Intereting Posts

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Clearly if $\lambda$ is an eigenvalue of $A$ then $p(\lambda)$ is an eigenvalue of $p(A)$ where $p$ is a polynomial. And there are cases where $A$ may have eigenvalues other than these.

Is there a general rule for finding all eigenvalues of $p(A)$? Does it depend on the field from which the elements of $A$ are taken? Are their any partial solutions to this problem?

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**If the field is $\mathbb{C}$ then $a$ is an eigenvalue of $p(A)$ if and only if $a = p(\lambda)$, with $\lambda$ an eigenvalue of $A$.**

**If the field is $\mathbb{R}$ then the previous result is false** and a counterexample is given by the rotation of ninety degrees, namely take $A$ to be the martix associated to the application $T(x,y) = (-y,x)$ and $p(x) = x^2$. This is a counterexample since (as you can check) $T$ has no eigenvalue, while $T^2 = -I$ has $-1$ as an eigenvalue.

The proof of the first result (as you can imagine) relies heavily on the fundamental theorem of algebra. Once you realize this the proof is straightforward.

**Sketch of the proof:**

- [$\Longrightarrow$] write $p(z) – a = c\prod(z – \lambda_i)$. This gives you $$p(T) – aI = c(T – \lambda_1I)\dots(T – \lambda_mI).$$

If $a$ is an eigenvalue of $p(T)$, then $p(T) – aI$ is not injective, and hence one of the $(T – \lambda_iI)$ is not injective. - [$\Longleftarrow$]If $a = p(\lambda)$, where $\lambda$ is an eigenvalue of $T$ with eigenvector $v$, then $T^kv = \lambda^kv$. From this, a one second thought gives $p(T)v = p(\lambda)v$, and hence the thesis.

I am sure that if you are interested in the proof it would be easy to fill this sketch with all the details needed to make it a rigorous proof. ðŸ˜€

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