Finding $(B^2+I)^{-1}$ using eigenvalues and size.

Question

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):

The rank of $B$
The determinant of $BT B$
The eigenvalues of $BT B$
The eigenvalues of $(B^2 + I)^{−1}$

What is the $T$ in $BTB$? Should that be $B^TB$? – Ben Grossmann Dec 15 '16 at 16:01 — Ben Grossmann, Dec 15 '16 at 16:01

score 2 · Answer 1 · answered Dec 15 '16 at 10:36

2

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 $\frac{1}{5}$.

answered Dec 15 '16 at 10:36

Mark Schultz-Wu

6,986

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

Finding $(B^2+I)^{-1}$ using eigenvalues and size.

1 Answers1

Linked