0
$\begingroup$

We have a real symmetric matrix $X$ such that $X^m = I$ for some natural number m, and I need to show that $X^2 = I$. I don't know where to start on this one, so any hints?

  • 1
    Hint: what can you say about eigenvalues of $X$? – 2017-01-17
  • 0
    Do you know spectral theorem https://en.wikipedia.org/wiki/Spectral_theorem? – 2017-01-17
  • 0
    @Evgeny They are $\pm1$? – 2017-01-17
  • 0
    @sigmabe Yes, I do – 2017-01-17
  • 0
    @JohnDoe Can you explain why they only can be $\pm 1$? – 2017-01-17

2 Answers 2

3

From $X^m=I$ we get for every eigenvector $v$ to a eigenvalue $\lambda$ that $$X^mv=\lambda^mv=Iv=v.$$ So we get $\lambda^m=1$ and from $\lambda\in\mathbb{R}$ we get $\lambda\in \{\pm 1\}$.

By spectral theorem ($X$ is symmetric) we get that $X$ is similar to a diagonalmatrix $D$ with diagonal entries out of $\{\pm 1\}$ and so $D^2=I$. In other words, there is an invertible matix $T$ with $$TXT^{-1}=D$$ and $D^2=I$. So we get $$X^2=(T^{-1}DT)(T^{-1}DT)=I.$$

1

If $ \mu$ is an eigenvalue of $X$, then $\mu^m=1$, hence $\mu= \pm 1$

Let $Y=X^2-I$, then $Y$ is symmetric and $ \sigma(Y)=\{0\}$

( $\sigma$ = set of eigenvalues).

We have that the spectral radius of $Y$ is $=0$. Since the norm of $Y$ = spectral radius of $Y$, we get

$||Y||=0$, hence $X^2=I$