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I solved this exercise, that is the exercise 5.B.11 of Linear algebra done right, third edition of Axler

Let $V$ a complex vector space and $T\in\mathcal L(V)$, $p\in\mathcal P(\Bbb C)$ a polynomial and $\alpha\in\Bbb C$. Prove that $\alpha$ is an eigenvalue of $p(T)$ if and only if $\alpha=p(\lambda)$ for some eigenvalue $\lambda$ of $T$.

for any polynomial of degree at least one. My problem is that I cannot see if the result holds (vacuously or not) when the polynomial is constant or zero.

If $V$ would be finite-dimensional I know that $T$ have at least one eigenvalue. My problem is that if $V$ is infinite-dimensional and $T$ injective then I dont see clearly if $T$ must have necessarily at least an eigenvalue.

By the other side it doesnt seems correct to assert that $\alpha=p(\lambda)$ when such $\lambda$ doesnt exists, that is, the statement dont seems to hold vacuously.

Can someone help me to clarify this question?

P.S.: I dont know exactly what tags I must use for this question.

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  • $\begingroup$ Doesn't the problem say $p$ is a polynomial of degree at least $1$? $\endgroup$ – egreg Jun 8 '17 at 23:02
  • $\begingroup$ @egreg no. It just say any polynomial with complex coefficients. Specifically it says for $p\in\mathcal P(\Bbb C)$, that is the set of polynomials with complex coefficients (including the zero and the constant ones). $\endgroup$ – Masacroso Jun 8 '17 at 23:02
  • $\begingroup$ If $p$ is a constant (zero or not), then the statement says "$\alpha$ is an eigenvalue of (the linear map given by multiplying by) $p$ if and only if $\alpha$ equals (the number) $p$". $\endgroup$ – Arthur Jun 8 '17 at 23:04
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    $\begingroup$ @Masacroso: You are correct. I missed your point. Indeed my proof below depends on $\partial p \ge 1$. $\endgroup$ – copper.hat Jun 9 '17 at 0:05
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    $\begingroup$ Thank you for pointing out this issue, which is an error on my part. For the next edition of the book, I have changed the hypothesis so that $p$ is a nonconstant polynomial. There is no need to assume that $V$ is finite-dimensional for this exercise. $\endgroup$ – Sheldon Axler Jun 9 '17 at 18:09
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Having reviewed the chapter in question, I think you're right: this holds if the polynomial has degree $>0$, or you know there is an eigenvalue, but is not true in general if there is no eigenvalue and $p$ is constant: then we do have $\{ p(\lambda) \mid \lambda \text{ is an eval of } T \} = \emptyset $.

So the result holds if at least one of the following is true:

  • $\deg{p}>0$.
  • $V$ is finite-dimensional,
  • or more generally, you know $T$ has an eigenvalue.

If none of these is true, the result may not hold.

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  • $\begingroup$ The book proved that if $V$ is a finite-dimensional complex vector space then $T$ have at least an eigenvalue. But the exercise doesnt state that $V$ is finite dimensional, it just state that $V$ is a complex vector space. $\endgroup$ – Masacroso Jun 8 '17 at 23:29
  • $\begingroup$ @Chappers It's false that every endomorphism has an eigenvalue if the vector space is not finite dimensional. $\endgroup$ – egreg Jun 8 '17 at 23:40
  • $\begingroup$ @egreg Yes, I know. The forward shift is the typical example. $\endgroup$ – Chappers Jun 8 '17 at 23:46
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    $\begingroup$ @Masacroso Hm. Having reviewed the chapter in question, I think you're right: this holds if the polynomial has degree $>0$, or you know there is an eigenvalue, but is not true in general if there is no eigenvalue and $p$ is constant. $\endgroup$ – Chappers Jun 8 '17 at 23:49
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    $\begingroup$ Looking in the solutions manual, Axler has assumed the existence of an eigenvalue in his proof. $\endgroup$ – Chappers Jun 8 '17 at 23:57

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