Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $T$ be an invertible linear operator on the finite dimensional $F$-vector space $V$. Prove that if $\lambda$ is an eigenvalue of $T$, then $\lambda^{-1}$ is an eigenvalue of $T^{-1}$ and that the eigenspaces $E_{\lambda}(T)=E_{\lambda^{-1}}(T^{-1})$.

I have shown the first part of the proof, but it remains to show that $E_{\lambda}(T)=E_{\lambda^{-1}}(T^{-1})$ is true. I'm thinking that this needs to be shown by proving the dimension of each of the eigenspaces, or can it be done through just by showing that all elements of one eigenspace are in the other, and vice versa.

share|cite|improve this question
What does the symbol $E_\lambda$ mean? – user1551 Nov 29 '12 at 17:27
Sorry I should have been clear. It is the eigenspace associated with a particular eigenvalue – tk2 Nov 29 '12 at 17:28
Nope, you've said that in the text. I just have overlooked it. – user1551 Nov 29 '12 at 17:31
up vote 2 down vote accepted

How about $Tv = \lambda v $ iff $\frac{1}{\lambda} v = T^{-1} v$ (Multiply the first equation by $T^{-1}$ and divide through by $\lambda$. $T$ is invertible, so all eigenvalues are $\neq 0$).

Explicitly, if $v \in E_\lambda(T)$ then $Tv = \lambda v $, hence $v = \lambda T^{-1} v $ and $\frac{1}{\lambda} v = T^{-1} v$. Hence $v \in E_{\frac{1}{\lambda}}(T^{-1})$. This gives $E_\lambda(T) \subset E_{\frac{1}{\lambda}}(T^{-1})$.

Since the same applies to $E_{\frac{1}{\lambda}}$ (and $(T^{-1})^{-1} = T$), we also have $E_{\frac{1}{\lambda}}(T^{-1}) \subset E_\lambda(T)$.

Hence they are equal.

share|cite|improve this answer
I have done what you are saying. – tk2 Nov 29 '12 at 17:25
@tkrm: copper.hat has answered both parts of your question in one line. Think about the "iff" in his proof. – user1551 Nov 29 '12 at 17:34
I made the answer more explicit. – copper.hat Nov 29 '12 at 17:34
Thanks. I made the same argument, after you posted the first time. – tk2 Nov 29 '12 at 17:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.