# 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.

• I should also add, you're on the right track there. If $T$ takes $span\{ h \}$ to $span \{ Th\}$, $T^*$ goes the other way, with the same norm when restricted to those subspaces. This is essentially the singular value decomposition. Commented Dec 18, 2013 at 4:50

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).

• Thank you for your answer，it's very usefull! Commented Dec 18, 2013 at 6:14
• I think it is directly to use the fact that $ranT^{*}=(kerT)^{\bot}$ Commented Dec 18, 2013 at 6:19
• It's easily to show that $(kerT)^{\bot}\cong ranT$, that is $$T:(kerT)^{\bot} \longrightarrow ranT$$ is an isomorphism. Commented Dec 19, 2013 at 3:08
• @MartinArgerami: If $T$ defines an injective map $\ker(T)^{\perp} \to ran(T)$, and $ran(T)$ is finite dimensional, $\ker(T)^{\perp}$ must be finite dimensional. Commented Nov 16, 2015 at 4:58
• T* is an extension of S, why? Commented May 30, 2023 at 20:29

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 $F$.

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.