Let $A\in M_n(\mathbb{R})$ be symmetric, such that $A^{10}=I.$ Prove $A^2=I$

My thoughts:

Since $A$ is symmetric, $A^2$ is symmetric, so there exists an orthogonal $P\in M_n(\mathbb{R})$ such that $D=P^{-1}A^2P$ is a diagonal matrix.

I tried to work with that in order to find the "right" diagonal matrix such that after power manipulations, I could prove that $A^{10}$ is similar to $A^2$ and conclude the result, but got stuck.

Any help is appreciated.

  • $\begingroup$ Curious question: Can't we take the fifth root of both sides to immediately give the result? Can't we take fifth roots of an identity matrix? $\endgroup$ – Twenty-six colours Jul 4 '17 at 15:23
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    $\begingroup$ @Twenty-sixcolours fifth root of identity matrix in general case may not be a identity matrix.. $\endgroup$ – Widawensen Jul 4 '17 at 15:26
  • $\begingroup$ Thanks. Also, would the result hold true if we were to say that $A$ is symmetric matrix, under the scalar field of complex numbers instead? $\endgroup$ – Twenty-six colours Jul 5 '17 at 3:25
  • $\begingroup$ @Twenty-sixcolours I have doubts in this case...the fifth rooth of number 1 in complex domain is not unique.. $\endgroup$ – Widawensen Jul 5 '17 at 8:49

You have shown that $A$ is similar to $D=diag(\lambda_1,\ldots ,\lambda_n)$, so that $A^{10}=I$ is equivalent to $\lambda_i^{10}=1$ for all $i$. Since the $\lambda_i$ are real, this is equivalent to $\lambda_i^{2}=1$ for all $i$, which in turn is equivalent to $A^2=I$.

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    $\begingroup$ Your equality $(\lambda_1\cdots \lambda_n)^{10}=1$ means, I think, $\lambda_i^{10}=1$ for all $i$. Similarly for $(\lambda_1\cdots \lambda_n)^2=1$. $\endgroup$ – Pierre-Yves Gaillard Jul 4 '17 at 14:29
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    $\begingroup$ @Pierre-YvesGaillard Yes, thank you! $\endgroup$ – Dietrich Burde Jul 4 '17 at 14:45
  • $\begingroup$ Because of which result do they have to be real? $\endgroup$ – mathreadler Jul 4 '17 at 15:47
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    $\begingroup$ It is commonly known that a symmetric matrix has real eigenvalues. $\endgroup$ – user370967 Jul 4 '17 at 15:50
  • $\begingroup$ @mathreadler A quick proof: consider the inner product on the inner product space with basis $B$. Then since $A$ is symmetric real, then the adjoint $A^\star$ is equal to $A^T$ (since $A$ is real w.r.t to bases $B$) which equals to $A$. Now let $\lambda$ be an eigenvalue of $A$. We have $\langle Av|v\rangle = \langle v|A^\star v \rangle = \overline{\lambda}\langle v|v \rangle$ by conjugate linearity in the second component. Also, by the linearity in the first component, we also have $\langle Av|v\rangle = \lambda \langle v|v \rangle$. Then the result follows by the cancellation property. $\endgroup$ – Twenty-six colours Jul 5 '17 at 3:28

The minimal polynomial of $A$ divides $x^{10}-1$.

$x^{10}-1=(x^2-1)q(x)$, where $q(x)$ has no real roots.

The eigenvalues of a real symmetric matrix are all real and so its minimal polynomial is a product of linear real factors.

Therefore, the minimal polynomial of $A$ divides $x^2-1$ and so $A^2=I$.


Your method was good.

Indeed $A^2$ is diagonalizable as symmetric.


$ A^2= PDP^{-1} $

$ A^{10}=PD^5P^{-1}=I$

$D^5=P^{-1}IP =I$

But if for diagonal $D$ we have $D^5=I$ then the only possibility $D=I$.

And consequently $ A^2= PIP^{-1} =I$.


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