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Once again, this is from a past qualifying exam I am trying to work on.

Here is the problem.

True or False? Let $A$ be a real $n\times n$ matrix such that $AA^T=A^TA$ and all eigenvalues of $A$ are real. Then $A$ must be symmetric.

Attempt. Well it looks as if this is related to the real spectral theorem. I know that any real symmetric matrix is diagonalizable and must have real eigenvalues. This is sort of a converse to that theorem right? I tried a few $2\times 2$ examples and they all point to the above statement being True. Am I right?. But I am unable to see a way to prove. We actually haven't learnt the spectral theorem nor is it on the qual syllabus. So I'm kind of unsure on this. Can you help?

Thank you in advance.

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Another handy theorem to know for your qual:

Let $A\in M_n(\mathbb{R})$ be normal. Then there exists an orthogonal matrix $Q\in M_n(\mathbb{R})$ such that:

$$Q^{\intercal}AQ=\begin{pmatrix}A_1&0&0&\ldots&0\\0&A_2&0&\ldots&0\\0&0&A_3&\ldots&0\\\vdots&\vdots&\vdots&\ldots&\vdots\\0&0&0&\ldots&A_m\end{pmatrix}$$

where $A_i$ is either a $1\times 1$ real matrix, or a $2\times 2$ real matrix of the form:

$$\begin{pmatrix}a&b\\-b&a\end{pmatrix}$$

Notice that a corollary of this theorem is that if all eigenvalues are real, then all of the $A_i$ are $1\times 1$ matrices, and it follows that $A$ is symmetric (just rearrange the above equation for $A$, using the fact that $Q^{-1}=Q^{\intercal}$, and then notice that the diagonal matrix of $A_i$'s is symmetric).

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  • $\begingroup$ Thanks for your answer. This helps. $\endgroup$ – minibuffer Aug 4 '13 at 19:23
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Yes, this is correct. Matrix $A$ is normal, and a normal matrix is self-adjoint iff all its eigenvalues are real. See e.g. http://www.math.msu.edu/~abbas/R3.pdf problem 4a.

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