I am a graduate student studying for a Linear Algebra qualifying exam and I have been going over sample problems from previous exams. The recommended text for these problems are Hoffman and Kunze "Linear Algebra", Chapter three of Jacobson "Algebra I" and The module theory section of Dummit and Foote. Most likely this problem is a variation of an exercise from one of the texts I just mentioned but I have had trouble placing it.

I think the background for the problem comes from a converse to the following lemma on page 186 of Hoffman and Kunze.

Lemma: Let $V$ be a vector space over the field $F$ and let $T$ be a linear operator on $V$. Suppose that $T \alpha = c \alpha$ for some vector $\alpha \in V$ and scalar $c \in F$. Then if $f$ is any polynomial, $f(T) \alpha = f(c) \alpha$.

Finally here is the question I am having problems with:

Let $T: \mathbb{C}^5 \rightarrow \mathbb{C}^5$ be a linear operator and let $g(x)$ be a polynomial in $\mathbb{C}[x]$. If $c$ is a characteristic value for $g(T)$, must there exist a characteristic value $a$ for $T$ such that $g(a) = c$? Explain why or why not

My guess is that the question is not true but I am having trouble constructing an example. Thank you for any advice you can give.

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    $\begingroup$ Hint: The answer is affirmative and in fact the result holds for all complex vector spaces ... $\endgroup$ Jul 19, 2011 at 12:32

1 Answer 1


Let $V$ be a complex vector space, let $T$ be an operator on $V$ and let $p\in \mathbb{C}[x]$. The following steps lead to a solution:

Exercise 1: Prove that the result holds if $p$ is a constant polynomial.

(1) (We can assume, without loss of generality, that $p$ is non-constant by Exercise 1.) Let $\lambda$ be an eigenvalue of $p(T)$. Note that $p(T)-\lambda I$ is not injective and we can factor the polynomial $p(z)-\lambda = c(z-c_1)\cdots (z-c_k)$ for some positive integer $k$ and scalars $c,c_1,\cdots,c_k\in\mathbb{C}$.

Exercise 2: Prove that $T-c_iI$ is not injective for some $1\leq i\leq k$. Deduce that $c_i$ is an eigenvalue of $T$.

Exercise 3: Prove that $p(c_i)=\lambda$.

You should now be able to solve your question. Let me give another couple of Exercises:

Exercise 4: Prove that there exists an operator $T:\mathbb{R}^2\to \mathbb{R}^2$ with characteristic polynomial $p(x)=x^2-1$. Is this operator unique? If $T$ is any such operator, prove that there is an eigenvalue of $p(T)$ that is not of the form $p(\lambda)$ for an eigevalue $\lambda$ of $T$. Therefore, the answer to your question is negative in the context of real vector spaces.

Exercise 5: Let $V$ be an odd-dimensional real vector space, let $T$ be an operator on $V$ and let $p\in\mathbb{R}[x]$. If $\lambda$ is an eigenvalue of $p(T)$, is it true that there is an eigenvalue $a$ of $T$ such that $p(a)=\lambda$? Prove or give a counterexample.

I hope this helps!

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    $\begingroup$ thank you I am in the process of working through your excersises $\endgroup$
    – user7980
    Jul 19, 2011 at 13:09
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    $\begingroup$ @user7980 May I ask if you have solved all the Exercises? (Exercises 1, 2, 3 are most relevant to your question but they are all routine.) $\endgroup$ Jul 20, 2011 at 10:43
  • $\begingroup$ Again my faulty intuition tells me excerise 5 is not true but I cannot come up with a counterexample. Am I wrong again? $\endgroup$
    – user7980
    Jul 21, 2011 at 8:39
  • $\begingroup$ For exercise two does it follow that $T-c_iI$ is not injective for some $c_i$ simply by contradiction i.e $T-c_i I$ injective for all $i$ implies $p(T) - \lambda I$ is injective $\endgroup$
    – user7980
    Jul 21, 2011 at 10:12

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