Are idempotent matrices always diagonalizable?

How to prove that any idempotent matrix is diagonalizable?

Solution 1:

I will give two proofs of the result, one using minimal polynomials and one without.

Proof 1: Let $A:V\rightarrow V$ be an idempotent operator on the $n$ dimensional space $V$. Suppose that the rank of the operator is $r$. Then there exists $r$ linearly independent vectors $\{\mathbf{u}_1,\ \cdots,\ \mathbf{u}_r\}$ in the image of $A$. Suppose that each $\mathbf{u}_i$ is in the image under $\mathbf{v}_i$: $$\mathbf{u}_i = A\mathbf{v}_i$$ Applying $A$ to the above gives $$A\mathbf{u}_i = \underbrace{A^2\mathbf{v}_i = A\mathbf{v}_i}_\text{idempotent} = \mathbf{u}_i$$ It follows that each $\mathbf{u}_i$ is an eigenvector under the eigenvalue $1$.

On the other hand, we also have $n-r$ vectors forming a basis for the kernel. Together, the basis for the image and the kernel forms a basis of eigenvectors for $V$, that is: $$V = \mathrm{im}(A) \oplus \ker(A)$$ Therefore $A$ is diagonalizable. $\square$

Using minimal polynomials, we can prove the result much quicker.

Proof 2: A linear operator is diagonalizable if and only if its minimal polynomial splits into distinct linear factors. If $A$ is idempotent then $$A^2 = A \implies A(A-I) =0$$ This means that $p(x) = x(x-1)$ is an annihilating polynomial for $A$. It follows that the minimal polynomial must divide $p$ and hence must split. Therefore $A$ must be diagonalizable. $\square$

Solution 2:

This is quite easy. For every vector $v$ one has $A^2v=Av$, so $Av$ is in the eigenspace for $\lambda=1$. Also $A(v-Av)=0$, so $v-Av$ is in the eigenspace for $\lambda=0$. (Define the eigenspace for$~\lambda$ as $\ker(A-\lambda I)$, even in case $\lambda$ should be no eigenvalue.) So $v=Av+(v-Av)$ is in the sum of those two eigenspaces; since $v$ was arbitrary, that sum is the whole space. This means $A$ is diagonalisable (eigenvalues $0,1$ only).