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Say I have a linear map $T : V → V$ such that $T^3 = T$.

Say it has three distinct eigenvalues. Then they must be $-1$, $0$ and $1$.

In an exercise they ask me to prove the following:

$$\frac12T^2 - \frac12T \ ∈ \ S_{-1}$$

$$Id - T^2 \ ∈ \ S_0$$

$$\frac12T + \frac12T^2 \ ∈ \ S_1$$

There's no problem with that.

The thing is that then they ask me to deduce that $V = S_{-1} ⊕ S_0 ⊕ S_1$.

Intuitively this seems obvious: if you had a subspace that wasn't one of these eigenspaces, then doing $T$ thrice can't be doing $T$.

However, I'd like to see if there's a stronger argument using the three things they ask you to prove first. Possibly it's not as obvious as I think it to be.

A rural reader
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Jule
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    The statements that you're supposedly being asked to prove don't make sense as written. Do you mean that $\left(\frac 12 T^2 - \frac 12 T\right)v \in S_{-1}$ for any vector $v$ (and something similar for the other statements)? – Ben Grossmann Aug 21 '24 at 02:53
  • Every vector then becomes a sum of eigenvectors, and the claim follows. I used this as a step here, and Ben Grossman does the same below. The more general form of the result is that if a linear transformation $T:V\to V$ satisfies a polynomial relation $f(T)=0$, where $f$ does not have repeated zeros, then $T$ is diagonalizable (and the underlying space decomposes into a direct sum of eigenspaces provided the eigenvalues are in the base field). – Jyrki Lahtonen Aug 21 '24 at 03:13
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    $T$ satisfies $x^3-x = x(x-1)(x+1)$. Assuming the characteristic is not $2$, that means the minimal polynomial of $T$ splits and is squarefree, so $T$ is necessarily diagonalizable. – Arturo Magidin Aug 21 '24 at 04:43
  • I guess this is same as this https://math.stackexchange.com/questions/2511311/proving-that-if-a-is-a-8-times-8-matrix-over-mathbbr-and-a3-a-then , because the $8\times 8$ in the linked question is unused. – Angae MT Aug 21 '24 at 06:56

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The trick is to notice that any vector $v \in V$ can be decomposed into a sum of eigenvectors: $$ v = \left(\frac 12 T^2 - \frac 12 T\right)v + \left(\text{Id} - T^2\right)v + \left(\frac 12 T^2 + \frac 12 T\right)v. $$

Ben Grossmann
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  • Thanks for the answer. Still I feel something is missing. Why is that enough? – Jule Aug 22 '24 at 16:43
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    @Jule I suggest you review the definition of diagonalizability. By demonstrating that every vector $V$ can be written as a linear combination of eigenvectors of $T$, we have demonstrated that $T$ is diagonalizable. If you prefer, you can build a basis for each of the eigenspaces and then put these bases together to build a basis of $V$; the fact that $V$ has a basis consisting entirely of eigenvectors of $T$ means that $T$ is diagonalizable. – Ben Grossmann Aug 22 '24 at 17:39