I have stumbled across this problem, but i'm stuck given my very short experience with matrix algebra. I would be glad if anyone can help me or tell me it's simply impossible.
I have 3 matrix and the following equation : $Y=AXA^t$ and would like to get $X$ and $Y$ on the same side and isolate $A$ or $AA^t$.
An example I'm working right now is:
$X = \begin{bmatrix}1 & .298 & .683 &.231 \\ .298 & 1 &.353 & .201 \\ .683 & .353&1&.311 \\ .231 & .201&.311&1\end{bmatrix}$,
which has the same properties of a covariance matrix.
$A = \begin{bmatrix}1 & 0 & 0 & 0 \\ -.298 & 1 & 0 & 0 \\ -.634 & -.164 & 1 & 0 \\ -.024 & -.102 &.-259 & 1 \end{bmatrix}$,
and
$Y = \begin{bmatrix}1 & 0 & 0 & .0 \\ 0 & .911 & 0 & 0 \\ 0 & 0&.509&0 \\ 0 & 0 & 0 & .893\end{bmatrix}$.
Thank you,
EDIT
@Wouter suggested vectorization:
$Y=AXA^t$,
$\text{vec}(Y)=(A\otimes A) \text{vec}(X)$,
however, this does not seem to work as their is some number off at the 4th and 5th decimal. $(A\otimes A) \text{vec}(X)= \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ .9114 \\ .0001 \\ -.0004 \\ 0 \\ .0001 \\ .5093 \\-.0001 \\ .0000 \\ -.0004 \\ -.0001 \\ .8924 \end{bmatrix}$.
Is this normal? Then, I'm not sure how to put $\text{vec}(X)$ with $\text{vec}(Y)$. Is it $\text{vec}(Y)[\text{vec}(X)^{-1}]^t=(A\otimes A)$, because this does not work either?
Again, thank for your help!
vectorization $$Y=A X A^T$$ $$\text{vec}(Y)=(A\otimes A) \text{vec}(X)$$ where $\otimes$ is the Kronecker product.