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Let $M$ be a square matrix with complex entries.

What is a characterization of $M$ such that $M = P^{T} D P$, where both $D$ and $P^{T} P$ are diagonal matrices?

For example, such a characterization includes all real symmetric matrices using only orthogonal matrices for $P$ (so that $P^{T} P$ is the identity matrix, which of course is diagonal).

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  • $\begingroup$ Do you mean the entrywise transpose or conjugate transpose? That is, would we consider $$\pmatrix{i\\ &i}$$ to be symmetric? $\endgroup$ Dec 31, 2014 at 0:27
  • $\begingroup$ In the former case, note Takagi's factorization $\endgroup$ Dec 31, 2014 at 0:34

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You can see that $M$ is symmetric (take its transpose!) so the characterization is "all real symmetric matrices" (at least for the real case). There's no surprise here, for $E = P^TP$, will have all entries nonnegative (they're the lengths of the columns of $P$), so they have square roots. Let $F$ be the matrix of square roots. Then letting $Q = F^{-1}P$, we have $$ Q^TQ = I$$, and $$P^T D P = (Q^T F^T) D (F Q) = Q^T (F^T D F) Q = Q^T ED Q,$$ which is a diagonalization of $M$ by the orthogonal matrix $Q$.

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  • $\begingroup$ Yes, thanks, but I already knew the real case. That is why I started my question by stating that $M$ has complex entries. Do you know a characterization for this case? $\endgroup$ Dec 30, 2014 at 3:27
  • $\begingroup$ Nope. Sorry for failing to note the "complex" part. Much of my answer almost stands, except that there are vectors $v$ with $v^T v = 0$, so you can't build $F^{-1}$. In a way, this is the main source of the difficulty: $P^t P = I$ is shortand (in the real case) for "columns of $P$ are orthonormal wrt standard inner product." But in the complex case, $\langle v, w \rangle = v^t w$ isn't a nice inner product at all, so there's less to say (for me, at least!). $\endgroup$ Dec 30, 2014 at 13:39
  • $\begingroup$ Yes, isotropic vectors. If that is the main difficulty, would it simplify things to further assume that $P$ is nonsingular? $\endgroup$ Dec 30, 2014 at 15:35
  • $\begingroup$ Knowing that $P^t P = I$ already means that $P$ is nonsingular. Beyond that trivial remark, I'm not sure I can think of anything I know that'd help with this problem...sorry! $\endgroup$ Dec 30, 2014 at 15:44
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If only $P^{T}P$ is to be diagonal, then all such matrices can be written as $P=U\Sigma$ where $U$ is a orthonormal matrix and $\Sigma$ is a diagonal matrix. However, more informative and useful is the case when $M$ is a normal matrix so that $MM^{T}=M^{T}M$. Then $M$ can be written in the way you are interested in.

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  • $\begingroup$ A normal matrix $M$ satisfies $M M^* = M^* M$ not $M M^T = M^T M$. $\endgroup$ Dec 30, 2014 at 6:14
  • $\begingroup$ "more...useful is the case when $M$ is a normal matrix". Not sure why you would say that. The "useful" case to me is the one I asked about, which does not assume normality. $\endgroup$ Dec 30, 2014 at 6:17
  • $\begingroup$ My Bad. I assumed you are interested only in real matrices. The measure of "usefullness" was from a general perspective. $\endgroup$ Dec 30, 2014 at 7:58
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It is a theorem due to Takagi that any complex (entrywise) symmetric matrix may be written as $M=PDP^T$, and $P$ may be chosen to be unitary.

As others have noted, it is clear that only symmetric matrices can be factored in this way, so that symmetry is both sufficient and necessary.

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