# Finding the eigenvalues of $A^2$ and $A^{-1}$ from the characteristic polynomial of $A$

In my question we are given a characteristic polynomial $x^3 - 4x^2 + x + 6$ for a matrix A and are told to find eigenvalues for both $A^{-1}$ and $A^2$. I just want to confirm that I am approaching this correctly. So I found the eigenvalues for matrix $A$, which are $x = -1, x = 2, x = 3$. Looking at the equation $A^n = P^{-1} D^n P$, where D is known because we have the eigenvalues, to find the eigenvalues of $A^2$ would we just have to do $(-1)^2, 2^2, 3^2$ ? and then a similar approach for $A^{-1}$.

Thank you.

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Looks OK....... – Marc van Leeuwen Mar 25 '13 at 15:39
Suppose $v$ is an EigenVector to, say, Eigenvalue 2. This means $Av = 2v$. Now, can you see what $A^{-1}v$ is, and what $A^2v$ is ? – Djaian Mar 25 '13 at 15:40
In general, if $\lambda$ is an eigenvalue, then for some vector $v$, $Av =\lambda v$. Then $A^2v = A(\lambda v) = \lambda Av = \lambda^2 v$. A similar argument can be made for $A^{-1}$. – Thomas Andrews Mar 25 '13 at 15:40
@anon: The question asks for the actual eigenvalues of $A^2$, $A^{-1}$, not for their characteristic polynomials. So you'll have to factor a polynomial somewhere, why not just do it for the given one? – Marc van Leeuwen Mar 25 '13 at 15:44
Oosp, I was confusing this with the many questions about traces of powers. – anon Mar 25 '13 at 15:45

HINTS Characteristic polynomial of $A^{-1 }$is the polynimial whose roots are recripocal to roots of the given polynomial.