Chapter E - Exercise C27

For Matrix \(A = \begin{bmatrix} 0 & 4 & 0 & -1 \\ -2 & 6 & -1 & 1 \\ -2 & 8 & -1 & -1 \\ -2 & 8 & -3 & 1 \end{bmatrix}\), the characteristic polynomial of \(A\) is \(p_A(x) = (x + 2)(x - 2)^2(x - 4)\).
Find the eigenvalues and corresponding eigenspaces of \(A\).

Solution:

To find the eigenvalues and eigenspaces of the matrix \(A\), we use the characteristic polynomial \(p_A(x) = (x + 2)(x - 2)^2(x - 4)\). The roots of this polynomial give us the eigenvalues of \(A\):

  1. \(x + 2 = 0\) gives the eigenvalue \(\lambda_1 = -2\)
  2. \(x - 2 = 0\) gives the eigenvalue \(\lambda_2 = 2\)
  3. \(x - 4 = 0\) gives the eigenvalue \(\lambda_3 = 4\)

Now we apply the eigenvalues in solving \((A - \lambda I)v = 0\) to find out the corresponding eigenvectors. \(I\) is the identity matrix and \(v\) is the eigenvector.

1. Eigenvalue \(\lambda_1 = -2\):

We solve \((A - (-2)I)v = 0\) or \((A + 2I)v = 0\):

\(A = \begin{bmatrix} 0 & 4 & 0 & -1 \\ -2 & 6 & -1 & 1 \\ -2 & 8 & -1 & -1 \\ -2 & 8 & -3 & 1 \end{bmatrix}\), we have:

\[ (A + 2I) = \begin{bmatrix} 2 & 4 & 0 & -1 \\ -2 & 8 & -1 & 1 \\ -2 & 8 & 1 & -1 \\ -2 & 8 & -3 & 3 \end{bmatrix} \]

Now we solve \((A + 2I)v = 0\) to get the eigenvectors for \(\lambda_1 = -2\). This involves finding the null space of the matrix \(A + 2I\).

Matrix \(A + 2I\)

a_plus_2I <- matrix(c(
  2, 4, 0, -1,
  -2, 8, -1, 1,
  -2, 8, 1, -1,
  -2, 8, -3, 3
), nrow = 4, byrow = TRUE)

Eigenspace for Eigenvalue -2

Finding the null space of the matrix \(A + 2I\).

# the null space of A + 2I
eigenspace1 <- nullspace(a_plus_2I)

eigenspace1
##            [,1]
## [1,] 0.22903933
## [2,] 0.05725983
## [3,] 0.68711800
## [4,] 0.68711800

The resulting vector span the eigenspace associated with the eigenvalue \(-2\).

2. Eigenvalue \(\lambda_2 = 2\):

We solve \((A - 2I)v = 0\):

\[ (A - 2I) = \begin{bmatrix} -2 & 4 & 0 & -1 \\ -2 & 4 & -1 & 1 \\ -2 & 8 & -3 & -1 \\ -2 & 8 & -3 & -1 \end{bmatrix} \]

Now we solve \((A - 2I)v = 0\) to get the eigenvectors for \(\lambda_2 = 2\). This involves finding the null space of the matrix \(A - 2I\).

Matrix A - 2I

a_minus_2I <- matrix(c(
  -2, 4, 0, -1,
  -2, 4, -1, 1,
  -2, 8, -3, -1,
  -2, 8, -3, -1
), nrow = 4, byrow = TRUE)

Eigenspace for Eigenvalue 2
Finding the null space of the matrix \(A - 2I\).

# the null space of A - 2I
eigenspace2 <- nullspace(a_minus_2I)

eigenspace2
##            [,1]
## [1,] -0.6804138
## [2,] -0.4082483
## [3,] -0.5443311
## [4,] -0.2721655

The resulting vector(s) span the eigenspace associated with the eigenvalue \(2\).

3. Eigenvalue \(\lambda_3 = 4\):

We solve \((A - 4I)v = 0\):

\[ (A - 4I) = \begin{bmatrix} -4 & 4 & 0 & -1 \\ -2 & 2 & -1 & 1 \\ -2 & 8 & -5 & -1 \\ -2 & 8 & -3 & -3 \end{bmatrix} \]

Now we solve \((A - 4I)v = 0\) to get the eigenvectors for \(\lambda_3 = 4\). This involves finding the null space of the matrix \(A - 4I\).

Matrix A - 4I

a_minus_4I <- matrix(c(
  -4, 4, 0, -1,
  -2, 2, -1, 1,
  -2, 8, -5, -1,
  -2, 8, -3, -3
), nrow = 4, byrow = TRUE)

Eigenspace for Eigenvalue 4
Finding the null space of the matrix \(A - 4I\).

# the null space of A - 4I
eigenspace3 <- nullspace(a_minus_4I)

eigenspace3
## NULL

We find that the null space of the matrix \(A - 4I\) is empty.

This means that there are no non-zero vectors \(x\) such that \((A - 4I)x = 0\), indicating that the eigenvalue \(4\) is not an eigenvalue of the original matrix \(A\).