Would symmetry positive semi-definite matrix always decomposable?

Question

Given symmetry positive semi-definite matrix $A \in R^{n\times n}$. And $Det(A) \geq 0$.

Would there always exist real matrix $B$, such that $A = B \cdot B^T$?

If so why? Or why not?

Answer 1

The answer is positive. There exists several decompositions like that.

See for example Cholesky_decomposition. Another way is to use the fact that a symmetric positive semi-definite real matrix is diagonalizable in an orthonormal basis with non negative eigenvalues. See symmetric matrices.