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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We give three proofs of this problem. The first one proves that $\R^n$ is a direct sum of eigenspaces of $A$, hence $A$ is diagonalizable.

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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