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Let`s have a square symmetric matrix A, and I wish to decompose it into two diagonalizable matrices so that: $$A=M^TM$$ where, $$M$$ and $$M^T$$ are; a diagonalizable matrix and its transpose. First, I tried decomposing A using PCA (principle component analysis); $$A=V\sigma V^T$$ This gives column vectors V and their corresponding eigen values. But then I can not be able to solve for M using Singular value Decomposition (SVD) because I do not have U: $$M=U\sqrt\sigma V^T$$ So what is the way out? Is there any sharp method of obtaining U and hence M or any numerical approximation method instead? Also I am interested to know whether such decomposition can be applied to any square symmetric matrix in general with no exception?

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    $\begingroup$ Why do you want to decompose $A$ into a product of two non-diagonalizable matrices? $\endgroup$
    – levap
    Dec 4 '15 at 18:44
  • $\begingroup$ Symmetric matrices represent inner products via the formula $\langle \mathbf u,\mathbf v\rangle_A=\mathbf v^TA\mathbf u$. Viewed from this perspective, $M^TM$ is the collection of all scalar products of the set of vectors given by the columns of $M$. That might help you find some candidates for the decomposition. $\endgroup$
    – amd
    Dec 4 '15 at 19:35
  • $\begingroup$ 1) To make the matter more sensible and probably easier, I want to decompose the square symmetrix matrix into two diagonalizable matrices as above. $\endgroup$
    – Isaacadel
    Dec 4 '15 at 19:45
  • $\begingroup$ 2) I dont understand your clue. M is the matrix which is I am looking for which I need to derive it from A. $\endgroup$
    – Isaacadel
    Dec 4 '15 at 19:46
  • $\begingroup$ I changed the topic title to mean a kind of decomposition into two diagonalizable matrices instead. $\endgroup$
    – Isaacadel
    Dec 4 '15 at 20:05
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I think I found the way out. I have mentioned in my first post that I would have calculated M if I had U because, $$M=U\sqrt\sigma V^T$$ In fact, U itself is an orthonormal matrix which means I can choose an arbitrarily orthonormal matrix and plug it in the SVD equation above to yield M. There are two notes here: 1) M will not be unique. 2) $$M^TM=A$$ which is my original square symmetric matrix because $$U^TU=I$$ So steps to decompose a square symmetric matrix A into $$M^TM$$. 1) Decompose the square symmetric matrix using prinicple components analysis. $$M^TM=V\sigma V^T$$. 2) Use $$V and \sqrt\sigma$$ matrices to construct M. 3) Multiply the matrices in step 2 by U which is an orthonormal matrix to yield; $$M=U\sqrt\sigma V^T$$. Done.

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