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!
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!
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.
This is answer to the original question when normality of $T$ were not assumed.
This is not true, consider right shift on $\ell_2$.
It is even isometric but its image is not dense in $H$, it is of codimension $1$ in $H$.