CUNY 605

EE.C26 For matrix

\[A = \left[\begin{array}{rrr} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \\ \end{array} \right] \]

The characteristic polynomial of A is \(p_A(x) = (4-x)(1-x)^2\). Find the eigenvalues and corresponding eigenspaces of A.

By 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 \(p_A(\lambda) = 0\).

\(\lambda\) is equal to 4 and 1, and are both eigenvalues of A< and these are the only eigenvalues of A*.

We will now take each eigenvalue in turn and compute its eigenspace. To do this, we row-reduce the matrix A - \(\lambda*I_3\) in order to determine solutions to the homogeneous system \(LS(A-\lambda*I_3,0)\) and then express the eigenspace as the null space of \(A - \lambda*I_3\). Theorem BNS then tells us how to write the null space as the span of a basis.

For \(\lambda\) = 4.

\[A - 4I_3 = \left[\begin{array}{rrr} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \\ \end{array} \right] - \left[\begin{array}{rrr} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \\ \end{array} \right] = \left[\begin{array}{rrr} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \\ \end{array} \right]\]

library(pracma)

A <- matrix(c(2,1,1,1,2,1,1,1,2), ncol = 3, byrow = FALSE) # Create Matrix A as described in Example
I <- matrix(c(1,0,0,0,1,0,0,0,1), ncol = 3, byrow = FALSE) # Creates Identify Matrix

new.A <- A - 4*I                                           # Creates the A - 4*I
rref.A <- rref(new.A)                                      # Row-reduced form

print("A:")
## [1] "A:"
A
##      [,1] [,2] [,3]
## [1,]    2    1    1
## [2,]    1    2    1
## [3,]    1    1    2
print("I:")
## [1] "I:"
I
##      [,1] [,2] [,3]
## [1,]    1    0    0
## [2,]    0    1    0
## [3,]    0    0    1
print("A - 4*I:")
## [1] "A - 4*I:"
new.A
##      [,1] [,2] [,3]
## [1,]   -2    1    1
## [2,]    1   -2    1
## [3,]    1    1   -2
print("Row Reduced Form of (A - 4*I):")
## [1] "Row Reduced Form of (A - 4*I):"
rref.A
##      [,1] [,2] [,3]
## [1,]    1    0   -1
## [2,]    0    1   -1
## [3,]    0    0    0

As we can see, variable \(x_3\) is a free variable and can be set up to any variable. To make calculations simple, we will choose \(x_3 = 1\). The Eigenspace is:

\[\epsilon_A(4) = \emptyset(A-FI_4) = span{\left[\begin{array}{r} 1 \\ 1 \\ 1 \\ \end{array} \right]}\]

We will now perform a similar task but now for\(\lambda\) = 1.

A <- matrix(c(2,1,1,1,2,1,1,1,2), ncol = 3, byrow = FALSE) # Create Matrix A as described in Example
I <- matrix(c(1,0,0,0,1,0,0,0,1), ncol = 3, byrow = FALSE) # Creates Identify Matrix

new.A <- A - 1*I                                           # Creates the A - 4*I
rref.A <- rref(new.A)                                      # Row-reduced form

print("A:")
## [1] "A:"
A
##      [,1] [,2] [,3]
## [1,]    2    1    1
## [2,]    1    2    1
## [3,]    1    1    2
print("I:")
## [1] "I:"
I
##      [,1] [,2] [,3]
## [1,]    1    0    0
## [2,]    0    1    0
## [3,]    0    0    1
print("A - 4*I:")
## [1] "A - 4*I:"
new.A
##      [,1] [,2] [,3]
## [1,]    1    1    1
## [2,]    1    1    1
## [3,]    1    1    1
print("Row Reduced Form of (A - 4*I):")
## [1] "Row Reduced Form of (A - 4*I):"
rref.A
##      [,1] [,2] [,3]
## [1,]    1    1    1
## [2,]    0    0    0
## [3,]    0    0    0

\[x_1 + x_2 + x_3 = 0 -> x_1 = -x_2 - x_3\]

For this particular eigenvalue, the free variables are \(x_1\) and \(x_2\). Again, we will arbitrarily assign \(x_1\) and \(x_2\) to equal to a and b respectively.

\[{\left[\begin{array}{r} x_1 \\ x_2 \\ x_3 \\ \end{array} \right]} = a{\left[\begin{array}{r} 1 \\ -1 \\ 0 \\ \end{array} \right]} + b{\left[\begin{array}{r} 1 \\ 0 \\ -1 \\ \end{array} \right]} \]

The eigenspace here is:

\[\epsilon_A(1) = \emptyset(A-FI_4) = span{\left[\begin{array}{r} 1 \\ -1 \\ 0 \\ \end{array} \right]},{\left[\begin{array}{r} 1 \\ 0 \\ -1 \\ \end{array} \right]}\]