Discussion Week 3 Eigenvector and Eigenvalues

Week 3 problem set for Discussion Board

• Approach is to first find the eigenvalues from the formula

\(\det(A - \lambda I_n) = 0\) (n in this case is 2)

that is \(\bigg\rvert\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} - \begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix}\bigg\rvert = 0\) (| |) bars means the determinant of the matrix.

simplfying the inside of the | | gives

\(\bigg\rvert\begin{pmatrix} 1 - \lambda & 1 \\ 1 & 1 - \lambda \end{pmatrix} \bigg\rvert = 0\)

the determinant of the above is

\((1 - \lambda) * (1 - \lambda) - 1 = 0\)

factoring the \((1 - \lambda)^2\) and simplfying gives

\(\lambda^2 - 2\lambda = 0\)

and the eigenvalues are \(\lambda = 0\) and \(\lambda = 2\)

• Finding the vectors in the eigenspace \(\xi_A(0)\) and \(\xi_A(2)\) involves solving the equation

\(\bigg(\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} - \begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix}\bigg) \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}= \begin{pmatrix} 0 \\ 0 \end{pmatrix}\)

for \(\lambda = 0\) we have

\(\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}= \begin{pmatrix} 0 \\ 0 \end{pmatrix}\)

that is we have \(x_1 + x_2 = 0\) and \(x_1 + x_2 = 0\) that is \(x_1 = -x_2\)

for a non-zero vector we can put \(x_1 = -1\) and \(x_2 = 1\) as vectors in the nullspace of \(A-\lambda I\)

for \(\lambda = 2\) we have

\(\begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}= \begin{pmatrix} 0 \\ 0 \end{pmatrix}\)

that is we have \(-x_1 + x_2 = 0\) and \(x_1 - x_2 = 0\) that is \(x_1 = x_2\)

a non zero vector we can put \(x_1 = 1\) and \(x_2 = 1\) as vectors in the nullspace of \(A-\lambda I\)

• Finding algebraic multiplicity is finding the number of times the eigenvalue \(\lambda\) appears as a root of the

characteristic polynomial

since each eigenvalue 0 and 2 appear only once as roots in the characteristic polynomial,

the algebraic multiplicity of those eigenvalues are both 1.

• The geometric multiplicity of an eigenvalue \(\lambda\) of A is the dimension of \(\xi_A(\lambda)\)

for the eigenvalues 0 and 2, the dimensions of the nullspace of \(A-\lambda I\) are simply 1.