Surjectivity and injectivity of Hilbert space operators

Question

Let $V, W$ be two Hilbert spaces, and $T : V \rightarrow W$ be a surjective bounded linear operator.

It is easy to check that $T^{\ast} : W \rightarrow V$ is an injective bounded linear operator.

Then is it true that $T T^{\ast} : W \rightarrow W $ is a positive self-adjoint isomorphism?

It is easy to check the positiveness, self-adjointness, and injectivity, but it is hard to check the surjectivity.

Does any of you have some idea for this?

Note that by definition, $<Tx, y>_{W}=<x,T^{\ast}y>_{V}$ for every $x \in V, y \in W$.

P.S. This question is from a paper I’m studying in. (I paraphrased the problem) If you need a reference, please read the 2nd line of page 586 of https://www.ams.org/journals/jams/2002-15-03/S0894-0347-02-00398-3/S0894-0347-02-00398-3.pdf (JINCHAO XU AND LUDMIL ZIKATANOV, THE METHOD OF ALTERNATING PROJECTIONS AND THE METHOD OF SUBSPACE CORRECTIONS IN HILBERT SPACE).

Answer 1

Here is another possibility to prove this, based on the closed-range theorem.

Since $T$ is surjective, it has closed range, and hence $R(T^*) = N(T)^\perp$.

Take $w\in W$. Then there is $v\in V$ with $Tv=w$. Write $v =v_1 + v_2$ with $v_1 \in N(T)^\perp$ and $v_2 \in N(T)$. Then $Tv_1=w$. But now $v_1 $ is in the range of $T^*$, so that there is $w_1$ with $T^*w_1=v_1$. This implies $TT^*w_1 = w$, and $TT^*$ is surjective.

Answer 2

Here is yet another explanation. The operator $T$ restricted to $(\ker T)^\perp=:\mathcal{H}_1$ is injective and surjective, hence invertible, by the Banach inverse mapping theorem. Thus $T^*$ is invertible as the operator from $\mathcal{H}$ onto $\mathcal{H}_1=(\ker T)^\perp. $ Hence $TT^*$ is invertible as a composition of two invertible operators.