$xABAx^T = xBx^T$ where $B\succeq A\succ0$

Question

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

Is there a relation between the matrices? their eigenvalues?

Answer 1

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.

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Example:

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:

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

Then:

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

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:

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