I would please like to check my work and see if it is a solution to Ex. 5a-35 in Axler's "Linear Algebra Done Right", 3rd ed.

Suppose $V$ is a finite dimensional vector space and $T\in \mathcal L(V)$ and $U$ is invariant (subspace was not mentioned, but I assume it is) under $T$. Prove each eigenvalue of $T/U$ is an eigenvalue of $T$.

My work so far:

For $v\notin U$ then $T/U(v +U):= Tv+U=\lambda (v +U)$. Or $Tv-\lambda v+U= 0+U$. And $\lambda$ is an eigenvalue of $T$.

Assuming this is correct so far, does this complete the proof?


EDIT Axler defines the Quotient Operator: Fot $T\in \mathcal L(V)$, $U$ an invariant subspace under $T$, and $T/U\in \mathcal L(T/U)$ and for $v\in V$,$$(T/U)(v+U)=Tv+U$$.

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    $\begingroup$ So $\;T/U\;$ means the linear map defined on the quotient $\;V/U\;$ and defined by means of $\;T\;$ ? Because if it is then the vectors in $\;V/U\;$ and in $\;V\;$ are different so I can't understand how the former are going to relate to the latter... you should explain this: this is not standard notation. $\endgroup$
    – DonAntonio
    Mar 9, 2017 at 22:39

1 Answer 1


Your proof shows the other implication: every eigenvalue of $T$ with eigenvector $v \notin U$ is an eigenvalue of $T/U$.

For the original statement, start with an eigenvalue $\lambda$ of $T/U$ and an eigenvector $v+U$ with $v \notin U$. Then $$ T/U(v+U) = \lambda v + U\,,$$ and by the definition of $T/U$, we have $$ Tv = \lambda v + u\,, $$ for some $u \in U$. Now consider some other $u' \in U$. Then $$ T(v+u') = \lambda v + u + Tu'\,,$$ and you want to find $u'$ such that $$ u + Tu' = \lambda u' \;\Leftrightarrow\; \lambda u' - T u' = u\,. $$ There are two cases:

  1. Note that if $U$ is invariant under $T$, then $U$ is invariant under $T - \lambda \operatorname{Id}$ for any $\lambda$. Hence, if $T-\lambda \operatorname{Id}$ is not invertible on $U$, then it is not invertible. Because $V$ is finite dimensional this implies that $\lambda$ is an eigenvalue of $T$.

  2. If $T-\lambda \operatorname{Id}$ is invertible on $U$, then you can define

    $$ u' = -(T - \lambda)^{-1}u\,, $$ and then $v+u'$ is an eigenvector with eigenvalue $\lambda$.

This problem turned out to be more interesting than anticipated.

  • $\begingroup$ I see you have visited the site a few times and perhaps did not have time to respond to my follow-up. So I have refined it. Why did you say I proved it in the other direction, as I began $T/U(v +U):= Tv+U=\lambda (v +U)$, only interspersing $Tv+U$? If that was proof in the direction you said, I think the question would be posed as "iff." Also when I get to $Tv-\lambda v + U=0+U$, or $Tv-\lambda v =0$ in $U$, why would one think $\lambda$ might not be an eigenvalue of $T$? $\endgroup$
    – user12802
    Mar 10, 2017 at 18:06
  • $\begingroup$ I think it was, because the proof you wrote worked in this direction, but not in the other one. It was not meant as a reflection as a reflection on your thought process. Sorry, it came out that way. If $v \notin U$ is an eigenvector of $T$, then $T/U(v+U) = \lambda v + U = \lambda(v+U)$. $\endgroup$ Mar 10, 2017 at 18:23
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    $\begingroup$ It is not quite an "iff" statement. Every eigenvalue of $T/U$ is an eigenvalue of $T$. However, only eigenvalues of $T$ with eigenvectors $v \notin U$ are eigenvalues of $T/U$. $\endgroup$ Mar 10, 2017 at 18:28
  • $\begingroup$ I think I see my omission regarding my second question. Initially I started with $v\notin U$ but I also have to consider a $u'$, as you say, $\in U$. Thanks for your patience. $\endgroup$
    – user12802
    Mar 10, 2017 at 18:31
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    $\begingroup$ You asked in your previous comment, where $u'$ comes from. It emerged after doodling on a piece of paper for half an hour as the right approach to take. After the fact, it seems the natural approach: you have a $v$, which is related to the eigenvalue, but you do not know, what is going on inside the space $U$, so you assume that you can find the right correction $u'$ and then see, what condition it has to satisfy. $\endgroup$ Mar 10, 2017 at 18:34

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