Assume $xABAx^T = xBx^T$ for some vector $x\neq0$ with $B\succeq A\succ0$.

Is there a relation between the matrices? their eigenvalues?

  • $\begingroup$ For some vector $x$? You need a more particular choice of $x$ if you hope to deduce anything about the relationship "between the matrices" or their eigenvalues. $\endgroup$ – hardmath Feb 17 at 4:46
  • $\begingroup$ Is $x\neq0$ sufficient?@hardmath $\endgroup$ – Morad Feb 17 at 5:48
  • $\begingroup$ @Morad What hardmath means is whether the assumption is true for one independent vector $x$ or for some particular subset of $X$, or, as you seem to imply in your comment, for all $x$. $\endgroup$ – Chrystomath Feb 17 at 8:05
  • 3
    $\begingroup$ The strongest assumption would be that the condition holds for all (nonzero?) $x$, but I have the impression you wanted to avoid imposing that condition. The problem statement is very terse, and expanding on the problem's setup, the goal, and what motivates your interest in the problem would improve your Question. $\endgroup$ – hardmath Feb 17 at 15:11
  • $\begingroup$ @hardmath I was thinking whether $A$ should have a unit-circle eigenvalue or some property of preserving the size in a certain direction but did not how to proceed. $\endgroup$ – Morad Feb 17 at 16:27

I will assume that "positive definite" here means symmetric (real) positive definite. The Loewner order can as easily be imposed on Hermitian matrices however.

The condition that a single nonzero vector $x$ satisfies:

$$ xABAx^T = xBx^T $$

is satisfied if the symmetric matrix $ABA - B$ is singular.

This can be arranged in many ways, and in particular if it were true of some pos. definite matrices $A,B$, then $B$ could as well be replaced by a scalar multiple $rB$. So the (positive real) eigenvalues of $B$ can be made arbitrarily large.

Thus it seems to me that the condition $B\succeq A\succ 0$ will not add much to restrict the choice of $A$. I'm not clear what the goal of the Question is, but I thought these few remarks might help the OP clarify thinking about what kind of inference about the matrices $A,B$ is desired.


A simple illustration of these ideas is possible with $2\times 2$ matrices. Any real symmetric matrix can be diagonalized by suitable choice of an orthogonal basis, though of course we cannot generally diagonalize both $A,B$ at the same time (that would require quite a special relationship). But there is no loss of generality in assuming $A$ is diagonal with positive entries $x,y$ and $B$ is an arbitrary symmetric positive definite matrix.

For simplicity I'll take:

$$ A = \begin{pmatrix} x & 0 \\ 0 & y \end{pmatrix} \\ B = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} $$


$$ ABA - B = \begin{pmatrix} 2x^2-2 & xy-1 \\ xy-1 & 2y^2 -2 \end{pmatrix} $$

Since all we want is for this matrix to be singular, it suffices to pick $x,y\gt 0$ such that its determinant is zero:

$$ (2x^2 - 1)(2y^2 - 1) - (xy - 1)^2 = 0 $$

To avoid having an example where the entries $x,y$ are equal, and then the eigenvectors of $A$ and $B$ and not necessarily distinct, I first picked $x = 2$ and solved for $y$:

$$ 7(2y^2 - 1) - (2y - 1)^2 = 0 $$ $$ 10y^2 + 4y - 8 = 0 $$

This has two real roots, opposite in sign, and the positive root is:

$$ y = \frac{-1 + \sqrt{21}}{5} \approx 0.716515139 $$

Now this makes $ABA - B$ singular, but it doesn't satisfy $B \succeq A$. So instead of $B$ as chosen above, we pick a sufficiently large multiple $rB$ so that $rB-A$ is positive definite.

In fact $3B - A$ is (real symmetric) positive definite because it's strictly diagonally dominant:

$$ 3B - A = \begin{pmatrix} 4 & 3 \\ 3 & \frac{31 - \sqrt{21}}{5} \end{pmatrix} $$

  • $\begingroup$ Can you please explain more about the the idea with the scaling $r$. $\endgroup$ – Morad Feb 17 at 21:03
  • 1
    $\begingroup$ Note that if $r\gt 0$, then $ABA-B$ is singular if and only if $r(ABA-B)=A(rB)A-(rB)$ is singular. In other words you can just as well replace $B$ by an arbitrarily large scalar multiple in order to make $rB\succeq A$. I can add more to my Answer to further explain that idea if it would help. $\endgroup$ – hardmath Feb 17 at 22:21
  • $\begingroup$ I see, it shows that the matrix ordering I gave is redundant. $\endgroup$ – Morad Feb 17 at 22:58

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.