How would you accomplish this:
Show that if $A$ is a positive semi definite matrix and $B$ is a positive definite matrix, both $n\times n$, then there is a matrix $C$, also $n\times n$, such that $CBC' = I$ and $CAC'$ is a diagonal matrix.
I tried using the Cholesky factorization $B = U'U$, where $U$ is an upper triangular matrix. Then if we let $C = (U')^{-1}$, the first condition works. However I don't think its true that the second part works. Any thoughts?
Thanks
On
This is called simultaneous diagonalisation by congruence. Presumably the notation $X'$ means the conjugate transpose of $X$ in the complex case, or simply transpose in the real case. If this is the case, consider a matrix of the form $C=UB^{-1/2}$, where $U$ is unitary, $B^{1/2}$ is the unique positive definite square root of $B$, and $B^{-1/2}$ is the inverse of $B^{1/2}$.
Consider $B = I$ and $A = \begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}$. If $CBC' = I$, then $CC' = I$, so $C$ is a 2 x 2 orthogonal matrix, i.e., a rotation or a rotation and reflection. Neither of these will conjugate the matrix $A$ into diagonal form. So I don't believe that you can prove the thing you want to prove.
Did you leave out some condition, like $AB = BA$?