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Let $V$ be a finite-dimensional inner product space, and assume given a self-adjoint linear transformation $T:V\to V$ such that $T^3 = T$.

How do I show that the only possible eigenvalues of $T$ are $\lambda = 0, 1,$ or $−1$? And how do I write the corresponding eigenspaces?

That is what I did so far:

We have $T^3 = T$.

So if $\lambda$ is an eigenvalue of $T$ then we have $\lambda^3 = \lambda$.

$\lambda^3 = \lambda$

$\lambda^3-\lambda=0$

$\lambda(\lambda^2 - 1) = 0$

so $λ = 0$ or $λ =1$ 0r $λ = -1$.

Now, how do we write the eigenvectors?

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  • $\begingroup$ Your reasoning is correct. You cannot explicitly compute the eigenspaces. Of course, these are the null space of $\lambda I-T$ $\endgroup$ – daw Dec 15 '17 at 18:00

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