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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!

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    Edit: Normal was not in the previous hypothesis2012-11-11

2 Answers 2

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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.

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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 $H$.

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    Yes Matt N. Sorry2012-11-11