# Let T be an element of L ( complex numbers^3 ) and suppose 0 is the only eigenvalue for T. Show that T^n = 0 for some n.

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Let T be an element of L ( complex numbers^3 ) and suppose 0 is the only eigenvalue for T. Show that T^n = 0 for some n.

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Proof:

We consider the characteristic function of T: f(x) = det(xI - T). We know that f(x) is a polynomial with degree 3 because T is an ...

#### Solution Summary

A proof involving complex numbers, eigenvalues and determinants is provided.

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