# If $B^3=B$, is $B$ diagonalizable?

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?

• yup, good proof! Apr 20 at 22:20
• Yes, at least for fields not of characteristic $2$, :) Apr 20 at 22:25
• 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 $1$. Apr 20 at 22:30
• @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? Apr 21 at 2:55
• 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. Apr 21 at 2:56