The Question:
Let $T:V \rightarrow W$ be a linear transformation where $V$ is a finite-dimensional vector space over the field $\Bbb F$.
(i) Prove that $\text{Im}(T')=(\text{Ker}(T)) ^ \circ$
(ii) Deduce that $\text{dim}(\text{Im}(T))=\text{dim}(\text{Im}(T'))$
(iii) Hence, deduce that, if $T:V\rightarrow V$ and $T':V' \rightarrow V'$, then $T$ and $T'$ have the same non-zero eigenvalues with the same multiplicities.
Notes:
$T':V' \rightarrow W'$ is the dual transformation of $T$ defined by $T'(f) = f(T)$
$(\text{Ker}(T)) ^ \circ$ is the annihilator of $\text{Ker}(T)$
My Attempt:
Individually, I am able to do these questions using whatever method available to me. Unfortunately, the words "hence" and "deduce" mean that I have to use previous parts of the question.
I was able to do (i), and deduce (ii) from (i).
Any hints on how to "hence, deduce" (iii)?
1 Answer
We let $I:V \to V$ be the identity on $V$ and $I’ : V’ \to V’$ be the identity on $V’$. Let $\lambda I: V \to V$ be the map that multiplies any element by $\lambda$. Note that the dual map $\lambda I’ : V’ \to V’$ is also given by multiplication by the same number. You can see this by noting $\lambda I’ (f) = f(\lambda I) = \lambda f(I)$. Now since we know $dim(Im(T)) = dim(Im(T’))$, it follows that $dim(Im(T- \lambda I)) = dim(Im((T - \lambda I)’))= dim(Im(T’ - \lambda I’))$. By rank-nullity, it follows that $dim(ker(T-\lambda I)) = dim(ker(T’-\lambda I’))$. This last equation expresses that the number $\lambda$ is an eigenvalue of $T$ if any only if it is an eigenvalue of $T’$, and moreover, the dimensions are equal so they have the same multiplicities.
