To understand that a linear map \(T\) converts solutions of \({\mathbf y}' = (T^{-1} A T) {\mathbf y}\) to solutions of \({\mathbf x}' = A {\mathbf x}\text{,}\) and, conversely, the inverse of a linear map \(T\) takes solutions of \({\mathbf x}' = A {\mathbf x}\) to solutions of \({\mathbf y}' = (T^{-1} A T) {\mathbf y}\text{.}\)
To understand that a change of coordinates converts the system \({\mathbf x}' = A {\mathbf x}\) to one of the following special cases,
In the previous sections of this chapter, we outlined procedures for solving systems of linear differential equations of the form
\begin{equation*}
\begin{pmatrix}
dx/dt \\ dy/dt
\end{pmatrix}
=
\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}
\begin{pmatrix}
x \\ y
\end{pmatrix}
=
A
\begin{pmatrix}
x \\ y
\end{pmatrix}
\end{equation*}
by determining the eigenvalues of \(A\text{.}\) In this section we will consider the following special cases for \(A\text{,}\)
Although it may seem that we have limited ourselves by attacking only a very small part of the problem of finding solutions for \(\mathbf x' = A \mathbf x\text{,}\) we are actually very close to providing a complete classification of all solutions. We will now show that we can transform any \(2 \times 2 \) system of first-order linear differential equations with constant coefficients into one of these special systems by using a change of coordinates.
Subsection5.6.1Linear Maps
First, we need to add a few things to our knowledge of matrices and linear algebra. A linear map or linear transformation on \({\mathbb R}^2\) is a function \(T: {\mathbb R}^2 \to {\mathbb R}^2\) that is defined by a matrix. That is,
\begin{equation*}
T
\begin{pmatrix}
x \\ y
\end{pmatrix}
=
\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}
\begin{pmatrix}
x \\ y
\end{pmatrix}.
\end{equation*}
When there is no confusion, we will think of the linear map \(T: {\mathbb R}^2 \to {\mathbb R}^2\) and the matrix
\begin{equation*}
\begin{pmatrix} a & b \\ c & d \end{pmatrix}
\end{equation*}
as interchangeable.
We will say that \(T: {\mathbb R}^2 \to {\mathbb R}^2\) is an invertible linear map if we can find a second linear map \(S\) such that \(T \circ S = S \circ T = I\text{,}\) where \(I\) is the identity transformation. In terms of matrices, this means that we can find a matrix \(S\) such that
If \(\det T = 0\text{,}\) then there are infinitely many nonzero vectors \({\mathbf x}\) such that \(T {\mathbf x} = {\mathbf 0}\text{.}\) Suppose that \(T^{-1}\) exists and \({\mathbf x} \neq {\mathbf 0}\) such that \(T {\mathbf x} = {\mathbf 0}\text{.}\) Then
which is a contradiction. On the other hand, we can certainly compute \(T^{-1}\text{,}\) at least in the \(2 \times 2\) case, if the determinant is nonzero.
Subsection5.6.2Changing Coordinates
In Subsection5.1.2, we discussed what a basis was along with the coordinates with respect to a particular basis. The vectors \({\mathbf e}_1 = (1, 0)\) and \({\mathbf e}_2 = (0, 1)\) form a basis for \({\mathbb R}^2\text{.}\) Indeed, if \({\mathbf z} = (-5, -4)\text{,}\) then we can write
We say that the coordinates of \(\mathbf z\) with respect to the basis \(\{ {\mathbf e}_1, {\mathbf e}_2 \}\) are \((-5,-4)\text{.}\) Now consider the vectors \({\mathbf v}_1 = (2,1)\) and \({\mathbf v}_2 = (3, 2)\text{.}\) Since
The coordinates of \({\mathbf z}\) with respect to the basis \(\{ {\mathbf v}_1, {\mathbf v}_2 \}\) are \((2, -3)\text{.}\)
Suppose we wish to convert the coordinates with repect to one basis to a new set of coordinates with respect to a different basis; that is, we wish to do a change of coordinates. Observe that
then the coordinates with respect to the basis \(\{ {\mathbf e}_1, {\mathbf e}_2 \}\) are given by \(\mathbf d = T \mathbf c\text{.}\) If we are given the coordinates with respect to the basis \(\{ {\mathbf v}_1, {\mathbf v}_2 \}\) for a vector, we simply need to multiply by the matrix \(T\text{.}\)
Now suppose that we wish to find the coordinates with respect to the basis \(\{ \mathbf v_1, \mathbf v_2\}\) if we know that a vector \(\mathbf z = d_1 \mathbf e_1 + d_2 \mathbf e_2\text{.}\) Since \(\mathbf d = T \mathbf c\text{,}\) we need only multiply both sides of the equation by \(T^{-1}\) to get \(\mathbf c = T^{-1} \mathbf d\text{.}\) In our example,
which are the coordinates of \(\mathbf z\) with respect to the basis \(\{ {\mathbf v}_1, {\mathbf v}_2 \}\text{.}\)
Subsection5.6.3Systems and Changing Coordinates
The idea now is to use a change of coordinates to convert an arbitrary system \({\mathbf x}' = A {\mathbf x}\) into one of the special systems mentioned at the beginning of the section (5.6.1), solve the new system, and then convert our new solution back to a solution of the original system using another change of coordinates.
Suppose that we consider a linear system
\begin{equation}
{\mathbf y}' = (T^{-1} A T) {\mathbf y}\label{linear06-equation-change-of-coordinates}\tag{5.6.2}
\end{equation}
where \(T\) is an invertible matrix. If \({\mathbf y}(t)\) is a solution of (5.6.2), we claim that \({\mathbf x}(t) = T {\mathbf y}(t)\) solves the equation \({\mathbf x}' = A {\mathbf x}\text{.}\) Indeed,
\begin{align*}
{\mathbf x}'(t) & = (T {\mathbf y})'(t)\\
& = T {\mathbf y}'(t)\\
& = T( (T^{-1} A T) {\mathbf y}(t))\\
& = A (T {\mathbf y}(t))\\
& = A {\mathbf x}(t).
\end{align*}
We can think of this in two ways.
A linear map \(T\) converts solutions of \({\mathbf y}' = (T^{-1} A T) {\mathbf y}\) to solutions of \({\mathbf x}' = A {\mathbf x}\text{.}\)
The inverse of a linear map \(T\) takes solutions of \({\mathbf x}' = A {\mathbf x}\) to solutions of \({\mathbf y}' = (T^{-1} A T) {\mathbf y}\text{.}\)
We are now in a position to solve our problem of finding solutions of an arbitrary linear system
\begin{equation*}
\begin{pmatrix}
x' \\ y'
\end{pmatrix}
=
\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}
=
\begin{pmatrix}
x \\ y
\end{pmatrix}.
\end{equation*}
Subsection5.6.4Distinct Real Eigenvalues
Consider the system \({\mathbf x}' = A {\mathbf x}\text{,}\) where \(A\) has two real, distinct eigenvalues \(\lambda_1\) and \(\lambda_2\) with eigenvectors \({\mathbf v}_1\) and \({\mathbf v}_2\text{,}\) respectively. Let \(T\) be the matrix with columns \({\mathbf v}_1\) and \({\mathbf v}_2\text{.}\) If \({\mathbf e}_1 = (1, 0)\) and \({\mathbf e}_2 = (0, 1)\text{,}\) then \(T {\mathbf e}_i = {\mathbf v}_i\) for \(i = 1, 2\text{.}\) Consequently, \(T^{-1} {\mathbf v}_i = {\mathbf e}_i\) for \(i = 1, 2\text{.}\) Thus, we have
for \(i = 1, 2\text{.}\) Therefore, the matrix \(T^{-1} A T\) is in canonical form,
\begin{equation*}
T^{-1} A T = \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}.
\end{equation*}
The eigenvalues of the matrix \(T^{-1} A T\) are \(\lambda_1\) and \(\lambda_2\) with eigenvectors \((1, 0)\) and \((0, 1)\text{,}\) respectively. Thus, the general solution of
The eigenvalues of \(A\) are \(\lambda_1 = 5\) and \(\lambda_2 = -1\) and the associated eigenvectors are \((1, 2)\) and \((1, -1)\text{,}\) respectively. In this case, our matrix \(T\) is
The linear map \(T\) converts the phase portrait of the system \({\mathbf y}' = (T^{-1}AT) {\mathbf y}\) (Figure5.6.3) to the phase portrait of the system \({\mathbf x}' = A {\mathbf x}\) (Figure5.6.4).
Activity5.6.1Distinct Real Eigenvalues and Transformation of Coordinates
Consider the system of linear differential equations \(d\mathbf x/dt = A \mathbf x\text{,}\) where
Now calculate \(T \mathbf y\) and compare this solution with the one that you obtained in Activity5.2.1.1Of couurse, we have much quicker ways of solving a system \(d\mathbf x/dt = A \mathbf x\) with distinct real eigenvalues. The goal of this section is show that we have covered all possible cases for \(2 \times 2\) systems of linear differential equations and not to invent new methods of solution.
Subsection5.6.5Complex Eigenvalues
Suppose the matrix
\begin{equation*}
A = \begin{pmatrix} a \amp b \\ c \amp d \end{pmatrix}
\end{equation*}
in system \({\mathbf x}' = A {\mathbf x}\) has complex eigenvalues. In this case, the characteristic polynomial \(p(\lambda) = \lambda^2 - (a + d)\lambda + (ad - bc)\) will have roots \(\lambda = \alpha + i \beta\) and \(\overline{\lambda} = \alpha - i \beta\text{,}\) where
The eigenvalues \(\lambda\) and \(\overline{\lambda}\) are complex conjugates. Now, suppose that the eigenvalue \(\lambda = \alpha + i \beta\) has an eigenvector of the form
\begin{equation*}
\mathbf v = {\mathbf v}_ 1 + i {\mathbf v}_2,
\end{equation*}
where \(\mathbf v_1\) and \(\mathbf v_2\) are real vectors. Then \(\overline{\mathbf v} = {\mathbf v}_ 1 - i {\mathbf v}_2\) is an eigenvector for \(\overline{\lambda}\text{,}\) since
If \({\mathbf v}_1\) and \({\mathbf v}_2\) are not linearly independent, then \({\mathbf v}_1 = c {\mathbf v}_2\) for some \(c \in \mathbb R\text{.}\) On one hand, we have
\begin{equation*}
A ({\mathbf v}_ 1 + i {\mathbf v}_2) = A (c {\mathbf v}_2 + i {\mathbf v}_2) = (c + i) A {\bf v}_2.
\end{equation*}
However,
\begin{align*}
A ({\mathbf v}_ 1 + i {\mathbf v}_2) & = (\alpha + i \beta) ( {\mathbf v}_ 1 + i {\mathbf v}_2)\\
& = (\alpha + i \beta) ( c + i) {\mathbf v}_2\\
& = ( c + i) (\alpha + i \beta) {\mathbf v}_2
\end{align*}
In other words, \(A {\mathbf v}_2 = (\alpha + i \beta) {\mathbf v}_2\text{.}\) However, this is a contradiction since the left-side of the equation says that we have real eigenvector while the right-side of the equation is complex. Thus, \({\mathbf v}_1\) and \({\mathbf v}_2\) are linearly independent.
Proposition5.6.6
Let \(A\) be a real matrix with eigenvalue \(\lambda = \alpha + i \beta\text{,}\) where \(\beta \neq 0\text{.}\) If
\begin{equation*}
{\mathbf v}_ 1 + i {\mathbf v}_2,
\end{equation*}
is an eigenvector for \(\lambda\text{,}\) then there exists a matrix \(T\) such that
The system \({\mathbf y}' = (T^{-1} AT ) {\mathbf y}\) is in one of the canonical forms and has a phase portrait that is a spiral sink (\(\alpha \lt 0\)), a center (\(\alpha = 0\)), or a spiral source (\(\alpha \gt 0\)). After a change of coordinates, the phase portrait of \({\mathbf x}' = A {\mathbf x}\) is equivalent to a sink, center, or source.
Since \(A\) has eigenvalue \(\lambda = (-1 + i)/2\) with an eigenvector \(\mathbf v = (2, -1 + i)\text{,}\) we can apply Euler's formula and write the solution as
which agrees with the solution that we found by transforming coordinates.
Subsection5.6.6Repeated Eigenvalues
Now suppose that \(A\) has a single real eigenvalue \(\lambda\text{.}\) Then the characteristic polynomial of \(A\) is \(p(\lambda) = \lambda^2 - (a + d)\lambda + (ad - bc)\text{,}\) then \(A\) has an eigenvalue \(\lambda = (a + d)/2\text{.}\)
Proposition5.6.11
If \(A\) has a single eigenvalue and a pair of linearly independent eigenvectors, then \(A\) must be of the form
Suppose that \({\mathbf u}\) and \({\mathbf v}\) are linearly indeendent eigenvectors for \(A\text{,}\) and let \(T\) be the matrix whose first column is \({\mathbf u}\) and second column is \({\mathbf v}\text{.}\) That is, \(T {\mathbf e}_1 = {\mathbf u}\) and \(T{\mathbf e}_2 = {\mathbf v}\text{.}\) Since \({\mathbf u}\) and \({\mathbf v}\) are linearly independent, \(\det(T) \neq 0\) and \(T\) is invertible. So, it must be the case that
\begin{equation*}
AT = (A {\mathbf u}, A {\mathbf v}) = (\lambda {\mathbf u}, \lambda {\mathbf v}) = \lambda ({\mathbf u}, {\mathbf v}) = \lambda IT,
\end{equation*}
Suppose that \(A\) has a single eigenvalue \(\lambda\text{.}\) If \(\mathbf v\) is an eigenvector for \(\lambda\) and any other eigenvector for \(\lambda\) is a multiple of \(\mathbf v\text{,}\) then there exists a matrix \(T\) such that
\begin{equation*}
T^{-1} A T
=
\begin{pmatrix}
\lambda & 1 \\
0 & \lambda
\end{pmatrix}.
\end{equation*}
If \({\mathbf w}\) is another vector in \({\mathbb R}^2\) such that \({\mathbf v}\) and \({\mathbf w}\) are linearly independent, then \(A \mathbf w\) can be written as a linear combination of \(\mathbf v\) and \(\mathbf w\text{,}\)
We can assume that \(\alpha \neq 0\text{;}\) otherwise, we would have a second linearly independent eigenvector. We claim that \(\beta = \lambda\text{.}\) If this were not the case, then
and \(\beta\) would be an eigenvalue distinct from \(\lambda\text{.}\) Thus, \(A {\mathbf w} = \alpha {\mathbf v} + \lambda {\mathbf w}\text{.}\) If we will let \({\mathbf u} = (1/ \alpha) {\mathbf w}\text{,}\) then
The characteristic polynomial of \(A\) is \(\lambda^2 - 6 \lambda + 9 = (\lambda - 3)^2\text{,}\) we have only a single eigenvalue \(\lambda = 3\) with eigenvector \(\mathbf v = (1, -2)\text{.}\) Any other eigenvector for \(\lambda\) is a multiple of \(\mathbf v\text{.}\) If we choose \(\mathbf w = (1, 0)\text{,}\) then \(\mathbf v\) and \(\mathbf w\) are linearly independent. Furthermore,
\begin{align*}
\mathbf x(t) \amp = T \mathbf y(t)\\
\amp = c_1 e^{3t} T \begin{pmatrix} 1 \\ 0 \end{pmatrix} + c_2 e^{3t} T \begin{pmatrix} t \\ 1 \end{pmatrix}\\
\amp = c_1 e^{3t} \begin{pmatrix} 1 \\ -2 \end{pmatrix} + c_2 e^{3t} \begin{pmatrix} 1/2 + t \\ -2t \end{pmatrix}.
\end{align*}
This solution agrees with the solution that we found in Example5.5.5.
In practice, we find solutions to linear systems using the methods that we outlined in Sections5.25.4. What we have demonstrated in this section is that those solutions are exactly the ones that we want.
Subsection5.6.7Important Lessons
A linear map \(T\) is invertible if and only if \(\det T \neq 0\text{.}\)
A linear map \(T\) converts solutions of \({\mathbf y}' = (T^{-1} A T) {\mathbf y}\) to solutions of \({\mathbf x}' = A {\mathbf x}\text{.}\)
The inverse of a linear map \(T\) takes solutions of \({\mathbf x}' = A {\mathbf x}\) to solutions of \({\mathbf y}' = (T^{-1} A T) {\mathbf y}\text{.}\)
A change of coordinates converts the system \({\mathbf x}' = A {\mathbf x}\) to one of the following special cases,
Given a \(2 \times 2\) linear system, what are the possible types of solutions?
Subsection5.6.8Exercises
For each of the matrices \(A\) in Exercise Group5.6.8.16, find (1) the eigenvalues, \(\lambda_1\) and \(\lambda_2\text{;}\) (2) for each eigenvalue \(\lambda_1\) and \(\lambda_2\text{,}\) find an eigenvector \(\mathbf v_1\) and \(\mathbf v_2\text{,}\) respectively; and (3) construct the matrix \(T = (\mathbf v_1, \mathbf v_2)\) and calculate \(T^{-1}AT\text{.}\)
For each of the matrices \(A\) in Exercise Group5.6.8.710, find (1) an eigenvalue, \(\lambda\text{;}\) (2) find an eigenvector \(\mathbf v = \mathbf v_{\text{Re}} + i \mathbf v_{\text{Im}}\) for \(\lambda\text{;}\) and (2) construct the matrix \(T = (\mathbf v_{\text{Re}} , \mathbf v_{\text{Im}})\) and calculate \(T^{-1}AT\text{.}\) Compare your result to \(\lambda\text{.}\)
For each of the matrices \(A\) in Exercise Group5.6.8.1116, find (1) the eigenvalue, \(\lambda\) and an eigenvector \(\mathbf v\) for \(\lambda\text{;}\) (2) choose a vector \(\mathbf w\) that is linearly independent of \(\mathbf v\) and compute \((A - \lambda I)\mathbf w\text{.}\) You should find that
\begin{equation*}
(A - \lambda I)\mathbf w = \alpha \mathbf v
\end{equation*}
for some real number \(\alpha\text{.}\) (3) Let \(\mathbf u = (1/\alpha) \mathbf w\) and form the matrix \(T = (\mathbf v, \mathbf u)\text{.}\) (4) Calculate \(T^{-1}AT\text{,}\) which should be in the form