Skip to main content

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

David Austin, Erica Bernstein

Section 10.3 Finding eigenvectors

Now that we can find the eigenvalues of a square matrix \(A\) by solving the characteristic equation \(\det(A-\lambda I) = 0\text{,}\) we will turn to the question of finding the eigenvectors associated to an eigenvalue \(\lambda\text{.}\) The key, as before, is to note that an eigenvector is a nonzero solution to the homogeneous equation \((A-\lambda I)\vvec = \zerovec\text{.}\) In other words, the eigenvectors associated to an eigenvalue \(\lambda\) form the null space \(\nul(A-\lambda I)\text{.}\)
This shows that the eigenvectors associated to an eigenvalue form a subspace of \(\real^n\text{.}\) We will denote the subspace of eigenvectors of a matrix \(A\) associated to the eigenvalue \(\lambda\) by \(E_\lambda\) and note that
\begin{equation*} E_\lambda = \nul(A-\lambda I)\text{.} \end{equation*}
We say that \(E_\lambda\) is the eigenspace of \(A\) associated to the eigenvalue \(\lambda\text{.}\)

Activity 10.3.1.

In this activity, we will find the eigenvectors of a matrix by finding all solutions to homogeneous equation \((A-\lambda I)\vvec = \zerovec\text{.}\)
  1. Let’s begin with the matrix \(A = \left[\begin{array}{rr} 1 \amp 2 \\ 2 \amp 1 \\ \end{array}\right] \text{.}\) We have seen that \(\lambda = 3\) is an eigenvalue. Find the eigenvectors by finding a all solutions to \((A-3I)\vvec = \zerovec\text{.}\) Since there are infinitely many soltions, there are infinitely many eigenvectors. Pick one and verify that it is an eigenvector by showing that \(A\vvec = 3\vvec\text{.}\)
  2. We also saw that \(\lambda = -1\) is an eigenvalue. Find the eigenvectors by finding a all solutions to \((A-(-1)I)\vvec = \zerovec\text{.}\) Since there are infinitely many soltions, there are infinitely many eigenvectors. Pick one and verify that it is an eigenvector by showing that \(A\vvec = -\vvec\text{.}\)
  3. Now consider the matrix \(A = \left[\begin{array}{rr} 3 \amp 0 \\ 0 \amp 3 \\ \end{array}\right] \text{.}\) Write the characteristic polynomial for \(A\) and use it to find the eigenvalues of \(A\text{.}\) For each eigenvalue, find the eigenvectors by finding all solutions to \((A-\lambda I)\vvec = \zerovec\text{.}\)
  4. Next, consider the matrix \(A = \left[\begin{array}{rr} 2 \amp 1 \\ 0 \amp 2 \\ \end{array}\right] \text{.}\) Write the characteristic polynomial for \(A\) and use it to find the eigenvalues of \(A\text{.}\) For each eigenvalue, find the eigenvectors by finding all solutions to \((A-\lambda I)\vvec = \zerovec\text{.}\)
  5. Finally, find the eigenvalues and eigenvectors of the diagonal matrix \(A = \left[\begin{array}{rr} 4 \amp 0 \\ 0 \amp -1 \\ \end{array}\right] \text{.}\) Explain your result by considering the geometric effect of the matrix transformation defined by \(A\text{.}\)
Solution.
  1. We have
    \begin{equation*} A - 3I = \mattwo{-2}{2}{2}{-2} \sim \mattwo{1}{-1}{0}{0}\text{.} \end{equation*}
    The solutions to \((A-3 I)\vvec = \zerovec\) gives eigenvectors, \(\vvec=\twovec{x_1}{x_2}=x_2\twovec{1}{1}\text{.}\)
  2. We have
    \begin{equation*} A - (-1)I = \mattwo{2}{2}{2}{2} \sim \mattwo{1}{1}{0}{0}\text{.} \end{equation*}
    The solutions to \((A- (-1) I)\vvec = \zerovec\) gives eigenvectors, \(\vvec=\twovec{x_1}{x_2}=x_2\twovec{-1}{1}\text{.}\)
  3. The characteristic equation is \((3-\lambda)^2 = 0\text{,}\) which means that there is a single eigenvalue \(\lambda=3\text{.}\)
    \begin{equation*} A - 3I = \mattwo{0}{0}{0}{0} \sim \mattwo{0}{0}{0}{0}\text{.} \end{equation*}
    The solutions to \((A-3 I)\vvec = \zerovec\) gives eigenvectors, \(\vvec=\twovec{x_1}{x_2}=x_1\twovec{1}{0}+ x_2\twovec{0}{1}\text{.}\) Note there are two linearly independent eigenvectors, \(\left\{\twovec{1}{0},\twovec{0}{1}\right\}\text{.}\)
  4. The characteristic equation is \((2-\lambda)^2 = 0\) so there is again a single eigenvalue \(\lambda=2\text{.}\)
    \begin{equation*} A - 2I = \mattwo{0}{1}{0}{0} \sim \mattwo{0}{1}{0}{0}\text{.} \end{equation*}
    The solutions to \((A-2 I)\vvec = \zerovec\) gives eigenvectors, \(\vvec=\twovec{x_1}{x_2}=x_1\twovec{1}{0}\)
  5. We have eigenvectors \(\twovec10\) with associated eigenvector \(\lambda=4\) and \(\twovec01\) with associated eigenvector \(\lambda=-1\text{.}\)
PrevTopNext