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

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 $(B^2 + I)^{−1}$
• What is the $T$ in $BTB$? Should that be $B^TB$? – Omnomnomnom Dec 15 '16 at 16:01

## 1 Answer

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 $\frac{1}{5}$.

• 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