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$T:V\rightarrow V$ is over $\mathbb{R}$ and $V$ of finite dimension $n$, and I know that it is orthogonally diagonalizable.

The Matrix that represents it - call it $A$ ,in orthonormal basis is orthogonally diagonalizable as well. So $A=PDP^*$ where $P^*$ is a unitary matrix, and $D$ a diagonal matrix.

Is it true that $A$, $D$, and $P$ are in $M_{n \times n}(\mathbb{R})$ (real matrices) because the transformation is over $\mathbb{R}$? I want to conclude that $P^*$ = $P^t$.

Thanks a lot!!

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No it is not in general true.

Let $A=\left(\begin{matrix}0&1\\ -1&0\end{matrix}\right)$.

The $A$ is anti-symmetric, and therefore it is orthogonally diagonalizable, and its eigenvalues are $\pm i$. Therefore, $D$ contains purely imaginary elements in the diagonal. Clearly, $P$ is not a real matrix either. In fact $$ P=\left(\begin{matrix} 1& i\\ i&1\end{matrix}\right) $$

Nevertheless it is true is $A$ is symmetric (real symmetric). In such case, the eigenvalues are real (and so is $D$) and a diagonalization is achieved via a real unitary matrix $P$.

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  • $\begingroup$ Is it true for symmetric matrices (if A is symmetric)? $\endgroup$ – CnR Feb 10 '14 at 7:15
  • $\begingroup$ See updated version. $\endgroup$ – Yiorgos S. Smyrlis Feb 10 '14 at 7:19
  • $\begingroup$ You really help me to pass my final exam. thanks :) $\endgroup$ – CnR Feb 10 '14 at 7:29

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