# eigenspaces/diagonalizability

• Nov 12th 2009, 11:12 AM
dannyboycurtis
eigenspaces/diagonalizability
Hey y'all I got a question regarding eigenspaces which is stumping me. Heres the problem, any tips or help would be greatly appreciated.

Let T be an invertible linear operator on a finite-dimensional vector space V.
Prove that the eigenspace of $\displaystyle T$ corresponding to $\displaystyle \lambda$ is the same as the eigenspace of $\displaystyle T^{-1}$ corresponding to $\displaystyle \lambda ^{-1}$

Also prove that if $\displaystyle T$ is diagonalizable, then $\displaystyle T^{-1}$ is diagonalizable.

Thanks!
• Nov 12th 2009, 11:52 AM
tonio
Quote:

Originally Posted by dannyboycurtis
Hey y'all I got a question regarding eigenspaces which is stumping me. Heres the problem, any tips or help would be greatly appreciated.

Let T be an invertible linear operator on a finite-dimensional vector space V.
Prove that the eigenspace of $\displaystyle T$ corresponding to $\displaystyle \lambda$ is the same as the eigenspace of $\displaystyle T^{-1}$ corresponding to $\displaystyle \lambda ^{-1}$

Also prove that if $\displaystyle T$ is diagonalizable, then $\displaystyle T^{-1}$ is diagonalizable.

Thanks!

$\displaystyle Tv=\lambda v\Longrightarrow (apply\,\,T^{-1}\,\,on\,\,both\,\,sides)\,v=T{-1}(\lambda v)$ , and now just use that the inverse of a linear map is also linear.