Are rank of $T$ and $T^*$ equal? Let $H$ be a infinite dimensional Hilbert space and $T:H\to H$ be an operator on $H$.Is it true that $\operatorname{rank}T=\operatorname{rank}T^*$. We know that this is true for finite dimension. Is it true for infinite dimensional Hilbert space or is there any example of $T$ and $T^*$ that their rank are different?
Thanks!
 A: Suppose that the bounded linear operator $T: H \to H$ is a finite rank operator of rank $n$. We can write $T$ in the form 
$$
T(x) = \sum_{i=1}^n \alpha_i ( x, u_i) v_i,
$$
where $u_i$ and $v_i$ are vectors of norm 1 in $H$ and $\alpha_i \in \mathbb C$ (or $\mathbb R$ if you are working with a real Hilbert space).
For more details on this point, see http://en.wikipedia.org/wiki/Finite-rank_operator. 
But then, we have
$$
T^*(x)= \sum_{i=1}^n \overline \alpha_i (x, v_i) u_i,
$$
and so $T^*$ is also of finite rank and of rank at most $n$. As every such representation for $T$ leads to a related representation for $T^*$ and vice versa, the two ranks must in fact be the same.
A similar argument also shows that if $T$ is of infinite rank, then so is $T^*$. Suppose $T$ had infinite rank but $T^*$ had finite rank. Then $T^*$ would have a representation like above, which leads to $(T^*)^*$ being of finite rank. But $T^{**}=T$, which is meant to be of infinite rank, a contradiction.
A: The operator $Tv \mapsto T^*v$ is actually a partial isometry. The closure of the ranges are Murray-von Neumann equivalent.
