for this multiple choice question. – angryavian Feb 17 at 17:29
• @angryavian - you are correct. As Chris Custer pointed out below his answer, the two eigenvectors given are not orthogonal. Therefore their eigenvalues have to be the same. I mis-interpreted the problem. – Paul Sinclair Feb 17 at 17:37