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As title says: Find all real symmetric matrices such that $A^5 = I_5$.

Obviously identity matrix and rotations of $\pm72$ degrees ($72 * 5 = 360$ degrees) - although I am not sure these rotation matrices are symmetric.

But can't think of general way how to find such symmetric matrices, or are the the only ones? If they are the only ones, how do I prove there are no other?

I started with calculating the inverse, i.e. $A^4 = A^{-1}$, but can't move beyond that point.

Thanks.

meerkat
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  • Do you know about the characteristic polynomial? If yes then my hint is: How can it look like? Or in different words what can you say about the eigenvalues? –  May 18 '24 at 17:23
  • @underflow yes I was thinking about this, by Cayley–Hamilton must be equal to $\lambda^5 - 1$ - i.e. there is only single real root = 1. Also not sure if it helps here, but determinant of A must be 1, because det($A^5$) = det($I_5$) = 1. – meerkat May 18 '24 at 17:28
  • That's not quite right. Also symmetric real matrices have ... eigenvalues. What is ...? –  May 18 '24 at 17:29
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    The spectral theorem states that every real symmetric matrix is diagonalizable. That is, we can write $A = R^{T} D R$ where $R$ is a unitary matrix ($R^T = R^{-1}$, they can be thought of as rotation matrices) and $D$ is a real diagonal matrix. We compute $A^5 = R^{T} D^5 R = I$, from which we get $D^5 = I$, The diagonal entries of $D$ are eigenvalues $\lambda_i$ which satisfy $\lambda_i^5 = 1$. But this has only one real solution, $\lambda_i = 1$. We get $D = I$, and $A = R^{T} D R = I$.

    https://en.wikipedia.org/wiki/Spectral_theorem#Hermitian_maps_and_Hermitian_matrices

     – user3257842 May 18 '24 at 17:32
  • @underflow sorry, maybe my application of Cayley-Hamilton was not the right one, I was thinking in reverse where such implication might not hold (https://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem). But also by spectral theorem, symmetric matrices are orthogonally diagonalizable, thus has to have 5 real eigenvalues? – meerkat May 18 '24 at 17:33
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    @meerkat Yes exactly. Symmetric real matrices have real eigenvalues. Now how to connect the eigenvalues of $A$ with those of $A^k$? –  May 18 '24 at 17:34
  • @user3257842 got it, thanks! – meerkat May 18 '24 at 17:48

1 Answers1

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By the spectral theorem, all eigenvalues of $A$ are real, and there exists an orthonormal basis of $\mathbb{R}^5$ consisting of eigenvectors of $A$. Every eigenvalue of $A$ must be a fifth root of unity (if $Av = \lambda v$ then $v = A^5v = \lambda^5 v$), and the only real fifth root of unity is $1$. It follows that $Av = v$ for all of the orthonormal basis vectors, and therefore for all $v \in \mathbb{R}^5$. Thus, $A$ can only be the identity.

sbares
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