Any positive semidefinite matrix can be written as $AA^{\ast}$ I know that for any matrix $A$, $AA^{\ast}$ is positive semidefinite (where $A^\ast$ is $\overline{A}^T$). Please help me show the following statement

Any positive semidefinite can be written as $AA^{\ast}$.

 A: If $A \ge 0$ and $A = A^*$, then $A$ is unitarily diagonalizable and all eigenvalues are real and non-negative. That is, for some unitary $U$, we have $U^*AU = \Lambda$, where $\Lambda = \operatorname{diag}(\lambda_1,...,\lambda_n)$, and $\lambda_k \ge 0$.
If we let $\Lambda^{\frac{1}{2}} = \operatorname{diag}(\sqrt{\lambda_1},...,\sqrt{\lambda_n})$, then we note that $(\Lambda^{\frac{1}{2}})^* = \Lambda^{\frac{1}{2}}$ and see that
$A = U \Lambda U^* = U \Lambda^{\frac{1}{2}} \Lambda^{\frac{1}{2}} U^*= U \Lambda^{\frac{1}{2}} (\Lambda^{\frac{1}{2}})^* U^* = U \Lambda^{\frac{1}{2}} ( U \Lambda^{\frac{1}{2}})^*$, as desired.
Aside:
The Hermitian assumption is necessary. The matrix $A= \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ satisfies $\langle x, Ax \rangle \ge 0$ for all $x$, but cannot be written as $B B^*$ for any matrix $B$. (This follows since $B B^*$ is self-adjoint, but $A$ is not.)
A: You cannot prove that any positive semidefinite matrix may be written $AA^*$ because it is not true.  A complete explication for the real case is found in my answer to this question; the argument in the complex case involves only the usual modifications:  transpose is replaced by Hermitian adjoint, orthogonal by unitary, and so forth.
Hope this helps!  Cheerio, 
and as always,
Fiat Lux!!!
