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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 hypothesis –  P. M. O. Nov 11 '12 at 19:59

2 Answers 2

up vote 5 down vote accepted

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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My mistake, I Missed a hypothesis. –  P. M. O. Nov 11 '12 at 19:45
    
@P.M.O. Did you mean Norbert? –  Matt N. Nov 11 '12 at 19:50
    
Yes Matt N. Sorry –  P. M. O. Nov 11 '12 at 19:52

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