3
$\begingroup$

For a skew symmetric block $n \times n$ matrix $B$, prove that matrix $M$ has two double eigenvalues.

$$ M = \begin{bmatrix} I & B \\ B & I\end{bmatrix} $$

For a proof, I was using the determinants, but got stuck as I do not know how to take into the account the skew symmetric matrices.

\begin{align} \det(M-\lambda I) &= \det((1-\lambda)I)\det((1-\lambda)I-(1-\lambda)^{-1}BB)\\ &=(1-\lambda)^{n}\det((1-\lambda)(I-(1-\lambda)^{-2}BB)) \\ &=(1-\lambda)^{n}(1-\lambda)^{n}\det((1-\lambda)^{-2}((1-\lambda)^{2}I-BB)) \\ &=(1-\lambda)^{n}(1-\lambda)^{n}(1-\lambda)^{-2n}\det((1-\lambda)^{2}I-BB) \\ &=\det((1-\lambda)^{2}I-BB) \end{align}

At the last step I got stuck as I am not sure how to proceed to prove that there are two double eigenvalues.

$\endgroup$
2
  • 2
    $\begingroup$ Think of the characteristic polynomial of $B^2$. $\endgroup$ Commented Sep 6, 2021 at 11:08
  • $\begingroup$ @RodrigodeAzevedo got it, thank you! $\endgroup$ Commented Sep 6, 2021 at 11:13

1 Answer 1

Reset to default
1
$\begingroup$

Following the @RodrigodeAzevedo comment, I found the solution to be: $$ \det\left((1-\lambda)^{2}I-BB\right) \implies \det\left(B^2-\tilde{\lambda}^2I\right) $$ $$ \det\left(B^2-\tilde{\lambda}^2I\right) = \det\left(B-\tilde{\lambda}I\right)\det\left(B+\tilde{\lambda}I\right) $$ considering the skew symmetric matrix $$ B=-B^{\mathrm T} $$ $$ \det\left(B^{\mathrm T}+\tilde{\lambda}I\right)\det\left(B+\tilde{\lambda}I\right) $$ And because the eigenvalues of a matrix and its transpose are the same, we have double eigenvalues.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.