Prove that $ \text {(null}~T^*)^\perp \subseteq \text{range}~T $

Question

Let $T \in L(V,W)$ where $L(V,W)$ denotes the set of linear maps from $V$ to $W$. Prove that $ \text {(null}~T^*)^\perp \subseteq \text{range}~T $ where $T^*$ is the adjoint operator ( not related to the adjoint matrix) and $A^\perp $ refers to the orthogonal complement of $A$. Reference source is linear algebra done right by sheldon axler

Attempt: Let $w_2 \in \text{(null}~T^*)^\perp$.

Our aim is to show that $\exists v \in V$ such that $Tv=w_2$.

We can express $W = \text {null} ~T ~\oplus~\text{(null}~T^*)^\perp$. Thus, Let $w = w_1+w_2$ where $w_1 \in \text {null} ~T$.

$T^*(w)=T^*(w_2)=v$ for some $v \in V$

If we somehow prove that $<(Tv-w_2),~(Tv-w_2)>=0$, then we are done

But expanding the above seems to make it only more complicated.

Any ideas on how to move ahead? I have browsed through similar questions and tried to employ similar techniques but in vain.

Thanks a lot!

Answer 1

Assuming they are finite dimension vector spaces, we have that $rank(T^*) = rank(T)$. Therefore $\frac{W}{null(T^*)} \cong (nullT^*)^\perp \cong imT^* \cong imT$. Since $imT \subset (nullT^*)^\perp$ we have an equality by the dimensions.