# Idempotent Matrix and its Eigenvalues

## Problem 176

Let $A$ be an $n \times n$ matrix. We say that $A$ is idempotent if $A^2=A$.

(a) Find a nonzero, nonidentity idempotent matrix.

(b) Show that eigenvalues of an idempotent matrix $A$ is either $0$ or $1$.

(The Ohio State University, Linear Algebra Final Exam Problem)

## Proof.

### (a) Nonzero, nonidentity idempotent matrix

Let $A=\begin{bmatrix} 0 & 1\\ 0 & 1 \end{bmatrix}$. Then $A$ is a nonzero, nonidentity matrix and $A$ is idempotent since we have
$A^2=\begin{bmatrix} 0 & 1\\ 0 & 1 \end{bmatrix}\begin{bmatrix} 0 & 1\\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 0 & 1\\ 0 & 1 \end{bmatrix}=A.$

### (b) Eigenvalues of an idempotent matrix $A$ is either $0$ or $1$

Let $\lambda$ be an eigenvalue of the idempotent matrix $A$ and let $\mathbf{x}$ be an eigenvector corresponding to the eigenvalue $\lambda$.
Namely we have
$A\mathbf{x}=\lambda \mathbf{x}, \mathbf{x}\neq \mathbf{0}. \tag{*}$ Then we compute $A^2\mathbf{x}$ in two ways.
First, since $A$ is idempotent we have $A^2=A$ and thus we have
$A^2\mathbf{x}=A\mathbf{x}\stackrel{(*)}{=} \lambda \mathbf{x}.$

Next, we compute as follows.
\begin{align*}
A^2\mathbf{x}=A(A\mathbf{x})\stackrel{(*)}{=}A(\lambda \mathbf{x})=\lambda (A\mathbf{x})\stackrel{(*)}{=}\lambda (\lambda \mathbf{x})=\lambda^2\mathbf{x}.
\end{align*}

Comparing these two computations, we obtain
$\lambda \mathbf{x}=\lambda^2 \mathbf{x}.$ Since $\mathbf{x}$ is a nonzero vector (because $\mathbf{x}$ is an eigenvector), we must have
$\lambda=\lambda^2.$ Hence solving $\lambda(\lambda-1)=0$, the possible values for $\lambda$ is either $0$ or $1$.
Thus, the idempotent matrix $A$ only have eigenvalues $0$ or $1$.

## Final Exam Problems and Solution. (Linear Algebra Math 2568 at the Ohio State University)

This problem is one of the final exam problems of Linear Algebra course at the Ohio State University (Math 2568).

The other problems can be found from the links below.

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### 7 Responses

1. 01/04/2017

[…] Since $A$ has three distinct eigenvalues, $A$ is diagonalizable. (Or, algebraic multiplicities are the same as geometric multiplicities.) Since the matrix $A$ has $0$ as an eigenvalue. Thus $A$ is not invertible. If $A$ is idempotent matrix, then the eigenvalues of $A$ is either $0$ or $1$. Thus $A$ is not an idempotent matrix. (For a proof of this fact, see the post Eigenvalues of an idempotent matrix.) […]

2. 05/25/2017

[…] 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 Amathbf{x}=mathbf{0}} text{ and } […]

3. 08/02/2017

[…] the post ↴ Idempotent Matrix and its Eigenvalues for solutions of this […]

4. 08/02/2017

[…] the post ↴ Idempotent Matrix and its Eigenvalues for […]

5. 09/13/2017

[…] Idempotent Matrix and its Eigenvalues […]

6. 09/28/2017

[…] Idempotent Matrix and its Eigenvalues […]

7. 10/16/2017

[…] Idempotent Matrix and its Eigenvalues […]

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