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In operator theory, a bounded operator ''T'' on a Hilbert space is said to be nilpotent if ''Tn'' = 0 for some ''n''. It is said to be quasinilpotent or topological nilpotent if its spectrum ''σ''(''T'') = . ==Examples== In the finite-dimensional case, i.e. when ''T'' is a square matrix with complex entries, ''σ''(''T'') = if and only if ''T'' is similar to a matrix whose only nonzero entries are on the superdiagonal, by the Jordan canonical form. In turn this is equivalent to ''Tn'' = 0 for some ''n''. Therefore, for matrices, quasinilpotency coincides with nilpotency. This is not true when ''H'' is infinite-dimensional. Consider the Volterra operator, defined as follows: consider the unit square ''X'' = () × () ⊂ R2, with the Lebesgue measure ''m''. On ''X'', define the (kernel) function ''K'' by : The Volterra operator is the corresponding integral operator ''T'' on the Hilbert space ''L''2(''X'', ''m'') given by : The operator ''T'' is not nilpotent: take ''f'' to be the function that is 1 everywhere and direct calculation shows that ''Tn f'' ≠ 0 (in the sense of ''L''2) for all ''n''. However, ''T'' is quasinilpotent. First notice that ''K'' is in ''L''2(''X'', ''m''), therefore ''T'' is compact. By the spectral properties of compact operators, any nonzero ''λ'' in ''σ''(''T'') is an eigenvalue. But it can be shown that ''T'' has no nonzero eigenvalues, therefore ''T'' is quasinilpotent. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Nilpotent operator」の詳細全文を読む スポンサード リンク
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