Decomposing $\mathbb{R}^{8}$ Using an $8\times 8$ matrix

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Let $A$ an $8\times 8$ real matrix such that $A^{21}=I$. Prove that $\mathbb{R}^{8}$ can be written as the direct sum of $4$ two dimensional vector subspaces invariant under $A$, that is $\mathbb{R}^{8}=\mathbb{V}_{1} \oplus \mathbb{V}_{2} \oplus \mathbb{V}_{3} \oplus \mathbb{V}_{4}$ where $A(\mathbb{V}_{j}) \subset \mathbb{V}_{j}$.

Now, the easy way of doing this would be to show that $A$ is normal and then apply the structure theorem for normal operators, but I wasn't able to do it.

The other way I see is to write out $A$ in Jordan normal form, and show everything by hand, but that seems extremely time consuming.

So, what is the fastest way to do this?

Many thanks!

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Note that $A$ need not be normal.

Counterexample: Take the block matrix $$ A = \pmatrix{B&I_2\\&I_2\\&&I_4} $$ where $I_n$ is the size-$n$ identity matrix and $$ B = \frac 12 \pmatrix{-1 & \sqrt 3\\ -\sqrt 3 & -1} $$

Hint: Note that the minimal polynomial of $A$ must divide $x^{21} - 1$. What does this tell you about the Jordan form of $A$?

In particular, note that $A$ must be diagonalizable over $\Bbb C$. What can its eigenvalues complex be?