# (Fast) eigen decomposition of $DXD$ where $D$ is diagonal, $X$ is symmetric with known eigen decomposition

Assuming that I already know the eigen-decomposition of a real symmetric matrix $X$, is there any way to use it to retrieve efficiently the eigen-decomposition of $DXD$, where $D$ is a diagonal matrix? (I mean, more efficiently than computing $DXD$ and decomposing it).

Modulo a base change, this is the same as asking for an efficient way to eigen-decompose $A\Sigma A$ with $\Sigma$ diagonal and $A$ symmetric with known eigen-decomposition.

Maybe some iterative algorithm can use the known eigenvectors as a starting point and converge really fast?

In fact, I strongly doubt there is anything, but I will feel better if someone can provide an definitive argument to close the question.

• On this site questions get closed even without providing a definitive argument ;-) Jan 21 '15 at 8:52
• well, if $D^2 = I,$ then the decomposition of $DXD = (DU)\Omega (DU)^{-1}, A = U\Omega U^{-1}$
– abel
Jan 21 '15 at 11:52
• Have a look at: math.stackexchange.com/a/1929763/88146 May 18 '19 at 2:13

The thing you care about is $$DXD$$. Let's rename that $$A$$. And let's say $$U$$ is the eigen-vector matrix of $$X$$ and $$V$$ is the eigen-value matrix of $$X$$.
\begin{align} A &= DXD\\ &= DU'VUD\\ &= D'U'VUD\\ &= (UD)'V(UD) \end{align}
I would say this is the eigen-decomposition of $$A$$. $$UE$$ is the eigen-vector matrix and $$V$$ is the eigen-value matrix.