# Idempotent Matrices are Diagonalizable

## Problem 429

Let $A$ be an $n\times n$ idempotent matrix, that is, $A^2=A$. Then prove that $A$ is diagonalizable.

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The second proof proves the direct sum expression as in proof 1 but we use a linear transformation.

The third proof discusses the minimal polynomial of $A$.

### Range=Image, Null space=Kernel

In the following proofs, we use the terminologies **range** and **null space** of a linear transformation. These are also called **image** and **kernel** of a linear transformation, respectively.

## Proof 1.

Recall that only possible eigenvalues of an idempotent matrix are $0$ or $1$.

(For a proof, see the post “Idempotent matrix and its eigenvalues“.)

Let

\[E_0=\{\mathbf{x}\in \R^n \mid A\mathbf{x}=\mathbf{0}\} \text{ and } E_{1}\{\mathbf{x}\in \R^n \mid A\mathbf{x}=\mathbf{x}\}\]
be subspaces of $\R^n$.

(Thus, if $0$ and $1$ are eigenvalues, then $E_0$ and $E_1$ are eigenspaces.)

Let $r$ be the rank of $A$. Then by the rank-nullity theorem, the nullity of $A$

\[\dim(E_0)=n-r.\]

The rank of $A$ is the dimension of the range

\[\calR(A)=\{\mathbf{y} \in \R^n \mid \mathbf{y}=A\mathbf{x} \text{ for some } \mathbf{x}\in \R^n\}.\]

Let $\mathbf{y}_1, \dots, \mathbf{y}_r$ be basis vectors of $\calR(A)$.

Then there exists $\mathbf{x}_i\in \R^n$ such that

\[\mathbf{y}_i=A\mathbf{x}_i,\]
for $i=1, \dots, r$.

Then we have

\begin{align*}

A\mathbf{y}_i&=A^2\mathbf{x}_i\\

&=A\mathbf{x}_i && \text{since $A$ is idempotent}\\

&=\mathbf{y}_i.

\end{align*}

It follows that $y_i\in E_1$.

Since $y_i, i=1,\dots, r$ form a basis of $\calR(A)$, they are linearly independent and thus we have

\[r\leq \dim(E_1).\]

We have

\begin{align*}

&n=\dim(\R^n)\\

&\geq \dim(E_0)+\dim(E_1) && \text{since } E_0\cap E_1=\{\mathbf{0}\}\\

&\geq (n-r)+r=n.

\end{align*}

So in fact all inequalities are equalities, and hence

\[\dim(\R^n)=\dim(E_0)+\dim(E_1).\]

This implies

\[\R^n=E_0 \oplus E_1.\]
Thus $\R^n$ is a direct sum of eigenspaces of $A$, and hence $A$ is diagonalizable.

## Proof 2.

Let $E_0$ and $E_1$ be as in proof 1.

Consider the linear transformation $T:\R^n \to \R^n$ represented by the idempotent matrix $A$, that is, $T(\mathbf{x})=A\mathbf{x}$.

Then the null space $\calN(T)$ of the linear transformation $T$ is $E_0$ by definition.

We claim that the range $\calR(T)$ is $E_1$.

If $\mathbf{x}\in \calR(T)$, then we have $\mathbf{y}\in \R^n$ such that $\mathbf{x}=T(\mathbf{y})=A\mathbf{y}$.

Then we have

\begin{align*}

\mathbf{x}&=A\mathbf{y}=A^2\mathbf{y} =A(A\mathbf{y})

=A\mathbf{x}.

\end{align*}

(The second equality follows since $A$ is idempotent.)

This implies that $\mathbf{x}\in E_1$, and hence $\calR(T) \subset E_1$.

On the other hand, if $\mathbf{x}\in E_1$, then we have

\[\mathbf{x}=A\mathbf{x}=T(\mathbf{x})\in \calR(T).\]
Thus, we have $E_1\subset \calR(T)$. Putting these two inclusions together gives $E_1=\calR(T)$.

By the isomorphism theorem of vector spaces, we have

\[\R^n=\calN(A)\oplus \calR(T)=E_0\oplus E_1.\]
Thus, $\R^n$ is a direct sum of eigenspaces of $A$ and hence $A$ is diagonalizable.

## Proof 3.

Since $A$ is idempotent we have $A^2=A$.

Thus we have $A^2-A=O$, the zero matrix, and so $A$ satisfies the polynomial $x^2-x$.

If $x^2-x=x(x-1)$ is not the minimal polynomial of $A$, then $A$ must be either the identity matrix or the zero matrix.

Since these matrices are diagonalizable (as they are already diagonal matrices), we consider the case when $x^2-x$ is the minimal polynomial of $A$.

Since the minimal polynomial has two distinct simple roots $0, 1$, the matrix $A$ is diagonalizable.

## Another Proof

A slightly different proof is given in the post “Idempotent (Projective) Matrices are Diagonalizable“.

The proof there is a variation of Proof 2.

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