C27 (LA, chapter E, pg 389)

For matrix

\[\mathbf{A} = \begin{bmatrix} 0 & 4 & -1 & 1 \\ -2 & 6 & -1 & 1 \\ -2 & 8 & -1 & -1 \\ -2 & 8 & -3 & 1 \end{bmatrix} \]

the characteristic polynomial of A is

\[ P_A(x) = (x+2)(x-2)^2(x-4) \,\]

Find the eigenvalues and corresponding eigenspaces of A.

Solving the characteristic polynomial for 0, the eigenvalues of matrix A are \(\lambda = -2\), \(\lambda = 2\), and \(\lambda = 4\).

For \(\lambda_1 = -2\)

\[\left(\left[\begin{array} {rrr} -2 & 0 & 0 & 0\\ 0 & -2 & 0 & 0\\ 0 & 0 & -2 & 0 \\ 0 & 0 & 0 & -2 \end{array}\right] - \left[\begin{array} {rrr} 0 & 4 & -1 & 1 \\ -2 & 6 & -1 & 1 \\ -2 & 8 & -1 & -1 \\ -2 & 8 & -3 & 1 \end{array}\right]\right) \begin{pmatrix} -2-0 & 0-4 & 0+1 & 0-1\\ 0-0 & -2-6 & 0+1 & 0-1\\ 0+2 & 0-8 & -2-1 & 0+1 \\ 0+2 & 0-8 & 0+3 & -2-1 \end{pmatrix} = \begin{pmatrix} -2 & -4 & 1 & -1\\ 0 & -8 & 1 & -1\\ 2 & -8 & -3 & 1 \\ 2 & -8 & 3 & -3 \end{pmatrix} \]

# Compose the RREF for lambda1
A_lambda1 <- matrix(c(-2, -4, 1, -1, 0, -8, 1, -1, 2, -8, -3, 1, 2, -8, 3, -3), nrow = 4, ncol = 4, byrow = TRUE)
rref(A_lambda1)
##      [,1] [,2] [,3] [,4]
## [1,]    1    0    0    0
## [2,]    0    1    0    0
## [3,]    0    0    1    0
## [4,]    0    0    0    1

For \(\lambda = -2\), the determinant at 0 reduces to the identity matrix, so \(v_1 = v_2 = v_3 = v_4 = 0\), or the zero vector. The set is linearly dependent and there is no non-trivial solution.

\[\varepsilon_A(-2) = N(A + 2I_3) = \Bigg\langle \begin{Bmatrix} \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \end{Bmatrix} \Bigg\rangle \]

For \(\lambda_1 = 2\)

\[ \left(\left[\begin{array} {rrr} 2 & 0 & 0 & 0\\ 0 & 2 & 0 & 0\\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \end{array}\right] - \left[\begin{array} {rrr} 0 & 4 & -1 & 1 \\ -2 & 6 & -1 & 1 \\ -2 & 8 & -1 & -1 \\ -2 & 8 & -3 & 1 \end{array}\right]\right) \begin{pmatrix} 2-0 & 0-4 & 0+1 & 0-1\\ 0-0 & 2-6 & 0+1 & 0-1\\ 0+2 & 0-8 & 2-1 & 0+1 \\ 0+2 & 0-8 & 0+3 & 2-1 \end{pmatrix} = \begin{pmatrix} 2 & -4 & 1 & -1\\ 0 & -4 & 1 & -1\\ 2 & -8 & 1 & 1 \\ 2 & -8 & 3 & 1 \end{pmatrix} \]

# Compose the RREF for lambda1
A_lambda2 <- matrix(c(2, -4, 1, -1, 0, -4, 1, -1, 2, -8, 1, 1, 2, -8, 3, 1), nrow = 4, ncol = 4, byrow = TRUE)
rref(A_lambda2)
##      [,1] [,2] [,3] [,4]
## [1,]    1    0    0    0
## [2,]    0    1    0    0
## [3,]    0    0    1    0
## [4,]    0    0    0    1

For \(\lambda = -2\), the determinant at 0 also reduces to the identity matrix, so \(v_1 = v_2 = v_3 = v_4 = 0\), or the zero vector. The set is linearly dependent and there is no non-trivial solution.

\[\varepsilon_A(2) = N(A - 2I_3) = \Bigg\langle \begin{Bmatrix} \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \end{Bmatrix} \Bigg\rangle \]

Lastly, for \(\lambda_1 = 4\)

\[ \left(\left[\begin{array} {rrr} 4 & 0 & 0 & 0\\ 0 & 4 & 0 & 0\\ 0 & 0 & 4 & 0 \\ 0 & 0 & 0 & 4 \end{array}\right] - \left[\begin{array} {rrr} 0 & 4 & -1 & 1 \\ -2 & 6 & -1 & 1 \\ -2 & 8 & -1 & -1 \\ -2 & 8 & -3 & 1 \end{array}\right]\right) \begin{pmatrix} 4-0 & 0-4 & 0+1 & 0-1\\ 0-0 & 4-6 & 0+1 & 0-1\\ 0+2 & 0-8 & 4-1 & 0+1 \\ 0+2 & 0-8 & 0+3 & 4-1 \end{pmatrix} = \begin{pmatrix} 4 & -4 & 1 & -1\\ 0 & -2 & 1 & -1\\ 2 & -8 & 3 & 1 \\ 2 & -8 & 3 & 3 \end{pmatrix} \]

# Compose the RREF for lambda1
A_lambda3 <- matrix(c(4, -4, 1, -1, 0, -2, 1, -1, 2, -8, 3, 1, 2, -8, 3, 3), nrow = 4, ncol = 4, byrow = TRUE)
rref(A_lambda3)
##      [,1] [,2] [,3] [,4]
## [1,]    1    0    0    0
## [2,]    0    1    0    0
## [3,]    0    0    1    0
## [4,]    0    0    0    1

Again for \(\lambda = 4\), the determinant at 0 reduces to the identity matrix, so \(v_1 = v_2 = v_3 = v_4 = 0\), or the zero vector. The set is linearly dependent and there is no non-trivial solution.

\[\varepsilon_A(4) = N(A - 4I_3) = \Bigg\langle \begin{Bmatrix} \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \end{Bmatrix} \Bigg\rangle \]