Decomposition of A Symmetric Matrix

Question

Can every symmetric matrix $A$ be written as a product $B^T B$ for some (not necessarily square) matrix $B$?

Answer 1

Obviously not. If the field is real, $B^TB$ is always positive semidefinite, but $A$ is not.

On the other hand, if the field is complex, the answer is yes. Every complex symmetric matrix $A$ admits a Takagi factorisation $USU^T$ where $U$ is unitary and $S$ is a nonnegative diagonal matrix. It follows that you may take $B=S^{1/2}U^T$.

Answer 2

Counter example : $$A= \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} $$ over $\mathbb{R}$.

If the matrix is a real matrix with negative determinant, and require B to be squared, then you cannot do that.

Suppose $A=B^TB$ and $B$ to be a square matrix.

Then $$\det(A)=\det(B^TB)=\det(B)^2\geq0$$

It shows that your claim requires $\det(A)\geq0$ when we particularly assumed $B$ to be a square matrix.

So, not all matrices can be decomposed into this form

Answer 3

The answer is no in general, but becomes yes under the additional assumption that $A$ is positive definite. When $A$ is symmetric positive definite, $A$ can be factored as $A = LL^T$ where $L$ is lower triangular. See this paper for more info.