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Given an $m \times n$ matrix $A$, it's easy to show that the matrix products $B = A^TA$ and $C = AA^T$ are both symmetric. I was wondering if any symmetric matrix with positive eigenvalues could be expressed as the product of some matrix and its transpose?

Note: It's obvious that if the eigenvalues are not all positive then this cannot be the case, as $A^TA$ and $AA^T$ both have to be positive semidefinite. However this is all I can think of regarding this problem.

Bluefire
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2 Answers2

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Nevermind, I just got this. I'll leave my answer up in case anyone else finds it useful.

Any symmetric matrix $A$ has to be orthogonally diagonalisable, that is, $A = PDP^T$ for some diagonal matrix $D$ and orthogonal matrix $P$. In this case by the diagonalisation theorem, the diagonal values of $D$ are the eigenvalues of $A$, all of which are given to be $\geq 0$. Therefore, one can "square-root" $D$ as $D = M^2$, where $M$ is a diagonal matrix with the square roots of the diagonal values of $D$ on its diagonal, that is, $m_{ii} = \sqrt{d_{ii}}$ and $m_{ij} = 0$ if $j \neq i$. Therefore:

$$ A = PM^2P^T = (PM)(MP^T) = (PM)(M^TP^T) = (PM)(PM)^T $$

Bluefire
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    Your answer is perfectly correct. But there is a more direct way (and more efficient on the algorithmic point of view): see my answer. – Jean Marie May 17 '17 at 03:04
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    Simply a complement to your last line: $A = PM^2P^T = (PMP^T)(PMP^T) = (PMP^T)^2$, which also expresses $A$ as the product of a matrix and its transpose. The matrix $A^{1/2}:=PMP^T$ is the (unique) positive square root of $A$. All this continues to hold if $A$ is merely positive semidefinite. – Hanno May 17 '17 at 14:23
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A direct answer is that, if a matrix is symmetric and has $>0$ eigenvalues, thus is a symmetric positive definite matrix, it has a Cholesky decomposition $A=C^TC$ (https://en.wikipedia.org/wiki/Cholesky_decomposition), with an upper triangular $C$.

Jean Marie
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    That's interesting, I hadn't heard of that However my problem with that is that it assumes the proof of another theorem. Is there simple(-ish) proof of the existence of a Cholesky decomposition for any symmetric matrix? – Bluefire May 17 '17 at 03:18
  • You must have all positive eigenvalues for that. – Jean Marie May 17 '17 at 03:23
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    sorry, you are correct, I meant for any symmetric matrix with positive eigenvalues. Is there a relatively simple proof? – Bluefire May 17 '17 at 03:30
  • There's a constructive, "if and only if" proof of Cholesky, @Blue; you might want to look it up. – J. M. ain't a mathematician May 17 '17 at 03:37
  • @Bluefire: I think you can use Gram-Schmidt for the corresponding inner product to find a basis showing that a positive-definite matrix is of the form $PAP^T$ for diagonal $A$ and triangular $P$, and then you can just take $C=P\sqrt A$. But I did not check it. – tomasz May 17 '17 at 03:39
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    Constructive proof of the existence of the Cholesky factorization: the other answer shows that $A=B^TB$ for some matrix $B$ (not necessarily triangular). Now take a QR factorization of $B$. – Federico Poloni May 17 '17 at 06:27
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    Do you realize that you need more underlying steps to prove Cholesky decomposition than ones present in OP's answer, right? – Danijel May 17 '17 at 14:13
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    @Danijel Yes, for the proof, you are right. I was thinking, without having been explicit enough, to the computational complexity of Cholesky decomposition which is rather low : $\dfrac13n^3$ (https://math.stackexchange.com/q/217738) for an operation on matrices. – Jean Marie May 17 '17 at 15:02
  • @Danijel, in practice, Cholesky is way cheaper to do than eigendecomposition, so it is good to have it here. – J. M. ain't a mathematician May 18 '17 at 08:41
  • @J.M.isn'tamathematician Indeed it is. I've got an impression that OP was asking for a proof, not for a technique to make an explicit construction. – Danijel May 18 '17 at 16:49