1
$\begingroup$

I want to prove that the absolute value of the eigenvalues of a matrix A are smaller than 1 for $$A=\left(\begin{array}{cc} 0 & -H_{11}^{-1} H_{12} \\ -H_{22}^{-1} H_{21} & 0 \end{array}\right)$$ I know that H is positive definite with $$H = \left[\begin{array}{ll} H_{11} & H_{12} \\ H_{21} & H_{22} \end{array}\right]$$ I am not sure how to proceed here

Improve this question
$\endgroup$
4
  • $\begingroup$ Why do you think it's true? $\endgroup$
    – Anthony Quas
    Commented Jun 23, 2021 at 14:37
  • $\begingroup$ I showed this to my professor and asked if this is true, as I need it for my thesis. He said it does indeed hold but did not explain the proof. $\endgroup$
    – anonymousguyfromtheworld
    Commented Jun 23, 2021 at 14:42
  • 1
    $\begingroup$ This is not the right site for this kind of question. Please use math.stackexchange.com instead. $\endgroup$
    – Christian Remling
    Commented Jun 23, 2021 at 15:11
  • $\begingroup$ As for the question itself, write $A=D+B$, with $D$ the (block) diagonal part. You want to show that $-D^{-1}B$ has ev's $\le 1$. If $Bv=-\lambda Dv$, then $0=v^*(\lambda D+B)v= (\lambda-1)v^*Dv+v^*Av \ge (\lambda -1)v^*Dv$ and $v^*Dv\ge 0$ as well. $\endgroup$
    – Christian Remling
    Commented Jun 23, 2021 at 15:13

1 Answer 1

Reset to default
2
$\begingroup$

Here is the proof. Please let me use lighter notations: $$H=\begin{pmatrix} B & C \\ C^T & D \end{pmatrix},$$ so that $$A=\begin{pmatrix} 0_p & B^{-1}C \\ D^{-1}C^T & 0_q \end{pmatrix}.$$ From Schur complement formula, the characteristic polynomial of $A$ equals \begin{eqnarray*} \chi_A(X) & = & X^{p-q}\det(X^2I_q-D^{-1}C^TB^{-1}C) \\ & = & X^{p-q}\det(X^2I_q-D^{-1/2}C^TB^{-1}CD^{-1/2}). \end{eqnarray*} It amounts therefore to proving that the positive definite symmetric matrix $D^{-1/2}C^TB^{-1}CD^{-1/2}$ is less than $I_q$. Equivalently, we want to now whether $C^TB^{-1}C\prec D.$ But this is true because $D-C^TB^{-1}C$ is the Schur complement of $B$ in $H$, and the positivity of $H$ is equivalent to the positivity of both $B$ and its complement.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thank you Denis for your answer! I should have added that H is not symmetric, however I believe that this should not matter if I al correct? $\endgroup$
    – anonymousguyfromtheworld
    Commented Jun 23, 2021 at 15:43

Not the answer you're looking for? Browse other questions tagged .