I've two small questions I'm quite stuck on, and I'd appreciate your help.

1) Any real square matrix, all whose eigenvalues are real, having an orthonormal basis of eigenvectors, is symmetric.

Now I know that due to The Spectral Theorem, if a matrix is symmetric then we can find an orthonormal basis of its eigenvectors. However, this is like trying to prove A -> B only knowing that B -> A, which isn't much good. I've tried to find a contradiction, but to no avail. I'm inclined to say that it is true.

2) If A is a symmetric matrix, and F is the matrix whose columns are an orthonormal basis of eigenvectors of A, then F is symmetric.

Now for this, I tried using the inner product but didn't get anywhere. I know of orthogonal matrices that aren't symmetric (like the Sin theta Cos theta one), but I don't think that's sufficient, as who's to say that matrices with that property always contain the orthonormal basis of eigenvectors of some matrix A.


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    $\begingroup$ Thus $A=PDP^{-1}$ with $D$ real diagonal. What does the hypothesis that the basis of eigenvectors is orthonormal tells you about $P$? $\endgroup$ – Did Apr 13 '12 at 18:45
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    $\begingroup$ The first is answered above by Didier. The second is false; take A=I and F any orthogonal (non simmetric) matrix. $\endgroup$ – leonbloy Apr 13 '12 at 19:06
  1. Let $\{u_i$} be the orthonormal basis of eigenvectors. Let $U$ denote the matrix with $u_i$ as the columns. We have, $ A u_i = d_i u_i $ for $ i=1 $ to $n$. Therefore, $A U = DU$ where D is the diagonal matrix with $\{d_i\}$ as its diagonal elements. $A=U^T D U \Rightarrow A^T=(U^TDU)^T=U^TDU=A$.
  2. Well, this is just not true. For example, consider the matrix $A=[[1, 1], [-1, 1]]$. Note that the set of orthonormal vectors are unique upto degeneracy.
  • $\begingroup$ Hi, Welcome to the site. Looking forward to seeing you often. :-) $\endgroup$ – user21436 Apr 13 '12 at 20:38

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