If $A$ is normal show that $A+A^*$ bijective implies $A$ is bijective Suppose $A$ is a normal ($||Ax||=||A^*x||$ for all $x \in H$ ) bounded linear operator on a Hilbert space $H$. We want to show that if $A+A^*$ is a bijection then so is $A$.
A is injective is very easy since $kerA=kerA^*$ we have that  if $Ax=0$ then $x \in ker(A)\cap ker(A^*)$ so $x \in ker(A+A^*)$ so $x=0_H$.
I'm stuck on surjective. My hint is to use the open mapping theorem (so $A+A^*$ is open); but I can't see how this helps me at all.
I thought maybe I could prove $Im(A)$ was dense and closed but I can't relate either to $A+A^*$.
EDIT
In
Since $A$ and $A^\ast$ commute, we have
$$
\frac 1 4(A+A^\ast)^2-\frac 14(A-A^\ast)^2=A^\ast A.
$$
Since $A+A^\ast$ and $i(A-A^\ast)$ are self-adjoint operators, it follows that
\begin{align*}
\|A x\|^2&amp;=\langle A^\ast Ax,x\rangle\\
&amp;=\frac 1 4\langle (A+A^\ast)^2x,x\rangle-\frac 1 4\langle (A-A^\ast)^2x,x\rangle\\
&amp;=\frac 1 4\|(A+A^\ast)x\|^2+\frac 1 4 \|i(A-A^\ast)x\|^2.
\end{align*}
By the open mapping theorem (or the bounded inverse theorem if youlike), there exists a constant $c&gt;0$ such that $\|(A+A^\ast)x\|\geq c\|x\|$. Thus
$$
\|Ax\|\geq \frac{c}{2}\|x\|.
$$
Thus $A$ is a bijective operator from $H$ to $\operatorname{im}A$ with bounded inverse. In particular, $\operatorname{im}A$ is closed. Finally, if $y\perp\operatorname{im}A$, then
$$
0=\langle y,AA^\ast y\rangle=\|A^\ast y\|^2,
$$
hence $y\in \ker A^\ast=\{0\}$. Therefore, $\operatorname{im}A$ is also dense in $H$.
Here is a shorter answer that requires a little more technology: By the spectral theorem, the unital $C^\ast$-algebra generated by $A$ is $\ast$-isomorphic to $C(K)$ for some compact $K$, so this question boils down to the invertibility of $f\in C(K)$ under the assumption that $\operatorname{Re}f$ is invertible. Of course, an element of $C(K)$ is invertible if and only if it has no zero. Furthermore, the zeros of $f$ coincide with the zeros of $|f|$. Now all we have to do is to notice that
$$
|f|^2=(\operatorname{Re}f)^2+(\operatorname{Im}f)^2\geq (\operatorname{Re}f)^2&gt;0.
$$
For finite dimensions:
Isnt $A$ and $A^H$ simulataneously diagonalizable by unitary matrix cant u use that to prove this by simply saying $A+A^H = U (D+D^H) U^H$ and hence bijective iff $\lambda_i+conj(\lambda_i) \neq 0$ and hence $\lambda_i \neq 0$ and hence $A$ is bijective .