Question comes from “A First Course in Linear Algebra” by Robert A. Beezer.

Question T10, Page 389

Problem Description:

A matrix \(A\) is idempotent if \(A^2 = A\). Show that the only possible eigenvalues of an idempotent matrix are \(\lambda = 0, 1\). Then give an example of a matrix that is idempotent and has both of these two values as eigenvalues.

Proof:

Consider a matrix \(A\) and vector \(x\) such that \(\mathbf{x}\) is an eigenvector of \(A\) with associated eigenvalue \(\lambda\). In other words:

\[ \begin{equation} A\mathbf{x} = \lambda\mathbf{x} \end{equation} \]

Given that \(A^2 = A\), we also know that:

\[ \begin{equation} A^2\mathbf{x} = \lambda\mathbf{x} \end{equation} \]

Based off the first equation and the properties of scalar multiplication, we can modify the expression \(A^2\mathbf{x}\) as follows:

\[ \begin{equation} A^2\mathbf{x} = A(Ax) = A(\lambda\mathbf{x}) = \lambda(A\mathbf{x}) = \lambda(\lambda\mathbf{x}) = \lambda^2\mathbf{x} \end{equation} \] Thus combining the result of these above three equations we have that:

\[ \begin{align} A\mathbf{x} &= \lambda\mathbf{x} \\ A\mathbf{x} &= \lambda^2\mathbf{x} \end{align} \]

The only way that this can be true is if \(\lambda^2 = \lambda\). Thus:

\[ \begin{align} \lambda^2 &= \lambda \\ \lambda^2 - \lambda &= 0 \\ \lambda(\lambda - 1) &= 0 \\ \end{align} \]

The solutions to the above equation are \(\lambda = 0\) and \(\lambda = 1\). Thus, we conclude that is \(A\) is an idempotent matrix, its only possible eigenvalues are 0 and 1. \(\blacksquare\)

Example:

The following matrix \(A\) is idempotent:

\[ A = \begin{pmatrix} 3 & -6 \\ 1 & -2 \\ \end{pmatrix} \]

This is confirmed in the code cell below:

A <- rbind(c(3, -6), c(1, -2))
A == A %*% A
##      [,1] [,2]
## [1,] TRUE TRUE
## [2,] TRUE TRUE

To find the eigenvalues \(\lambda\) of \(A\), we can solve the following equation:

\[ \begin{align} \text{det} \begin{pmatrix} 3 - \lambda & -6 \\ 1 & -2 - \lambda \\ \end{pmatrix} &= 0 \\ (3-\lambda)(-2-\lambda) + 6 &= \\ \lambda^2 -3\lambda + 2\lambda - 6 + 6 &= \\ \lambda^2 -\lambda = 0 \end{align} \]

We see from the above that the eigenvalues of this idempotent matrix are indeed equal to 0 and 1. The math above is confirmed via the R eigen function:

round(eigen(A)[[1]])
## [1] 1 0