4
$\begingroup$

I have two matrices A, B. A relates to a physical system and all of its eigenvalues have negative real parts. B is a diagonal matrix with nonzero and unequal elements on its diagonal, which are in fact its eigenvalues. Let's say A and B are 5 by 5 and I want to know about eigenvalues of A*B. If for example 2 elements of B are negative and the rest are positive, can I say that 2 eigenvalues of A*B have positive real parts, and other eigenvalues of A*B have negative real parts ? I tried with different choices of A and B, and this thing held for all of them [seems like sign of eigenvalues of A and B are multiplied by each other in A*B], so I thought that there might be a proof or theory about it ! Can anybody help me ?

$\endgroup$

1 Answer 1

6
$\begingroup$

What would be true is the following: suppose $A$ is symmetric and negative definite, and let $(-A)^{1/2}$ be the positive definite square root of $-A$. Then by Sylvester's Law of Inertia, the numbers of positive and negative eigenvalues of $(-A)^{1/2} B (-A)^{1/2}$ are equal to the corresponding numbers for $B$; but the eigenvalues of $(-A)^{1/2} B (-A)^{1/2}$ are equal to the eigenvalues of $-AB$, so the number of positive (resp. negative) eigenvalues of $AB$ is equal to the number of negative (resp. positive) eigenvalues of $B$.

EDIT: Here is a $3 \times 3$ counterexample where $A$ is not symmetric. Take $$ A = \pmatrix{ -8.2 & 1 & 1\cr -45 & -3 & 0\cr 0 & 8.3 & 5.3\cr},\ B = \pmatrix{2 & 0 & 0\cr 0 & -1 & 0\cr 0 & 0 & -2\cr}$$ Then $A$ has eigenvalues $-3, -2.2, -0.7$, and $B$ has one positive and two negative eigenvalues, but $AB$ has one negative real eigenvalue (approximately $-23.9$) and two complex eigenvalues with negative real part (approximately $-.0432 \pm .878 i$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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