18.2 Computing eigenvalues and eigenvectors

Let’s understand a little on solving \(\det (A - \lambda I ) =0\) to determine the eigenvalues.

First consider the 2 by 2 matrix:

\[\begin{equation*} A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, \end{equation*}\]

So then \(A - \lambda I\) is the matrix:

\[\begin{equation*} A - \lambda I=\begin{pmatrix} a - \lambda & b \\ c & d-\lambda \end{pmatrix} \end{equation*}\]

The determinant of a 2 by 2 matrix is the product of the diagonal entries (for \(A - \lambda I\) they are \((a-\lambda)\cdot (d-\lambda)\)) less the product of the off-diagonal entries (in this case \(bc\)). So \(\det(A-\lambda I)=0\) is the equation \((a-\lambda)(d-\lambda)-bc=0\). When we multiply this equation out we obtain a quadratic equation to solve (for \(\lambda\) - see Exercise 18.7). The equation \(\det(A - \lambda I)=0\) even has a special name - it is called the characteristic equation. (We will find out why later.)

Example 18.1 Compute the eigenvalues for the matrix \(\displaystyle A = \begin{pmatrix} -1 & 1 \\ 0 & 3 \end{pmatrix}\).

Solution. The matrix \(A-\lambda I\) is: \(\displaystyle A-\lambda I = \begin{pmatrix} -1-\lambda & 1 \\ 0 & 3-\lambda \end{pmatrix}\). So we have:

\[\begin{equation} \det(A-\lambda I) = (-1-\lambda)(3-\lambda) - 0 = 0 \end{equation}\]

Solving the equation \((-1-\lambda)(3-\lambda)=0\) yields two eigenvalues: \(\lambda = -1\) or \(\lambda = 3\).
Once we have determined the eigenvalues we next compute the eigenvectors associated with each eigenvalue. Remember that an eigenvector is a vector \(\vec{v}\) consistent with \(A \vec{v} = \lambda \vec{v}\) or \(A \vec{v} - \lambda \vec{v} =\vec{0}\). Note that the eigenvectors have the form \(\displaystyle \vec{v} = \begin{pmatrix} x \\ y \end{pmatrix}\).
Example 18.2 Compute the eigenvectors for the matrix \(\displaystyle A = \begin{pmatrix} -1 & 1 \\ 0 & 3 \end{pmatrix}\).

Solution. So if we consider the right hand side of \(A \vec{v} - \lambda \vec{v} =\vec{0}\) we have:

\[\begin{equation*} ( A \vec{v} - \lambda \vec{v} ) = \begin{pmatrix} -x +y - \lambda x \\ 3y - \lambda y \end{pmatrix}. \end{equation*}\].

Examining this setup gives us two equations to work with: \(- x + y - \lambda x = 0\) and \(3y-\lambda y =0\). We will need to analyze them for each of our eigenvalues.

  • Case 1: \(\lambda = -1\). In the first equation we have \(-x+y+x =0\), which just yields \(y=0\). For the second equation we also have \(3y+y =0\), so that tells us again that \(y=0\). Notice how we solved for \(y\), but we didn’t uniquely determine \(x\). While this seems a little unsatisfying, that is ok. The form of this particular straight line solution is \(\displaystyle s_{1}(t)=e^{-t} \begin{pmatrix} x \\ 0 \end{pmatrix}\), where \(x\) is a free variable.
  • Case 2: \(\lambda = 3\). For the second equation we have \(3y - 3y=0\), which is always true. However in the first equation we have \(- x + y - 3x = 0\), or \(y = 4x\). In this case, \(x\) can be anything as well, but this condition means that \(y\) will have to be 4 times that value. Hence, this particular straight line solution is \(\displaystyle s_{2}(t)=e^{3t} \begin{pmatrix} x \\ 4x \end{pmatrix}\).
Notice that in both of our cases we had a free variable \(x\). Given an initial condition to the differential equation we would specify this free variable (a good choice would be \(x=1\)), or just leave this as a constant.

Once we have computed the eigenvalues and eigenvectors, we are now ready to express the most general solution for a system of differential equations. For a two-dimensional system of linear differential equations (\(\displaystyle \frac{d}{dt} \vec{x} = A \vec{x}\), the most general solution is \(\vec{x}(t) = c_{1} e^{\lambda_{1}t} \vec{v}_{1} + c_{2} e^{\lambda_{2}t} \vec{v}_{2}\)

Example 18.3 What is the solution to the differential equation \(\displaystyle \frac{d}{dt} \vec{x} = \begin{pmatrix} -1 & 1 \\ 0 & 3 \end{pmatrix} \vec{x}\)?.

Solution. Since we have already computed the eigenvalues and eigenvectors, our most general solution is:

\[\begin{equation*} \vec{x} = c_{1} e^{- t} \begin{pmatrix} 1 \\ 0 \end{pmatrix} + c_{2} e^{3t} \begin{pmatrix} 1 \\ 4 \end{pmatrix} \end{equation*}\]

18.2.1 Eigenvalues computed with demodelr

While computing eigenvalues and eigenvectors is a good algebraic exercise, we can also program this in R using the function eigenvalues from the demodelr package. The syntax works where \(\displaystyle A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) is entered in as eigenvalues(a,b,c,d,matrix_rows) where matrix_rows is the number of rows6. What gets returned from the function will be the eigenvalues and eigenvectors for any square matrix.

# For a two dimensional equation the code assumes the default is a 2 by 2 matrix, .
eigenvalues(matrix_entries = c(-1, 1, 0, 3),
            matrix_rows = 2)
## eigen() decomposition
## $values
## [1]  3 -1
## 
## $vectors
##          X1 X2
## 1 0.2425356  1
## 2 0.9701425  0
# This is equivalent because we have a two dimensional equation:
#eigenvalues(c(-1, 1, 0, 3))

Notice that the eigenvalues and the eigenvectors get returned. How you read the output for the eigenvector is that X1 is the eigenvector associated with the first eigenvalue (\(\lambda = 3\)) and X2 is the eigenvector associated with the second eigenvalue (\(\lambda = -1\)). The eigenvector associated with \(\lambda=3\) is a little different from what we computed - R will normalize the vector, which means that its total length will be one.7 However upon closer inspection in our solution we found that the second component was 4 times the first for the eigenvector, which is indeed the case.

  1. If you have a 2 by 2 matrix, you can leave out matrix_rows (so just eigenvalues(a,b,c,d)) as the default is a 2 by 2 matrix.↩︎

  2. The length of a vector \(\vec{v}\) is denoted as \(||\vec{v}||\) and is computed the following way: \(||\vec{v}||=\sqrt{v_{1}^{2}+v_{2}^{2}+...+v_{n}^{2}}\). We normalize a vector to a length of 1 by dividing each component by its length.↩︎