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Suppose A has eigenvalues of $-1$, $2$, and $4$. Find eigenvalues of the following matrices:

$B= A^{-1} + A$ and $C= A^2 + A -3I$

I am aware that I use quadratics to solve this, but thought I'd ask, since I haven't done something like this before. Thanks awesome math peoples.

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Suppose that $Av = 2v$ for some $v$. Then, multiplying both sides by $A^{-1}$ and dividing by $2$ gives us $A^{-1}v = \frac{1}{2}v$. So,

$$Bv = A^{-1}v + Av = \frac{1}{2}v + 2v = \frac{5}{2}v.$$

All the other cases can be solved similarly.

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Since $B$ and $C$ are polynomials of $A$, then the eigenvalues of those matrices are the result of evaluating those polynomials at the eigenvalues of $A$.

Edit:

Since $$B(A) = A^{-1}+A$$ then the polynomial correspond to $B$ is $p_B(x) = x^{-1}+x$. Therefore, you can plug in all the eigenvalues into $p_B(x)$ to get the eigenvalues of $B$. The reason why this works is that if you assume $A$ has an eigenvalue decomposition $$ A = T\Lambda T^{-1}$$ then $A^n = T\Lambda^n T^{-1}$, and $A^m + A^n = T(\Lambda^m+\Lambda^n) T^{-1}$, so any polynomial function of $A$ maps the eigenvalues according to the polynomial.

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  • $\begingroup$ $B$ is not a polynomial in $A$. $\endgroup$
    – mookid
    Nov 7 '14 at 23:48
  • $\begingroup$ $A^{-1}$ can be written as a polynomial of $A$ by Cayley-Hamilton, but we would need the algebraic multiplicities of the eigenvalues $\endgroup$
    – datalava
    Nov 7 '14 at 23:53
  • $\begingroup$ I meant polynomial in a looser sense where the exponents need not be positive. It still works. $\endgroup$
    – Victor Liu
    Nov 8 '14 at 0:38
  • $\begingroup$ So for B(A) = A^-1 + A, the eigenvalues of B would be -2, 5/2, and 17/4? $\endgroup$
    – JON oMEGA
    Nov 8 '14 at 1:21

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