Let \(A\in M_n(F)\) satisfy \(A^2=A\). Prove that \(A\) is diagonalizable over \(F\), and determine its possible diagonal form.
Solution
1. The minimal polynomial of \(A\) divides\[ x^2-x=x(x-1). \]
The relation \(A^2=A\) is exactly \[ A(A-I)=0, \] so the polynomial \(x(x-1)\) annihilates \(A\). Hence the minimal polynomial divides it.
2. The polynomial \(x(x-1)\) has no repeated root over any field.
Its roots are \(0\) and \(1\), which are distinct in every field. Thus every divisor of \(x(x-1)\) is squarefree.
3. Therefore \(A\) is diagonalizable.
A linear operator is diagonalizable over \(F\) if and only if its minimal polynomial splits over \(F\) as a product of distinct linear factors. By <1>1–<1>2, the minimal polynomial of \(A\) has this form.
4. In a suitable basis,\[ A\sim\operatorname{diag}(\underbrace{1,\ldots,1}_{r},\underbrace{0,\ldots,0}_{n-r}) \]for some \(0\le r\le n\).
By 3 the only possible eigenvalues are the roots \(0\) and \(1\) of the annihilating polynomial. Hence a diagonal form consists only of \(0\)’s and \(1\)’s. The number \(r\) of \(1\)’s is \[ r=\operatorname{rank}(A)=\dim\operatorname{im}(A), \] while the number of \(0\)’s is \(n-r=\dim\ker A\).