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$\newcommand{\Ker}{\operatorname{Ker}}$

Let $X$ be Hilbert space and let $T:X\to X$ be a bounded operator. Define the operator $S: X/\Ker T \to X/\Ker T$ via $S(x+\Ker T)=Sx+\Ker T$. I can show that $||S||\leq ||T||$, and I am wondering whether the norms are actually equal. Is this true or false in general?

Edit: Considering the answer below, I am adding the conditions that $\Ker(T)$ is finite dimensional and $\operatorname{Im}(T)$ is not (say $T$ is surjective). Could you please show a counterxample in this situation as well.

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up vote 2 down vote accepted

Suppose $T^2 = 0$. Then $\text{im}(T) \subseteq \text{ker}(T)$, so...

Edit: The extra conditions do not really change anything. Here is an example where $T$ is surjective and $\text{ker}(T)$ is one-dimensional. Let $T$ act on a separable Hilbert space with orthonormal basis $e_1, e_2, ...$ by $$T e_1 = 0$$ $$T e_2 = 2 e_1$$ $$T e_n = \frac{1}{2} e_{n-1}, n \ge 3.$$

The point here is that $\text{ker}(T)$ may still contain the part of the image of $T$ which determines its norm.

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Just to add to Qiaochu's answer, if $ker(T)$ and $im(T)$ are orthogonal then the norms are equal. Otherwise, even if $ker(T)$ and $im(T)$ don't intersect, the norms could be different. –  Theo Aug 19 '12 at 4:27
    
@Theo: right. In particular this is the case if $T$ is self-adjoint. –  Qiaochu Yuan Aug 20 '12 at 0:35
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