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This question originated from my interest in a bigger question: to what extent are spectral or other kinds of matrix decompositions necessary to prove some properties about Hermitian or positive (semi)definite matrices? Here are some example statements that can be proved without using any matrix decompositions, although the proofs are much easier if we are allowed to use spectral decomposition or take matrix square roots:

  1. If $P\succeq0$ and $\langle Px,x\rangle=0$, then $Px=0$.
  2. If $P\succeq0$ and $A=A^\ast$, all eigenvalues of $AP$ are real.
  3. If $P\succ0$ and $P^n=I$ for some integer $n\ge2$, then $P=I$.

Yesterday, I was thinking about a variant of statement 2 above:

  • If $P\succ0$ and $A=A^\ast$, then $A\succ0$ if and only if $AP$ has a real positive spectrum.

The usual proof relies on the fact that $AP$ is similar to $P^{1/2}AP^{1/2}$. Strictly speaking, the existence of $P^{1/2}$ does not involve spectral decomposition — just think about the proofs for the existence of positive square roots of bounded positive linear operators in Hilbert spaces — but for positive definite matrices, spectral decomposition is undoubtedly a shortcut for proof. Other decompositions of $P$ can also be used. E.g. $AP$ is similar to $L^\ast AL$ where $L$ is the Cholesky factor of $P$.

Can we prove the bullet statement above without using any sort of matrix decomposition?

The forward implication ($A\succ0$ only if $AP$ has a positive spectrum) is easy: let $(\lambda, x)$ be an eigenpair of $AP$. Then $\langle Px,APx\rangle=\lambda \langle Px,x\rangle$. Therefore $\lambda=\frac{\langle Px,APx\rangle}{\langle Px,x\rangle}$ is real and $>0$.

However, I have trouble to prove the backward implication ($A\succ0$ if $AP$ has a positive spectrum). Since I need to infer positive definiteness from a spectral condition here, I am willing to compromise and accept uses of spectral decomposition on $A$ or Jordan form of $AP$, but I still want to avoid decompositions of $P$. Any idea?

user1551
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  • @user8675309 The $\succeq0$ assumption was coincidental. For proof purpose one may of course assume that $AP$ has a strictly positive spectrum. Anyway I will change it right away. – user1551 Nov 09 '24 at 17:05
  • how about using the theorem of Lyapunov in https://math.stackexchange.com/questions/670650/square-root-of-positive-definite-nonsymmetric-matrix ? i.e. define $T:M_n\big(\mathbb C\big)\longrightarrow M_n\big(\mathbb C\big)$ given by
    $T\big(X\big) = (AP)X + X(AP)^*$ and consider $T(A)$. The Horn proof is decomposition based but the integral based inverse (which I'm less familiar with) may do it. You can get the result via $T_\tau(A)$ using the path $(1-\tau)\cdot (AP) + \tau \cdot A$ then finish with winding numbers since $T_\tau(A)$ can never have an eigenvalue in $i\cdot \mathbb R$.     – user8675309 Nov 09 '24 at 20:27
  • @user8675309 Yeah, if you have a proof I would appreciate it. I was pondering about the same tricks too, but haven’t reached anything yet. Frankly I can’t quite concentrate now. My work machine is an old MacBook. I decided to upgrade the software yesterday (“brew upgrade” if you know what I mean) but after a whole day, the process is still not finished. No idea why it doesn’t install binaries but opts to compile stuffs from source. The processor is working so hard that the fan is now spinning like crazy. This is so frustrating. – user1551 Nov 09 '24 at 20:55
  • haha, and here I thought it was difficult to do ubuntu upgrades – user8675309 Nov 09 '24 at 21:11
  • @user8675309 You are right. Thanks a lot! I did have thought of going Lyapunov, but I kept thinking about $X\mapsto AX+XA$ or $X\mapsto PX+XP$ rather than $T:X\mapsto APX+XPA$ and so I came to a dead end. In hindsight the right choice of $T$ should be obvious. – user1551 Nov 10 '24 at 19:51

1 Answers1

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As pointed out by user8675309 in a comment, all we need is to view $A$ as the solution to an appropriate Lyapunov equation.

It is well-known that if $A_0$ Hurwitz (a.k.a. stable) and $Q$ is any complex square matrix, the Lyapunov equation $A_0X+XA_0^\ast=-Q$ always has a solution $X=\int_0^\infty e^{tA_0}Qe^{tA_0^\ast}dt$. This means the linear endomorphism $T:X\mapsto A_0X+XA_0^\ast$ is surjective. In turn it must be invertible and the solution to the Lyapunov equation $T(X)=-Q$ is unique. From the integral representation of $X$ above, it is also obvious that when $Q$ is positive or negative definite, $X$ must be Hermitian and it has the same definiteness as $Q$.

Now, when $A_0=-AP$ and $Q=2APA$, the Lyapunov equation is solved by $X=A$. Since $Q\succ0$, so is $A$.

user1551
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