library(pracma)

A:

According to theorem EMRCP, Eigenvalues of a Matrix are Roots of Characteristic Polynomials Suppose A is a square matrix. Then \(\lambda\) is an eigenvalue of A if and only if \(p_{A} (\lambda) = 0\)

By setting the characteristic polynomial equation equal to zero, \(p_{A}(x) =(x + 2)(x - 2)^2(x - 4) = 0\)

Roots are \(x = -2, x = 2\ and\ x = 4,\).

Eigenvalues of given matrix are \(\lambda = -2, \lambda = 2\) and \(\lambda = 4\)

Eigenspace of a Matrix(\({\cal E}_{A}\)): Suppose that A is a square matrix and \(\lambda\) is an eigenvalue of A. Then the eigenspace, \({\cal E}_{A}(\lambda)\), is the set of all the eigenvectors of A for \(\lambda\), together with the inclusion of the zero vector.

#initialize the matrix
A = matrix(c(0,-2,-2,-2, 4,6,8,8, -1,-1,-1,-3, 1,1,-1,1), nrow=4, byrow=TRUE)

#Make matrix similar to problem
A = t(A)

A
##      [,1] [,2] [,3] [,4]
## [1,]    0    4   -1    1
## [2,]   -2    6   -1    1
## [3,]   -2    8   -1   -1
## [4,]   -2    8   -3    1

Eigenvector of \(n \times n\) matrix A, where eigenvalue is \(\lambda\) is \({\cal E}_{A}(\lambda) = Null\ space(A - \lambda I_{n})\)

\(\lambda = -2, A - \lambda I_{4}\) \(= \left[\begin{array}{cc}0 & 4 & -1 & 1\\-2 & 6 & -1 & 1 \\-2 & 8 & -1 & -1 \\-2 & 8 & -3 & 1\end{array}\right] -(-2) \left[\begin{array}{cc}1 & 0 & 0 & 0\\0 & 1 & 0 & 0 \\0 & 0 & 1 & 0 \\0 & 0 & 0 & 1\end{array}\right]\)

lambda = -2
lambdaI = lambda * diag(4)

result = A - lambdaI

#using pracma library get reduced row echelon form for result
rrefMx = rref(result)
rrefMx
##      [,1] [,2] [,3] [,4]
## [1,]    1    0    0    0
## [2,]    0    1    0    0
## [3,]    0    0    1   -1
## [4,]    0    0    0    0

Null space, \({\cal E}_{A}(-2) = {\cal N}(A - (-2I_{4}))\)

\({\cal N}(A + 2I_{4}) = \left[\begin{array}{cc}1 & 0 & 0 & 0\\0 & 1 & 0 & 0 \\0 & 0 & 1 & -1 \\0 & 0 & 0 & 0\end{array}\right] \left[\begin{array}{cc}v_1 \\ v_2 \\ v_3 \\ v_4\end{array}\right] = \left[\begin{array}{cc}0 \\ 0 \\ 0 \\ 0\end{array}\right]\)

Solving the equations, \(v_1 = 0,\ v_2 = 0,\ (v_3 - v_4 = 0)\)

\(v_1 = 0,\ v_2 = 0,\ v_3 = v_4, if\ v_3 = t, then\ v_4 = t\)

\({\cal E}_{A}(-2) = \left\{\left[\begin{array}{cc} v_1 \\ v_2 \\ v_3 \\ v_4\end{array}\right] = t\left[\begin{array}{cc} 0 \\ 0 \\ 1 \\ 1\end{array}\right] | t\ \epsilon\ Real\ Numbers\right\}\)

Eigenspace for \(\lambda = -2\) is \({\cal E}_{A}(-2) = \left\langle\left[\begin{array}{cc} 0 \\ 0 \\ 1 \\ 1\end{array}\right]\right\rangle\)

\(\lambda = 2, A - \lambda I_{4}\) \(= \left[\begin{array}{cc}0 & 4 & -1 & 1\\-2 & 6 & -1 & 1 \\-2 & 8 & -1 & -1 \\-2 & 8 & -3 & 1\end{array}\right] -2 \left[\begin{array}{cc}1 & 0 & 0 & 0\\0 & 1 & 0 & 0 \\0 & 0 & 1 & 0 \\0 & 0 & 0 & 1\end{array}\right]\)

lambda = 2
lambdaI = lambda * diag(4)

result = A - lambdaI

#using pracma library get reduced row echelon form for result
rrefMx = rref(result)
rrefMx
##      [,1] [,2] [,3] [,4]
## [1,]    1    0 -0.5 -1.5
## [2,]    0    1 -0.5 -0.5
## [3,]    0    0  0.0  0.0
## [4,]    0    0  0.0  0.0

Null space, \({\cal E}_{A}(2) = {\cal N}(A - 2I_{4})\)

\({\cal N}(A - 2I_{4}) = \left[\begin{array}{cc}1 & 0 & -0.5 & -1.5\\0 & 1 & -0.5 & -0.5 \\0 & 0 & 0 & 0 \\0 & 0 & 0 & 0\end{array}\right] \left[\begin{array}{cc}v_1 \\ v_2 \\ v_3 \\ v_4\end{array}\right] = \left[\begin{array}{cc}0 \\ 0 \\ 0 \\ 0\end{array}\right]\)

Solving the equations, \(v_1 - 0.5v_3 -1.5v_4= 0,\ v_2 - 0.5v_3 -0.5v_4= 0\)

\(v_1 = 0.5v_3 +1.5v_4,\ v_2 = 0.5v_3 +0.5v_4,\ if\ v_3 = a\ and\ v_4 = b\)

\(v_1 = 0.5a +1.5b,\ v_2 = 0.5a +0.5b\)

\({\cal E}_{A}(2) = \left\{\left[\begin{array}{cc} v_1 \\ v_2 \\ v_3 \\ v_4\end{array}\right] = a\left[\begin{array}{cc} 0.5 \\ 0.5 \\ 1 \\ 0\end{array}\right] + b\left[\begin{array}{cc} 1.5 \\ 0.5 \\ 0 \\ 1\end{array}\right] | a,b\ \epsilon\ Real\ Numbers\right\}\)

Eigenspace for \(\lambda = 2\) is \({\cal E}_{A}(-2) = \left\langle\left[\begin{array}{cc} 0.5 \\ 0.5 \\ 1 \\ 0\end{array}\right] , \left[\begin{array}{cc} 1.5 \\ 0.5 \\ 0 \\ 1\end{array}\right]\right\rangle\)

\(\lambda = 4, A - \lambda I_{4}\) \(= \left[\begin{array}{cc}0 & 4 & -1 & 1\\-2 & 6 & -1 & 1 \\-2 & 8 & -1 & -1 \\-2 & 8 & -3 & 1\end{array}\right] -4 \left[\begin{array}{cc}1 & 0 & 0 & 0\\0 & 1 & 0 & 0 \\0 & 0 & 1 & 0 \\0 & 0 & 0 & 1\end{array}\right]\)

lambda = 4
lambdaI = lambda * diag(4)

result = A - lambdaI

#using pracma library get reduced row echelon form for result
rrefMx = rref(result)
rrefMx
##      [,1] [,2] [,3] [,4]
## [1,]    1    0    0   -1
## [2,]    0    1    0   -1
## [3,]    0    0    1   -1
## [4,]    0    0    0    0

Null space, \({\cal E}_{A}(4) = {\cal N}(A - 4I_{4})\)

\({\cal N}(A - 4I_{4}) = \left[\begin{array}{cc}1 & 0 & 0 & -1\\0 & 1 & 0 & -1 \\0 & 0 & 1 & -1 \\0 & 0 & 0 & 0\end{array}\right] \left[\begin{array}{cc}v_1 \\ v_2 \\ v_3 \\ v_4\end{array}\right] = \left[\begin{array}{cc}0 \\ 0 \\ 0 \\ 0\end{array}\right]\)

Solving the equations, \(v_1 - v_4 = 0,\ v_2 - v_4= 0,\ v_3 - v_4 = 0\)

\(v_1 = v_4,\ v_2 = v_4,\ v_3 = v_4\, if\ v_4 = t\)

\({\cal E}_{A}(4) = \left\{\left[\begin{array}{cc} v_1 \\ v_2 \\ v_3 \\ v_4\end{array}\right] = t\left[\begin{array}{cc} 1 \\ 1 \\ 1 \\ 1\end{array}\right] | t\ \epsilon\ Real\ Numbers\right\}\)

Eigenspace for \(\lambda = 4\) is \({\cal E}_{A}(4) = \left\langle\left[\begin{array}{cc} 1 \\ 1 \\ 1 \\ 1\end{array}\right]\right\rangle\)