For a symmetric matrices $A$ and $B$ with $AB=BA$, show that the property $X$ holds

Question

For any square matrix $C$ with real entries, define $\lambda_1(C)$ as the maximum of the absolute values of the eigen values of $C$. Two square matrix $A_{n \times n}$ and $B_{n \times n}$ with real entries satisfy property $X$ if

$$\lambda_1(AB)\leq \lambda_1(A) \lambda_1(B)$$

1) For a symmetric matrices $A$ and $B$ with $AB=BA$, show that the property $X$ holds.

2) If the condition of symmetry dropped, does the property $X$ still hold?

My work: From the given condition $AB$ is also symmetric and all $A, B, AB$ are diagonalisable. But after that how do I proceed? Please help. Thanks.

In the case that $A$ and $B$ are diagonalizable, it suffices to note that they are necessarily simultaneously diagonalizable — Ben Grossmann, Feb 19 '18 at 05:24
I suspect that property $X$ can be shown to hold in general using continuity arguments — Ben Grossmann, Feb 19 '18 at 05:25
https://math.stackexchange.com/questions/603375/norm-of-a-symmetric-matrix-equals-spectral-radius might be of some help? — Theo Bendit, Feb 19 '18 at 05:41

score 1 · Answer 1 · answered Feb 19 '18 at 05:50

1

The following post shows that $\lambda_1 (A)$ is same as the norm of the matrix A. It is well-known that $||AB|| \leq ||A|| ||B||$. Hence 1) holds even if $AB \neq BA$. 2) is false. Try $2X2$ matrices. Norm of a symmetric matrix equals spectral radius

answered Feb 19 '18 at 05:50

Kavi Rama Murthy

359,332

Ohh nice....... – 1256 Feb 19 '18 at 05:52
I thought about posting the same thing, but I'm just wondering about $AB$. It doesn't need to be symmetric even if $A$ and $B$ are, so its maximum modulus of eigenvalues need not be its spectral radius. – Theo Bendit Feb 19 '18 at 05:54
But of course, if $AB = BA$, then $AB$ is symmetric! – Theo Bendit Feb 19 '18 at 05:57

score 0 · Answer 2 · answered Feb 20 '18 at 19:33

According to Burnside Theorem for matrix algebras, see the bottom of page 3 of these notes, we can restrict to the case of upper triangular matrices $A$ and $B$, as $A$ and $B$ can be simultaneously upper-triangularized. This is a similarity operation, which does not disturb eigenvalues. See also this answer.

Now let $A$ and $B$ be upper-triangular. We can read-off the eigenvalues of $AB$ from its diagonal, whose entries are $a_{ii}b_{ii}$. Thus the eigenvalues of $AB$ are products of diagonal entries of $A$ and $B$, which are itself their eigenvalues. Hence the claim follows.

For a symmetric matrices $A$ and $B$ with $AB=BA$, show that the property $X$ holds

2 Answers2