# Prove each eigenvalue of $T/U$ is an eigenvalue of $T$.

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?

Thanks

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 $v\in V$,$$(T/U)(v+U)=Tv+U$$.

• 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. Mar 9, 2017 at 22:39

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.

• 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 $T$?
– user12802
Mar 10, 2017 at 18:06
• 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 $T/U(v+U) = \lambda v + U = \lambda(v+U)$. Mar 10, 2017 at 18:23
• 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 $T/U$. Mar 10, 2017 at 18:28
• 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.
– user12802
Mar 10, 2017 at 18:31
• 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. Mar 10, 2017 at 18:34