Square root of non-symmetric positive semi-definite matrix.

Question

Let $A$ be a $d \times d$ square positive semi-definite, not necessarily symmetric, real matrix in the sense that $v^TAv\ge 0$ for any vector $v\in\mathbb R^d$. How to show that there exists $B$ such that $$A=B^TB?$$

If $A$ is symmetric, this is easy to construct such $B$ by considering eigendecomposition. What about the non-symmetric

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Edit: Such $A$ is necessarrily symmetric, so how about changing the question to the existence of $B$ such that $$A=B^2$$?

Answer 1

If $A = B^TB$ then $A^T = B^T B^{TT} = B^T B = A$ so then $A$ is symmetric, so it can't be done.

edit: wikipedia has an explanation for the $A = B^2$ version of the question.