0

I am trying to solve the following problem:

Let $A=[a_{ij}]\in \mathbb{R}^{n\times n}$ is positive semidefinite and have positive diagonal entries. Show that the matrix $B=[b_{ij}]$ is positive semidefinite, where $b_{ij}=\frac{a_{ij}}{(a_{ii}a_{jj})^\frac{1}{2}}$?

At the beginning, I was trying to find the relation between the diagonal entries and eigenvalues of a symmetric to solve it, but I didn't find it. Then I think maybe there is a way to write the matrix $B$ into some form with $A$, but I failed to do it. Can you help me? Thank you!

Rookie
  • 99
  • This seems like it should be trivial just from the definition. Have you tried forgetting all that eigenvalue stuff and just showing that $\sum_i\sum_j\lambda_i\lambda_jb_{ij}\ge0$??? – David C. Ullrich Sep 29 '20 at 14:37

2 Answers2

3

Hint. $B=D^{-1}AD^{-1}$ where $D=\operatorname{diag}(a_{11}^{1/2},\ldots,a_{nn}^{1/2})$.

user1551
  • 139,064
  • How does that help??? – David C. Ullrich Sep 29 '20 at 14:35
  • 1
    @DavidC.Ullrich $B$ is PSD because it is congruent to the PSD matrix $A$. – user1551 Sep 29 '20 at 14:42
  • Assuming "congruent" is the same as "similar": Huh??? If you'd said $B=DAD^{-1}$ or $B=D^{-1}AD$ then yes... – David C. Ullrich Sep 29 '20 at 14:44
  • 1
    @DavidC.Ullrich No. Two matrices $A$ and $B$ are congruent if $B=P^\ast AP$ for some nonsingular matrix $P$. Congruent Hermitian matrices have the same indices of inertia, by Sylvester's law of inertia. – user1551 Sep 29 '20 at 14:47
  • @DavidC.Ullrich you should be aware that similarity transforms in general do not preserve positive (semi)definiteness while congruence transforms always do. See e.g. here https://math.stackexchange.com/questions/3798668/if-a-matrix-u-satisfy-uut-geq-0-positive-semidefinite-does-its-similar/ – user8675309 Sep 29 '20 at 17:36
  • Right. Once it was clear that what you wrote wasn't just a typo or something it didn't take long to figure that out... – David C. Ullrich Sep 29 '20 at 18:20
2

This is trivial from the definition of PSD. Since $A$ is PSD, for any choice of $(\lambda_j)$ we have $$\sum_j\sum_k\lambda_j\lambda_kb_{jk} =\sum\sum\omega_j\omega_ka_{jk}\ge0,$$where $\omega_j=\lambda_j/\sqrt{a_{jj}}$.

quid
  • 42,135
David C. Ullrich
  • 89,985