Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $A, B \in \mathbb{C^{n^2}}$ be hermitian matrices.

If $A$ is positive-definite, then there exists an invertible matrix $S\in \mathbb{C}^{n^2}$ such that $S^*AS=I_n$ and $S^*BS$ is a diagonal matrix.

I'm assuming we're supposed to use Sylvester's law of inertia (SLI). In fact finding $S$ which satisfies $S^*AS=I_n$ follows from SLI. But I see no reason why the same $S$ should diagonalize $B$.

Any ideas?


share|cite|improve this question
up vote 0 down vote accepted

What you mentioned is a standard result in matrix theory. If $A$ is positive definite, it can be unitarily diagonalized as $A=UDU^\ast$, where $D$ is a positive diagonal matrix. So, $A$ has a Hermitian square root $A^{1/2}:=UD^{1/2}U^\ast$, where $D^{1/2}$ is the entrywise square root of $D$.

Now consider $A^{-1/2}BA^{-1/2}$. This is Hermitian. Hence it can be unitarily diagonalized as $A^{-1/2}BA^{-1/2} = V\Lambda V^\ast$, where $\Lambda$ is a real diagonal matrix. Now take $S = A^{-1/2}V$ (hence $S^\ast=V^\ast A^{-1/2}$) and the result follows.

Both this result and Sylvester's law of inertia follow from unitary diagonalization of Hermitian matrices, but each of these two results does not depend on the other.

share|cite|improve this answer

This is more subtle than SLI. The fact in your question is equivalent to the spectral theorem for Hermitian matrices. I will use this formulation of this theorem:

If $X \in \mathbb{C}^{n^2}$ is a Hermitian matrix, then there exists a unitary matrix $Y \in \mathbb{C}^{n^2}$ such that $Y^{-1} X Y$ is a diagonal matrix.

You can see that the theorem can be obtained as a consequence from your statement if you take $A=I$ and $B=X$. It can also be reverted: you can prove your statement using this theorem. Can you see how?

As for the spectral theorem itself, it should be proved in the book you're reading (or the course you're attending, or whatever other resources you're using).

UPDATE: OK, here are a few details on how to derive the statement in the question from the spectral theorem.

First, there exists an invertible matrix $P$ such that $P^*AP=I$. Consider matrix $B'=P^*BP$. Clearly, $B'$ is Hermitian. By the spectral theorem, there is a unitary matrix $Q$ such that $Q^*B'Q=D$ is diagonal. Now we say that matrix $S=PQ$ has the required properties. Indeed: $$ S^*AS=Q^*P^*APQ = Q^*Q = I, $$ (here we used that $Q$ is unitary), and $$ S^*BS=Q^*P^*BPQ = Q^*B'Q = D. $$

share|cite|improve this answer
I know about the spectral theorem. To answer your question: no, I can't see how :( – Guy Dec 28 '12 at 23:11
@Guy $Y^*BY = D$ where $D$ is diagonal. This is the spectral theorem for Hermitian with $Y^*Y=I$. What now about $Y^*S^*ASY$? – adam W Dec 28 '12 at 23:29
@adamW What is $S$? – Guy Dec 28 '12 at 23:44
@Guy Sorry, I think I made a mistake somewhere. But if $S$ such that $S^*AS=I$ is found first, then next find $Y$ such that $Y^*BY = D$ with $Y^*Y=I$. Can those be combined in some way? For instance, $Y^*S^*ASY = Y^*IY = I$ shows that more than one congruence gives the identity. – adam W Dec 29 '12 at 0:12
Wait a minute. Why is it true that $(SY)^*B(SY)$ is diagonal? – Guy Dec 29 '12 at 1:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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