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I am trying to solve 3c from this released exam:

Determine all operators $T \in \mathcal{L}(V)$ such that $T^3 = T$ and $T^* = -T$. What can $T$ be?

From part b I have deduced that the eigenvalues must be imaginary. Using a similar argument to what I used in part b:

Consider any eigenvalue $\lambda$ for $T$ then:

$$Tx = \lambda x$$ and: $$\langle Tx, y \rangle = \langle x, -Ty \rangle$$ $$\langle T^3x, y \rangle = \langle x, -Ty \rangle$$ $$\langle \lambda^3x, y \rangle = \langle x, -\lambda y \rangle$$

Let $\lambda = ib$ and we see:

$$ \lambda^3 = -\overline{\lambda}$$ $$ (ib)^3 = ib$$ $$ -ib = ib$$ $$ b = 0$$

So the eigenvalues must be all zero then. I do not know how to make any further deductions, any further guidance would be appreciated.

Dair
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$T^3 = T$ implies that T is diagonalizable.

Hence T is diagonalizable and 0 is its only eigenvalue. What can you conclude ?

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  • This would imply that $T = P^{-1}DP$ for some diagonal matrix $D$. But the eigenvalues as I have shown must be $0$ so $D = 0$ hence $T = P^{-1}0P = 0$ so $T = 0$. Thanks. – Dair May 04 '16 at 08:12