T10 A matrix \(A\) is idempotent if \(A^2 = A\).
Show that the only possible eigenvalues of an idempotent matrix are \(\lambda = 0\) and \(\lambda = 1\).
Then give an example of a matrix that is idempotent and has both of these two values as eigenvalues.
Solution:
Let \(A\) be an idempotent matrix with eigenvalue \(\lambda\) and corresponding eigenvector \(x\) , such that \(x \neq 0\) .
Then \(Ax={\lambda}x\) .
Because \(A\) is idempotent, we have \(A^2=A\), so \(A^2 x = Ax = \lambda x\) .
Thus, \(A^2x = AA x = A(Ax) = A(\lambda x) = A \lambda x = \lambda A x = \lambda (Ax) = \lambda (\lambda x) = \lambda^2 x\) .
So, \(\lambda x = A x = A^2 x = \lambda^2 x\) , which means \(\lambda^2 x - \lambda x = \lambda (\lambda - 1) x = 0\) .
Since \(x \neq 0\), we must have \(\lambda \in \{0,1\}\) .
An example of such an idempotent matrix is \(A = \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}\) , as \(A^2=\begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}=\begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}=A\) .
A = c(
0,0,
1,1
)
A = matrix(A,2,2,T)
A## [,1] [,2]
## [1,] 0 0
## [2,] 1 1### A^2
A2 = A %*% A
A2## [,1] [,2]
## [1,] 0 0
## [2,] 1 1### check if A^2 == A
A2 == A## [,1] [,2]
## [1,] TRUE TRUE
## [2,] TRUE TRUE### Compute eigenvalues, (normalized) eigenvectors
eigen(A)## eigen() decomposition
## $values
## [1] 1 0
##
## $vectors
## [,1] [,2]
## [1,] 0 0.7071068
## [2,] 1 -0.7071068