Linear Algebra

Sup /sci/, here's the deal.

T is an invertible linear operator on a finite-dimensional vector space V.
I know that for any eigenvalue λ of T, λˉ¹ is an eigenvalue of Tˉ¹. How can I show that the eigenspace of T corresponding to λ is the same as the eigenspace of Tˉ¹ corresponding to λˉ¹?

Let S(λ) be the eigenspace corresponding to eigenvalue λ. 
You want to show that if any vector v is an element of S(λ)
then it is also an element of S(λˉ¹).

So:  if v is an element of S(λ), then Tv = λv by definition of an eigenspace.  Since T is invertible, apply Tˉ¹ to both sides and you get
Tˉ¹Tv = Tˉ¹λv
  λˉ¹v = Tˉ¹v  ----> means v is an eigenvector of Tˉ¹ w/ eigenvalue λˉ¹.

So v is also an element of S(λˉ¹).  Since v was arbitrary, any vector in S(λ) is also in S(λˉ¹).  This means S(λ) is a subset of S(λˉ¹).
Now you have to show that S(λˉ¹) is a subset of S(λ) by using the same reasoning.

Then use the theorem: if A is a subset of B and B is a subset of A, then A = B.

My other λ is an eigenvalue.

thank you /sci/ i did Tˉ¹(Tv) = λˉ¹Tv  but didn't go through with the rest. First time I asked, and such a good answer ;_; thank you