suppose $V$ is a finite dimensional vector space over $\mathbb C$ and $T:V\to V$ is a linear transformation. I want to find the dimension of subspace of $\ker(T)\cap T(V)$. Is it equal to $\dim T(V)-\dim T^2(V)$?
dimension of vector subspace of $\ker(T)\cap T(V)$
3
$\begingroup$
linear-algebra
linear-transformations
1 Answers
5
We indeed have the following: $$ \dim [\ker(T) \cap T(V)] = \dim T(V) - \dim T^2(V) $$ To see that this is the case: note that $\ker(T) \cap T(V)$ can be characterized as the kernel of $T|_{T(V)}$ (the restriction of $T$ to $T(V)$). Applying the rank-nullity theorem to $T|_{T(V)}:T(V) \to T(V)$, we have $$ \dim [\ker(T) \cap T(V)] = \dim \ker T|_{T(V)} = \dim T(V) - \dim T(T(V)) $$ as desired.