[Math] How to decompose a square symmetric matrix into two diagonalizable matrices provided that one of them is the transpose of the other

Question

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?

Answer 1

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.