Show that $\dim(\operatorname{ran}(T))=\dim( \operatorname{ran}(T^∗))$ My question is :

If $T \in B_{00}(H, K)$ show that $T^* \in B_{00}(K, H)$ and $\dim (\rm ran T) = \dim(\rm ran T^*)$. [$B_{00}(H, K)$ is the set of continuous finite rank operators]

First part is easy.
For second part of the question : (1) and (2) and (3) 'explains' why $\dim(\rm ran(T))=\dim(\rm ran(T^∗))$ but none is understandable:
In 1, how $H=\ker T\oplus (\ker T)^\perp$ implies $\dim(\mathop{\rm ran}T^*)\,=\, \dim(\mathop{\rm ran}T)\,$?
In 2, $\operatorname{dim} \operatorname{null} T + \operatorname{dim} (\operatorname{null} T)^\perp = \operatorname{dim} V$ implies $\dim(\mathop{\rm ran}T^*)\,=\, \dim(\mathop{\rm ran}T)\,$?
In 3, how $T^*P=T^*$ implies $\operatorname{rank}T^*\le \operatorname{rank}P=\operatorname{rank}T$?
Could someone explain one of the questions in the other answers I mentioned above, or give a different easy to understand proof for the main question?
 A: Let us start with a simple lemma:

Lemma: Let $H$ and $K$ be Hilbert spaces and let $T$ be a continuous linear function from $H$ to $K$. Then $\rm ran \,T^* \subseteq (\ker T)^\perp$.

Proof: Given any $y \in K$, we have, for all $x \in \ker T$,
$$ \langle T^*y, x \rangle_H = \langle y, Tx \rangle_K =  \langle y, 0 \rangle_K =0 $$
So, for all $y \in K$, $T^*y \in (\ker T)^\perp$. So, we have that $\rm ran \,T^* \subseteq (\ker T)^\perp$. $\square$
Now, let $B_{00}(H, K)$ be the set of continuous finite rank operators. Let us prove

If $T \in B_{00}(H, K)$, then $T^* \in B_{00}(K, H)$ and $\dim (\rm ran \, T) = \dim(\rm ran T^*)$.

Proof: Let $T$ be a continuous linear function from $H$ to $K$. It is easy to see that
$T|_{(\ker T)^\perp)}: (\ker T)^\perp \rightarrow \rm ran \, T$ is a isomorphism.
So,
$$ \dim (\ker T)^\perp  = \dim \rm ran \, T $$
Using the lemma above, we have
$$ \dim \rm ran \, T^* \leq \dim (\ker T)^\perp  = \dim \rm ran \, T $$
So, we have $\dim \rm ran \, T^* \leq \dim \rm ran \, T $. Applying it to $T^{*}$, we have $\dim \rm ran \, T = \dim \rm ran \, T^{**} \leq \dim \rm ran \, T^* $. So $\dim \rm ran \, T =  \dim \rm ran \, T^* $.
So, we have proved that, for any $T$ be a continuous linear function from $H$ to $K$, $\dim \rm ran \, T =  \dim \rm ran \, T^* $.
Now, if $T \in B_{00}(H, K)$, then $\dim (\rm ran \, T)$ is finite. So $\dim (\rm ran \, T^*)$ is finite. So, $T^* \in B_{00}(K, H)$. $\square$
