Normal operators are an important class of operators in operator theory and spectral theory.We encounter it a lot in functional analysis.The following is my definition for normal operators:
Defn. Let $H$ be a Hilbert space over $\mathbb R$ or $\mathbb C$ and let $T\in BL(H)$ i.e. $T$ is a bounded linear operator on $H$.Then $T$ is said to be normal if $TT^*=T^*T$.
With the above definition we can make the following observations $\|Tx\|^2=\langle Tx,Tx\rangle=\langle x,T^*Tx\rangle=\langle x,TT^*x\rangle=\langle T^*x,T^*x\rangle=\|T^*x\|^2$ which implies $\|Tx\|=\|T^*x\|,\forall x\in H$ for a normal operator $T$.Now it is clear that $Tx=0\iff T^*x=0$ and hence $N(T)=N(T^*)$ and we can say $T$ is injective $\iff$ $T^*$ is injective.
Now suppose I want to investigate when $T$ is invertible.If $H$ is a finite dimensional inner product space,then $T$ is one-one $\iff $ $T$ is onto by Rank-Nullity theorem.But we do not have this advantage in case of infinite dimensional spaces.
But we can observe that:
If $T\in BL(H)$ is normal.$T$ is onto $\implies T$ is injective.
I prove my claim in the following way:
If $T$ is onto then $R(T)=H \implies R(T)^\perp=\{0\}\implies N(T^*)=\{0\}\implies N(T)=\{0\}$ because $N(T)=N(T^*)$ for a normal operator $T\in BL(H)$.Thus $T$ is injective.
But I think,that $T$ is injective is not enough to conclude that $T$ is onto.
This is because $N(T)=\{0\}\implies N(T^*)=\{0\}$ by normality of $T$ and hence $\overline {R(T)}=N(T^*)^\perp =H$(i.e. $R(T)$ is dense) but from here we cannot conclude that $R(T)=H$ in which case we would get $T$ is onto.
However if we impose one extra condition that $R(T)$ has to be closed then we have the following:
If $T\in BL(H)$ is normal.Then $T$ is one-one and $R(T)$ closed $\implies T$ is onto.
In fact,combining with the previous result and noting that $T$ is onto $\implies R(T)=H$ which is closed,we have the following:
If $T\in BL(H)$ is normal.Then $T$ is onto $\iff T$ is one-one and $R(T)$ is closed.
In fact we can note that $T$ is invertible $\implies T$ is onto and that $T$ is onto $\implies T$ is injective and onto $\implies T$ is invertible.So,we can write:
If $T\in BL(H)$ is normal,then $T$ is invertible $\iff T$ is onto $\iff T$ is one-one and $R(T)$ is closed.
We have also another criterion for a normal operator to be invertible.This is the following:
If $T\in BL(H)$ is a normal operator,then $T$ is invertible $\iff $ there is a $\delta>0$ such that $\|Tx\|\geq \delta \|x\|$ for all $x\in H$.
The proof of the equivalence is as follows:
$(\implies )$ Let $T$ be invertible.Now $T:H\to H$ bounded linear operator and $H$ is Hilbert space( and hence Banach space) and $T$ is surjective hence $T$ is open.So,$T^{-1}$ is a bounded linear operator(we are actually using inverse function theorem).Now $T^{-1}$ being a bounded operator we have $\|T^{-1}y\|\leq C\|y\|$ for all $y\in H$ (for some $C>0$).But let $x=T^{-1}(y)$ so $Tx=y$ and hence we get $\|x\|\leq C\|Tx\|$ for all $x\in H$.Hence,dividing by $C$ on both sides and letting $\delta=\frac{1}{C}$ we get $\|Tx\|\geq \delta\|x\|$ for all $x\in H$.
$(\impliedby)$ Let $\|Tx\|\geq \delta \|x\|$ for all $x\in H$ (for some $\delta>0$)
Now $Tx=0\implies x=0$ easily from the inequality and hence $T$ is injective.We have already shown that $T$ is injective and $T$ is normal $\implies R(T)$ is dense in $H$ i.e. $\overline{R(T)}=H$.
Now,it suffices to show that $R(T)$ is closed.So,take $\{Tx_n\}$ a sequence in $R(T)$ such that $Tx_n\to y$ as $n \to \infty$.Then $\|T(x_m-x_n)\|\geq \delta \|x_m-x_n\|$ for all $m,n\in \mathbb N$.Now $\{Tx_n\}$ is convergent and hence Cauchy,so the LHS $\to 0$ as $m,n\to \infty$ and hence $\|x_m-x_n\|\to 0$ as $m,n\to \infty$ .So, $\{x_n\}$ is Cauchy in $H$ and $H$ is complete,so $\{x_n\}\to x\in H$ for some $x$.So,$Tx_n\to Tx$ as $n\to \infty$ because $T$ is continuous.So,$Tx=y$ and hence $y\in R(T)$ .So,we have shown that $R(T)$ is closed and hence $R(T)=H$ which is the same as saying that $T$ is onto.
Thus we have the following characterization of invertible normal operators in $BL(H)$:
Proposition: Let $H$ be a Hilbert space and $T\in BL(H)$ be a normal operator.Then the following are equivalent:
$(1)$ $T$ is invertible.
$(2)$ $T$ is onto.
$(3)$ $T$ is one-one and $R(T)$ is closed.
$(4)$ There is a $\delta>0$ such that $\|Tx\|\geq \delta \|x\|$ for all $x\in H$.
However,I could not come up with an example of a normal operator which is injective but not surjective.Can someone try to give an example?