The adjoint of finite rank operator is finite rank If $T \in \mathfrak{B}_{00}(\mathfrak{H},\mathfrak{K})$, show that $T^{*} \in \mathfrak{B}_{00}(\mathfrak{K},\mathfrak{H})$ and $dim(ran T) = dim(ran T^{*})$.
The $\mathfrak{B}_{00}(\mathfrak{H},\mathfrak{K})$ is the set of continuous finite rank operaters.
Suppose $ranT$ is finite dimension subspace in the $\mathfrak{K}$, choose $Th_{1}, \dots,Th_{n} $ as a base. I think maybe $h_{1}, \dots,h_{n}$ should be the base of $ranT^{*}$,but I don't know how to verify my idea.
Help? Thanks advance.
 A: Since you used "adjoint", I assume these are operators on Hilbert spaces.
We can restrict $T$ to $ker(T)^{\perp}$, which is finite dimensional. So 
$$
T : ker(T)^{\perp} \rightarrow ran(T)
$$
is an (invertible) operator between finite dimensional Hilbert spaces. It has an adjoint $S$, defined on $ran(T)$. $T^*$ is the extension of $S$ by zero to $ran(T)^{\perp}$, and therefore is finite rank.
The dimension claim follows from the fact that $ker(T)^{\perp}$ and $ran(T)$ have the same dimension (finite in this case).
A: This can be seen as a very general consequence of the fact that the dual of a finite dimensional space is finite dimensional.
A finite rank operator $T\colon X\to Y$ can be seen as a composition $i\circ S$ of two operators: $S\colon X\to F$ and $i\colon F\to Y$, where $F$ is finite-dimensional.
So we have a diagram $X\to F\to Y$, and passing to duals, we have a diagram $Y^*\to F^*\to X^*$ (corresponding to $T^*=S^*\circ i^*$). Since $F$ is finite dimensional, so is $F^*$, so $T^*$ has finite rank.
To obtain corresponding statement about equality of ranks, you need to know that if $S$ is onto and $i$ is injective, then $S^*$ is injective and $i^*$ is onto, and the dimension of $F^*$ is the same as dimension of $F$.
A: Since $T$ is a bounded linear operator defined on whole Hilbert space, $\mathcal{H}$, with finite dimensional range, n, we can choose an ONB(orthonormal basis), of size n, in $range(T)$. Let $\{x_j\}_{j=1}^n$ be an ONB of $range(T)$. Note that $T^*$ is also bounded linear operator defined on whole Hilbert space, $\mathcal{H}$. Now let's define $T^*x_j = y_j$ and $A =T^*|_{range(T)}$ , so we have
$$ A x_j =y_j .$$ Therefore for any $x \in range(T) = \mathcal{D}(A)$ we have
\begin{equation*}
Ax = A( \sum_{j=1}^{n} a_j x_j) =  \sum_{j=1}^{n} a_j y_j,
\end{equation*}
for some complex $a_j$'s.  Hence $\{y_j\}_{j=1}^n$ spans $range(A)$. Now if for some complex $b_j$'s, let $\sum_{j=1}^{n} b_j y_j = 0$ which implies
$$ \sum_{j=1}^{n} b_j x_j \in Ker(A) .$$
Since $\mathcal{D}(A) = range(T) = Ker(T^*)^\perp$, because $range(T)$ is closed, we get
$$ \sum_{j=1}^{n} b_j x_j = 0.$$
Therefore all $b_j$'s are zero, as $x_j$'s are linearly independent. Hence $y_j$'s are linearl independent. Thus, $y_j$'s are ONB of $range(A)$
after normalisation. Further see that
$$ range(A) = range(T^*|_{range(T)}) = range(T^*|_{Ker(T^*)^\perp}) = range (T^*),$$ which implies that $y_j$'s are ONB of $range(T^*)$. Hence, we conclude that $T^*$ is finite rank.
