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

David Austin, Erica Bernstein

Section 10.2 Finding eigenvalues

Definition 10.2.1.

Given a square \(n\times n\) matrix \(A\text{,}\) we say that a nonzero vector \(\vvec\) is an eigenvector of \(A\) if there is a scalar \(\lambda\) such that
\begin{equation*} A\vvec = \lambda \vvec\text{.} \end{equation*}
The scalar \(\lambda\) is called the eigenvalue associated to the eigenvector \(\vvec\text{.}\)

Example 10.2.2.

Consider the matrix \(A = \left[\begin{array}{rr} 7 \amp 6 \\ 6 \amp -2 \\ \end{array}\right]\) and the vector \(\vvec=\twovec{2}{1}\text{.}\) We find that
\begin{equation*} A\vvec = \left[\begin{array}{rr} 7 \amp 6 \\ 6 \amp -2 \\ \end{array}\right] \twovec{2}{1} = \twovec{20}{10} =10\twovec{2}{1} =10\vvec\text{.} \end{equation*}
In other words, \(A\vvec = 10\vvec\text{,}\) which says that \(\vvec\) is an eigenvector of the matrix \(A\) with associated eigenvalue \(\lambda = 10\text{.}\)
Similarly, if \(\wvec = \twovec{-1}{2}\text{,}\) we find that
\begin{equation*} A\wvec = \left[\begin{array}{rr} 7 \amp 6 \\ 6 \amp -2 \\ \end{array}\right] \twovec{-1}{2} = \twovec{5}{-10} =-5\twovec{-1}{2} =-5\wvec\text{.} \end{equation*}
Here again, we have \(A\wvec = -5\wvec\) showing that \(\wvec\) is an eigenvector of \(A\) with associated eigenvalue \(\lambda=-5\text{.}\)
We will first see that the eigenvalues of a square matrix appear as the roots of a particular polynomial. To begin, notice that we originally defined an eigenvector as a nonzero vector \(\vvec\) that satisfies the equation \(A\vvec = \lambda\vvec\text{.}\) We will rewrite this as
\begin{equation*} \begin{aligned} A\vvec \amp {}={} \lambda\vvec \\ A\vvec - \lambda\vvec \amp {}={} \zerovec \\ A\vvec - \lambda I\vvec \amp {}={} \zerovec \\ (A-\lambda I)\vvec \amp {}={} \zerovec\text{.} \\ \end{aligned} \end{equation*}
In other words, an eigenvector \(\vvec\) is a solution of the homogeneous equation \((A-\lambda I)\vvec=\zerovec\text{.}\) This puts us in the familiar territory explored in the next activity.

Activity 10.2.1.

The eigenvalues of a square matrix are defined by the condition that there be a nonzero solution to the homogeneous equation \((A-\lambda I)\vvec=\zerovec\text{.}\)
  1. If there is a nonzero solution to the homogeneous equation \((A-\lambda I)\vvec = \zerovec\text{,}\) what can we conclude about the invertibility of the matrix \(A-\lambda I\text{?}\)
  2. If there is a nonzero solution to the homogeneous equation \((A-\lambda I)\vvec = \zerovec\text{,}\) what can we conclude about the determinant \(\det(A-\lambda I)\text{?}\)
  3. Let’s consider the matrix
    \begin{equation*} A = \left[\begin{array}{rr} 1 \amp 2 \\ 2 \amp 1 \\ \end{array}\right] \end{equation*}
    from which we construct
    \begin{equation*} A-\lambda I = \left[\begin{array}{rr} 1 \amp 2 \\ 2 \amp 1 \\ \end{array}\right] - \lambda \left[\begin{array}{rr} 1 \amp 0 \\ 0 \amp 1 \\ \end{array}\right] = \left[\begin{array}{rr} 1-\lambda \amp 2 \\ 2 \amp 1-\lambda \\ \end{array}\right]\text{.} \end{equation*}
    Find the determinant \(\det(A-\lambda I)\text{.}\) What kind of equation do you obtain when we set this determinant to zero to obtain \(\det(A-\lambda I) = 0\text{?}\)
  4. Use the determinant you found in the previous part to find the eigenvalues \(\lambda\) by solving the equation \(\det(A-\lambda I) = 0\text{.}\)
  5. Consider the matrix \(A = \left[\begin{array}{rr} 2 \amp 1 \\ 0 \amp 2 \\ \end{array}\right]\) and find its eigenvalues by solving the equation \(\det(A-\lambda I) = 0\text{.}\)
  6. Consider the matrix \(A = \left[\begin{array}{rr} 0 \amp -1 \\ 1 \amp 0 \\ \end{array}\right]\) and find its eigenvalues by solving the equation \(\det(A-\lambda I) = 0\text{.}\)
  7. Find the eigenvalues of the triangular matrix \(\left[\begin{array}{rrr} 3 \amp -1 \amp 4 \\ 0 \amp -2 \amp 3 \\ 0 \amp 0 \amp 1 \\ \end{array}\right] \text{.}\) What is generally true about the eigenvalues of a triangular matrix?
Solution.
  1. The matrix \(A-\lambda I\) cannot be invertible.
  2. It must be the case that \(\det(A-\lambda I) = 0\text{.}\)
  3. We find that \(\det(A-\lambda I) = \lambda^2 - 2\lambda - 3 = 0\text{.}\)
  4. \(\lambda^2 - 2\lambda - 3= (\lambda-3)(\lambda+1) = 0\) so we find eigenvalues \(\lambda = 3\) and \(\lambda = -1\text{.}\)
  5. For this matrix, we have \(\det(A-\lambda I) = (\lambda-2)^2=0\) so there is one eigenvalue, \(\lambda = 2\text{.}\)
  6. \(\det(A-\lambda I) = \lambda^2 + 1 = 0\) so there are complex eigenvalues, \(\lambda = i\) and \(\lambda = -i\text{.}\)
  7. Because the determinant of a triangular matrix equals the product of its diagonal entries, The eigenvalues are equal to the entries on the diagonal.
This activity demonstrates a technique that enables us to find the eigenvalues of a square matrix \(A\text{.}\) Since an eigenvalue \(\lambda\) is a scalar for which the equation \((A-\lambda I)\vvec = \zerovec\) has a nonzero solution, it must be the case that \(A-\lambda I\) is not invertible. Therefore, its determinant is zero. This gives us the equation
\begin{equation*} \det(A-\lambda I) = 0 \end{equation*}
whose solutions are the eigenvalues of \(A\text{.}\)

Example 10.2.3.

If \(A = \left[\begin{array}{rr} -4 \amp 4 \\ -12 \amp 10 \\ \end{array}\right] \text{,}\) we see that
\begin{equation*} \begin{aligned} \det(A-\lambda I) \amp {}={} 0 \\ \\ \det \left[\begin{array}{rr} -4 - \lambda \amp 4 \\ -12 \amp 10 - \lambda \\ \end{array}\right] \amp {}={} 0 \\ \\ (-4-\lambda)(10-\lambda) + 48 \amp {}={} 0 \\ \lambda^2-6\lambda + 8 {}={} 0 \\ (\lambda-4)(\lambda-2) {}={} 0\text{.} \\ \end{aligned} \end{equation*}
This shows us that the eigenvalues are \(\lambda = 4\) and \(\lambda=2\text{.}\)
In general, the expression \(\det(A-\lambda I)\) is a polynomial in \(\lambda\text{,}\) which is called the characteristic polynomial of \(A\text{.}\) If \(A\) is an \(n\times n\) matrix, the degree of the characteristic polynomial is \(n\text{.}\) For instance, if \(A\) is a \(2\times2\) matrix, then \(\det(A-\lambda I)\) is a quadratic polynomial; if \(A\) is a \(3\times3\) matrix, then \(\det(A-\lambda I)\) is a cubic polynomial.

Example 10.2.4.

Consider the matrix \(A=\begin{bmatrix} 5 \amp -1\\ 4 \amp 1 \\ \end{bmatrix} \text{,}\) whose characteristic equation is
\begin{equation*} \lambda^2 - 6\lambda + 9 = (\lambda-3)^2 = 0. \end{equation*}
In this case, the characteristic polynomial has one real root, which means that this matrix has a single real eigenvalue, \(\lambda = 3\text{.}\)

Example 10.2.5.

To find the eigenvalues of a triangular matrix, we remember that the determinant of a triangular matrix is the product of the entries on the diagonal. For instance, the following triangular matrix has the characteristic equation
\begin{equation*} \begin{aligned} \det\left( \left[\begin{array}{rrr} 4 \amp 2 \amp 3 \\ 0 \amp -2 \amp -1 \\ 0 \amp 0 \amp 3 \\ \end{array}\right] -\lambda I\right) \amp {}={} \det \left[\begin{array}{rrr} 4-\lambda \amp 2 \amp 3 \\ 0 \amp -2-\lambda \amp -1 \\ 0 \amp 0 \amp 3-\lambda \\ \end{array}\right] \\ \\ \amp {}={}(4-\lambda)(-2-\lambda)(3-\lambda) = 0\text{,} \end{aligned} \end{equation*}
showing that the eigenvalues are the diagonal entries \(\lambda = 4,-2,3\text{.}\)
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