3
$\begingroup$

$A$ is reflection matrix $2\times2$. $$B=A^4-2A^3-A-5I.$$ Find numbers $k$, $t$ in $\mathbb R$ so that $B^{-1}=kB+tI$.

I know that reflection matrix have eigenvalues of $1$, $-1$ ($A^2=I$) I got this: $$B=A^4-2A^3-A-5I=(A^2)^2-2(A^2)A-A-5I$$ and then:

$$B=I-2A-A-5I=-3A-4I$$

But what I can do from here?

Thanks!

$\endgroup$
  • $\begingroup$ Could you expand $BB^{-1}=I$ and then solve for k and t? $\endgroup$ – JB King Feb 22 '13 at 17:21
1
$\begingroup$

One quick and dirty way to do this: As you said, $A^2=I$ gives $B= -3A - 4I$. Now, orthogonally diagonalize $A$ as $A=QDQ^T$ where $D=\begin{pmatrix}1\\&-1\end{pmatrix}$. So $\operatorname{trace}(B)=-8$ and $\det B=\det\left(-3\begin{pmatrix}1\\&-1\end{pmatrix}-4I\right)=7.$ Therefore the characteristic equation of $B$ is $B^2+8B+7I=0$. Hence $B(B+8I)=-7I$ and $B^{-1}=-(B + 8I)/7=(3A-4I)/7$.

Alternatively, by Cayley-Hamilton theorem, if $B$ is invertible, its inverse would be a degree-1 polynomial in $B$ and hence a degree-1 polynomial in $A$. So, you just need to find $p$ and $q$ such that $(pA+qI)(-3A-4I)=I$.

$\endgroup$

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.