Image of basis vectors of $\text{ran }T^*$ are basis vectors of $\text{ran }T$?

Question

Let $T: \mathscr{H} \rightarrow \mathscr{K}$ be a finite rank bounded linear operator, where $\mathscr{H}$ and $\mathscr{K}$ are Hilbert spaces. I am trying to prove that the dimension of the range of $T$ is the same as the dimension of the range of $T^*$. As an intermediate step, I am proving the following:

Let $\{k_1,...,k_m\}$ be a set of basis vectors for the range of $T^*$, the adjoint of $T$. Is it true that $\{Tk_1,...,Tk_m\}$ is a basis for the range of $T$?

Here is my attempt at a proof:

For any $j=1,...,m$, and any $v \in \text{ran} T$, such that $\langle Tk_j, v \rangle = 0$, then I will prove that $v=0$. This is essentially proving that the only vector in $\text{ran } T$ that is orthogonal to $\{Tk_1,...,Tk_j\}$ is zero, which is equivalent to proving that $\{Tk_1,...,Tk_j\}$ are basis vectors.

$$0=\langle Tk_j, v \rangle = \langle k_j, T^*v \rangle $$

But since $k_1,...,k_m$ are a basis for $\text{ran } T^*$, then there exists scalars $a_1,...,a_m$ such that $T^*v = a_1k_1 + ... + a_m k_m$.

$$0=\langle k_j, a_1k_1 + ... + a_m k_m \rangle = a_j$$

Since this is true for each $j=1,...,m$, we have that $v=0$.

My concern is that $\{Tk_1,...,Tk_m\}$ are not orthonormal. But at least they are linearly independent, so I could use Gram Schmidt, making them orthonormal.

Is this a reasonable approach?

Answer 1

Since the range is finite, it is a closed subspace of $\mathscr H,$ which allows us to argue as follows: if we restrict $T$ to $(\ker T)^{\perp}, $ we get an isomorphism of Hilbert spaces, namely, $\tilde T:(\ker T)^{\perp}\to \text {ran} T.\ $ By a standard result we also have $(\ker T)^{\perp}=\overline {ran T^{*}}$. But then $\overline {ran T^{*}}=ranT^* $ and the result follows.