TITLE: For normal $T\in\mathcal{B}(H)$ operator $T$ is injective iff its image is dense
QUESTION [5 upvotes]: Let $H$ a Hilbert space, $T \in \mathcal{B}(H)$ is normal. 
Show that:
$T$ is injective iff $\mathrm{Im}(T)$ is dense in $H$
Any help is appreciated!

REPLY [9 votes]: This should go as follows.
For any operator $T \in \mathcal{B}(H)$ one has that $\ker T^* = (\mathrm{Im} T)^{\perp}$.
This implies that $\ker T = \ker T^*T$ because restricted to the image of $T$, $T^*$ is injective. But now since the operator $T$ is normal you get
$$ \ker T  = \ker T^*T = \ker TT^* = \ker T^* = (\mathrm{Im} T)^{\perp}.$$
Now it follows that $T$ is injective if and only if the orthogonal complement of the image is trivial, which says that $T$ has dense image.