Skip to main content

Math 211 Activities Workbook: Adapted from: Understanding Linear Algebra by David Austin

David Austin, Erica Bernstein

Section 10.4 Eigenvalues and their multiplicities

Activity 10.4.1.

The number of times that \(\lambda_j - \lambda\) appears as a factor in the characteristic polynomial, is called the multiplicity of the eigenvalue \(\lambda_j\text{.}\)
  1. Find the eigenvalues of \(\left[\begin{array}{rr} 0 \amp -1 \\ 4 \amp -4 \\ \end{array}\right]\) and state their multiplicities. Can you find a basis of \(\real^2\) consisting of eigenvectors of this matrix?
  2. Consider the matrix \(A = \left[\begin{array}{rrr} -1 \amp 0 \amp 2 \\ -2 \amp -2 \amp -4 \\ 0 \amp 0 \amp -2 \\ \end{array}\right]\) whose characteristic equation is
    \begin{equation*} (-2-\lambda)^2(-1-\lambda) = 0\text{.} \end{equation*}
    1. Identify the eigenvalues and their multiplicities.
    2. For each eigenvalue \(\lambda\text{,}\) find a basis of the eigenspace \(E_\lambda\) and state its dimension.
    3. Is there a basis of \(\real^3\) consisting of eigenvectors of \(A\text{?}\)
  3. Now consider the matrix \(A = \left[\begin{array}{rrr} -5 \amp -2 \amp -6 \\ -2 \amp -2 \amp -4 \\ 2 \amp 1 \amp 2 \\ \end{array}\right]\) whose characteristic equation is also
    \begin{equation*} (-2-\lambda)^2(-1-\lambda) = 0\text{.} \end{equation*}
    1. Identify the eigenvalues and their multiplicities.
    2. For each eigenvalue \(\lambda\text{,}\) find a basis of the eigenspace \(E_\lambda\) and state its dimension.
    3. Is there a basis of \(\real^3\) consisting of eigenvectors of \(A\text{?}\)
  4. Consider the matrix \(A = \left[\begin{array}{rrr} -5 \amp -2 \amp -6 \\ 4 \amp 1 \amp 8 \\ 2 \amp 1 \amp 2 \\ \end{array}\right]\) whose characteristic equation is
    \begin{equation*} (-2-\lambda)(1-\lambda)(-1-\lambda) = 0\text{.} \end{equation*}
    1. Identify the eigenvalues and their multiplicities.
    2. For each eigenvalue \(\lambda\text{,}\) find a basis of the eigenspace \(E_\lambda\) and state its dimension.
    3. Is there a basis of \(\real^3\) consisting of eigenvectors of \(A\text{?}\)
Solution.
  1. There is one eigenvalue \(\lambda=-2\) having multiplicity two. Because the eigenspace \(E_{-2}\) is one-dimensional, however, we cannot find a basis for \(\real^2\) consisting of eigenvectors of \(A\text{.}\)
  2. For the \(3\times3\) matrix \(A\text{,}\)
    1. We have eigenvalues \(\lambda=-2\) with multiplicity \(2\) and \(\lambda=-1\) with multiplicity \(1\text{.}\)
    2. The eigenspace \(E_{-2}\) is two-dimensional with basis \(\left\{\threevec{-2}{0}{1},\threevec010\right\}\text{.}\) The eigenspace \(E_{-1}\) is one-dimensional with basis vector \(\threevec{-1}20\text{.}\)
    3. We are able to form a basis for \(\real^3\) consisting of eigenvectors of \(A\text{.}\)
  3. For the \(3\times3\) matrix \(A\text{,}\)
    1. We have eigenvalues \(\lambda=-2\) with multiplicity \(2\) and \(\lambda=-1\) with multiplicity \(1\text{.}\)
    2. The eigenspace \(E_{-2}\) is one-dimensional with basis vector \(\threevec{-2}01\text{.}\) The eigenspace \(E_{-1}\) is also one-dimensional with basis vector \(\threevec{-1}{2}{0}\text{.}\)
    3. It is not possible to form a basis for \(\real^3\) consisting of eigenvectors of \(A\text{.}\)
  4. For this matrix,
    1. There are three eigenvalues \(\lambda=-2\text{,}\) \(-1\text{,}\) and \(1\text{,}\) each having multiplicity \(1\text{.}\)
    2. A basis vector for the eigenspace \(E_{-2}\) is \(\threevec{-2}01\text{.}\) A basis vector for the eigenspace \(E_{-1}\) is \(\threevec{1}{-2}0\text{.}\) A basis vector for the eigenspace \(E_{1}\) is \(\threevec{-2}{3}{1}\text{.}\)
    3. We can form a basis for \(\real^3\) consisting of eigenvectors of \(A\text{.}\)
PrevTopNext