0
$\begingroup$

A $3\times 3$ matrix $B$ is known to have eigenvalues $0, 1$ and $2$. This information is enough to find three of these (give the answers where possible):

  1. The rank of $B$
  2. The determinant of $BT B$
  3. The eigenvalues of $BT B$
  4. The eigenvalues of $(B^2 + I)^{−1}$
$\endgroup$
  • $\begingroup$ What is the $T$ in $BTB$? Should that be $B^TB$? $\endgroup$ – Omnomnomnom Dec 15 '16 at 16:01
2
$\begingroup$

We have that $A$ has eigenvalue $\lambda\iff A^{-1}$ has eigenvalue $\lambda^{-1}$.

To see this, note that: \begin{align*} Av & = \lambda v \\ A^{-1}Av & = \lambda A^{-1}v \\ v & = \lambda A^{-1}v \\ A^{-1}v & = \frac{1}{\lambda}v \end{align*} Here, we're assuming $A$ is invertible. This will be fine, as while $B$ isn't invertible, $B^2+I$ is.

Now, we have that $B^2+I$ has eigenvalues $0^2+1,1^2+1$, and $2^2+1$, or $1,2,5$. It follows that $(B^2+I)^{-1}$ has eigenvalues $1,\frac{1}{2}$, and $\frac{1}{5}$.

$\endgroup$
  • $\begingroup$ Thank you! I wasnt sure if i could add the eigenvalues like that. This was invaluable for me thank you again! $\endgroup$ – Control systems engineer Dec 15 '16 at 10:48
  • $\begingroup$ @Controlsystemsengineer See this, for example, for eigenvalues of a polynomial of a matrix. $\endgroup$ – amd Dec 15 '16 at 18:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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