Are idempotent matrices always diagonalizable?

Question

How to prove that any idempotent matrix is diagonalizable?

Answer 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. $\square$

Answer 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).

Answer 3

I don't know if this will be helpful to the original poster (10 years late and likely using a different set of tools) but I found myself pondering this question myself while reading Fulton and Harris today. I think the following works:

Suppose that $A$ is an $n\times n$ matrix such that $A^k= I$. Let $G = \mathbb{Z}_k$ with generator $g\in G$. Then the map $\rho: G \to GL_n(\mathbb{C})$ given by $\rho(g) = A$ is a representation of $G$. Since $G$ is abelian, $V$ decomposes as a sum of 1-dimensional representations, say $$V = \bigoplus_{i=1}^n \mathrm{span}(v_i).$$ Now each of the subspaces $\mathrm{span}(v_i)$ is $G$-invariant and in particular, this means that each must be fixed by the matrix $A$ and thus $v_i$ is an eigenvector for $A$. Thus $V$ has a basis of eigenvectors and $A$ is diagonalizable.

Answer 4

More generally, given a (unital) ring $R$, (left) $R$-module $M$, and idempotent (left) $R$-module endomorphism $T\colon M\to M$, the short exact sequence $$0\to \text{ker}\left(T\right)\hookrightarrow M\twoheadrightarrow \text{im}\left(T\right)\to 0$$ is split by the canonical inclusion $\text{im}\left(T\right)\to M$. The splitting lemma then constructs (explicitly in terms of the above maps) an isomorphism $$\psi\ \colon\ \text{ker}\left(T\right)\oplus \text{im}\left(T\right)\overset{\simeq}{\to} M$$ and the commutativity of the relevant diagram moreover ensures that $\psi^{-1}\circ T\circ\psi = 0_{\text{ker}\left(T\right)}\oplus 1_{\text{im}\left(T\right)}$.

(And if $\text{ker}\left(T\right)$ and $\text{im}\left(T\right)$ are free of finite rank, $T$ will then diagonalize in an appropriate basis of $M$.)