Let $B\in M_n(\mathbb{F})$ such that $B^3=B$. Is $B$ diagonalizable?

If $B^3=B$, then $B^3-B=0$. Consider the polynomial $p(x)=x^3-x$. If $p(B)= B^3-B=0$. Since we know that the minimal polynomial of $B$ must divide any polynomial $g(x)\in \mathbb{F}[x]$ such that $g(B)=0$, then $m_B(x)|p(x)$ and $p(x)=x(x-1)(x+1)$. This means that $m_B(x)$ splits and each root of $m_B$ has multiplicity 1. So, $B$ is diagonalizable.

Is this correct?

  • 1
    $\begingroup$ yup, good proof! $\endgroup$
    – hunter
    Apr 20 at 22:20
  • 7
    $\begingroup$ Yes, at least for fields not of characteristic $2$, :) $\endgroup$ Apr 20 at 22:25
  • 5
    $\begingroup$ Depends on $\mathbb{F}$. If $1\neq -1$, your argument works. But if the characteristic is $2$, then the minimal polynomial divides $x(x+1)^2$, and it could have repeated factors. One example is $B=\left(\begin{array}{cc}0&1\\1&0\end{array}\right)$ over the field of $2$ elements. It has $B^2=I$, so $B^3=B$; but the only eigenvalue is $1$, and the eigenspace has dimension $1$. $\endgroup$ Apr 20 at 22:30
  • 2
    $\begingroup$ @RedFive Except for the field with two elements, no matrix has "unique eigenvectors". If $v$ is an eigenvector, so is $\alpha v$ for any $\alpha\neq 0$. For that matter, what are the "unique eigenvectors" of the identity matrix? $\endgroup$ Apr 21 at 2:55
  • 1
    $\begingroup$ Right... got it. Thanks. It has been over 20 years since I studied these things at a higher level - it is slowly coming back to me thanks to MSE. $\endgroup$
    – Red Five
    Apr 21 at 2:56


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