Diagonalization of Special Matrices (12.3.21-22)

Dummit and Foote Abstract Algebra, section 12.3, exercise 21-22:

21. Let $A$ be a matrix such that $A^2=A$. Show that $A$ can be diagonalized with $1$s and $0$s down the diagonal.

22. Let $A$ be a matrix such that $A^3=A$. Show that $A$ can be diagonalized over $\mathbb{C}$. Is this true over any field $F$?

Proof: (21) Let $f_i(x)^{α_i}$ be the $i^th$ invariant factor in the variant factor decomposition of $V$ over $F[x]$. Let $1$ be viewed as the $F[x]$ generator of this direct summand. Since $x^2(1)=x(1)$ we have $x(x-1)=0$. Since $-1 \neq 0$ and $f_i(x)^{α_i}$ is a power of a single prime power dividing $x(x-1)$ we must have $f_i(x)^{α_i}∈\{x,x-1\}$. Thus in the Jordan form of $A$ it is diagonal with $1$s and $0$s down the diagonal.

(22) Restarting as above we can see $f_i(x)^{α_i}$ divides $x^3-x=x(x-1)(x+1)$. When $F$ has characteristic greater than $2$ these are all distinct prime factors and thus $A$ is diagonalizable with $-1$s, $0$s, and $1$s down the diagonal. When $F$ has characteristic $2$ we note $x-1=x+1$ so $x^3-x=x(x-1)^2$ and thus by choosing the Jordan canonical form of the matrix with invariant factors $(x-1)^2$ we obtain a matrix that is not diagonalizable and is seen to satisfy $$\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}^3=\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}~\square$$