# How to find a basis of real eigenvectors for a real symmetric matrix?

While real symmetric matrices have real eigenvalues, their eigenvectors do not have to be real. If I am interpreting the conversation correctly, THIS math.stackexchange thread seems to suggest that it is, nonetheless, possible to choose a basis such that ALL of the eigenvectors are real. Is this true?

If so, how would I do this?

As a concrete example, I have a 448 x 448 real symmetric matrix. I compute the eigenvalues and eigenvectors using MATLAB and 24 of the eigenvectors that it returns are complex, all the rest are real. How can I make these 24 complex eigenvectors real?

• Eigenvectors are vectors, not real numbers. – Rick Sanchez Apr 19 '16 at 1:17
• @RickSanchez real vectors = vectors whose entries are all real. – Robert Israel Apr 19 '16 at 1:18
• I see. I've never heard this term before. – Rick Sanchez Apr 19 '16 at 1:20

Yes: if $\lambda$ is a real eigenvalue of a real matrix $A$, there will be a basis of $\text{ker}(A - \lambda I)$ consisting of real eigenvectors.