Cholesky decomposition for symmetric positive semi-definite matrices

Question

On page 5 here: https://stanford.edu/class/ee363/lectures/lmi-s-proc.pdf

$A$ and $B$ are decomposed into $A^{1/2} A^{1/2}$ and same for $B$.

Is this from Cholesky decomposition? Can someone prove that PSD symmetric matrices always admit this factorization?

Answer 1

No, $A^{1/2}$ does not refer to the lower triangular matrix obtained from Cholesky decomposition. It is the unique positive semidefinite square root of $A$. That is, if $A=QDQ^T$ is an (Schur) orthogonal diagonalisation, then $A^{1/2}$ is defined as $QD^{1/2}Q^T$, where $D^{1/2}$ is the entrywise square root of $D$. It is evident that $A^{1/2}$ is positive semidefinite (and symmetric in particular) and $(A^{1/2})^2=A$.

The identity in question follows from the tracial property $\operatorname{tr}(XY)=\operatorname{tr}(YX)$: \begin{aligned} \operatorname{tr}(AB) &=\operatorname{tr}(\underbrace{A^{1/2}}_X\,\underbrace{A^{1/2}B^{1/2}B^{1/2}}_Y)\\ &=\operatorname{tr}(\underbrace{A^{1/2}B^{1/2}B^{1/2}}_Y\,\underbrace{A^{1/2}}_X)\\ &=\operatorname{tr}\left((A^{1/2}B^{1/2})(A^{1/2}B^{1/2})^T\right)\\ &=\| A^{1/2}B^{1/2}\|_F^2. \end{aligned}