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Let $\mathcal{H}$ denote a separable Hilbert space. How would one construct a nonzero operator $T \in B(\mathcal{H})$ such that $T^2 = 0$? Here $B(\mathcal{H})$ is the set of bounded linear operators.

My current suggestion is to define $T$ such that $T(x_1,x_2,\dotsc) = (x_2,0,0,\dotsc)$, but I don't know if this even makes sense?

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    $\begingroup$ Such $T$ exists only if $\dim \mathcal H > 1$. $\endgroup$ – user251257 Dec 1 '16 at 22:08
  • $\begingroup$ Yes, this operator works in $\ell_2$, say. $\endgroup$ – Eduardo Longa Dec 1 '16 at 22:09
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Your example is perfectly fine. Also, if $dim\; \mathcal H = 1$, the only linear Operator is multiplication by a scalar, so then $T^2=0$ already implies that $T$ is the null operator.

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