1
$\begingroup$

I'm trying to find a proof of a supposedly simple property of symmetric positive definite matrices concerning simultaneous diagonalization.

Let $A$ and $B$ be real symmetric matrices, with A positive definite and B positive semidefinite. Call $C=A^{-1}B$.

I know from e.g. Theorem 4.5.15 in Horn-Johnson's Matrix Analysis that there exists a non-singular matrix $S$ such that $SAS^T$ and $SBS^T$ are diagonal if and only if there exists a non-singular matrix $R$ such that $R^{-1}CR$ is diagonal and $C$ is diagonalizable. In the problem I'm working on, I can suppose that $C$ satisfies these properties, then I'm sure that the matrix $S$ indeed exists.

Moreover, from Theorem 12.19 in Laub's Matrix Analysis for Scientist and Engineers I know that there exists a non-singular matrix $S$ such that $S^TAS=1$ and $S^TBS=D$, with D a diagonal matrix whose diagonal elements are the eigenvalues of $C$.

I specify that these theorems work with weaker hypotheses than mine: Horn-Johnson's one deals with symmetric $A$ and $B$, Laub's further assumes $A$ to be positive definite, with no remarks on $B$.

Now, is it possible to prove that in the situation I'm working on it is $S=R$? Maybe it is a stupid question, but I cannot prove this result in general. The only thing I can derive is that $R^{-1}SDS^{-1}R=D$, but I have examples that make me think that $S=R$ is a general property, while I'm not able to show why. I only suppose that this can be proven if one includes the positive-semidefiniteness of $B$ in the hypoteses of e.g. Laub's theorem.

Thank you all in advance for your help.

$\endgroup$
  • $\begingroup$ Please correct the variable names. E.g., $Q$ only appears in the existence assumption. $\endgroup$ – Tobias Nov 12 '16 at 11:30
0
$\begingroup$

Your guess of $S=R$ is right if $A$ is positive definite and $B$ is symmetric.

Eigen decomposition of $A$: $A = Q \Lambda Q^T$ with orthogonal $Q$ and diagonal $\Lambda>0$.

We obtain $A = P\,P^T$ with $P:=Q\Lambda^{-\frac12}$.

The matrix $P^T B P$ is symmetric. Therefore, it admits an eigenvalue decomposition $D = V^T P^T B P V$ with diagonal $D$ and orthogonal $V$.

We have $1 = V^T \underbrace{P^T A P}_{=1} V$ as required.

Thus $S:= PV = Q\Lambda^{-\frac12}V$ is the first wanted transformation matrix.

Now we test whether we can use $R:=S$ for $R^{-1} A^{-1} B R \overset{?}{=} D$.

\begin{align*} S^{-1}A^{-1}BS &= \underbrace{V^T \Lambda^{\frac12} Q^T}_{S^{-1}} \underbrace{Q\Lambda^{-1} Q^T}_{A^{-1}}\, B\, \underbrace{Q \Lambda^{-\frac12} V}_{S}\\ &= V^T \underbrace{\Lambda^{-\frac12}Q^T}_{P^{T}} B \underbrace{Q \Lambda^{-\frac12}}_{P} V\\ &= D \end{align*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.