Formulas for Matrix Exponentials

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We now complete our discussion of exact solutions to planar linear systems of ODEs $\dot {X}=CX$ . There are three different methods for finding closed form solutions to systems of ODEs. We have discussed two of these methods. In the first we find solutions by the direct method, that is, we find two linear independent solutions whose linear combinations form the space of solutions. See Section ??. In the second method we use similarity and normal form equations (whose solutions are obtained using matrix exponentials) to find closed form solutions. See Sections ?? and ??. In this section we present a third method based on computable formulas for matrix exponentials derived using the Cayley Hamilton theorem.

A Formula for the Matrix Exponential

For $2\times 2$ matrices $C$ with eigenvalues $\lambda _1$ and $\lambda _2$ there is a simple formula for the matrix exponential $e^{tC}$ whose derivation depends on the Cayley Hamilton theorem. When the eigenvalues $\lambda _1$ and $\lambda _2$ of $C$ are distinct, the formula is

$\begin{equation} \label{E:exdist} e^{tC} = \frac{1}{\lambda_2-\lambda_1}\left(e^{\lambda_1 t}(C-\lambda_2I_2) - e^{\lambda_2 t}(C-\lambda_1I_2)\right). \end{equation}$ When the eigenvalues are equal, the formula is $\begin{equation} \label{E:exeq} e^{tC} = e^{\lambda_1 t}(I_2 + tN) \end{equation}$ where $N = C - \lambda _1I_2$ .

Note that when computing the matrix exponential using either (??) or (??), it is not necessary to compute the eigenvectors of $C$ . This is a substantial simplification. But it is with the use of formula (??) that the greatest simplification occurs.

The Example with Equal Eigenvalues Revisited

Let us reconsider the system of differential equations (??) $\frac {dX}{dt} = \mattwo {1}{-1}{9}{-5} X = CX$ with initial value $X_0 = \vectwo {2}{3}.$ The eigenvalues of $C$ are real and equal to $\lambda _1=-2$ .

We may write $C = \lambda _1 I_2 + N = -2I_2+N,$ where $N = \mattwo {3}{-1}{9}{-3}.$ It follows from (??) that

$\begin{equation} \label{e:solntob} e^{tC} = e^{-2t}\left(I_2+t\mattwo{3}{-1}{9}{-3}\right) = e^{-2t}\mattwoc{1+3t}{-t}{9t}{1-3t}. \end{equation}$ Hence the solution to the initial value problem is: $\begin{align*} X(t) &= e^{tC}X_0 = e^{-2t}\mattwoc{1+3t}{-t}{9t}{1-3t}\vectwo{2}{3} \\ &= e^{-2t}\vectwo{2+3t}{3+9t}. \end{align*}$

The Cayley Hamilton Theorem

The Cayley Hamilton theorem states that a matrix satisfies its own characteristic polynomial. More precisely:

Cayley Hamilton Theorem Let $A$ be a $2\times 2$ matrix and let $p_A(\lambda ) = \lambda ^2 + a\lambda + b$ be the characteristic polynomial of $A$ . Then $p_A(A) = A^2 + aA + bI_2 = 0.$

Proof

Suppose $B=P\inv AP$ and $A$ are similar matrices. We claim that if $p_A(A)=0$ , then $p_B(B)=0$ . To verify this claim, recall from Lemma ?? that $p_A=p_B$ and calculate $\begin{align*} p_B(B) &= p_A(P\inv AP) = (P\inv AP)^2 + aP\inv AP + bI_2 \\ &= P\inv p_A(A)P= 0. \end{align*}$

Theorem ?? classifies $2\times 2$ matrices up to similarity. Thus, we need only verify this theorem for the matrices

$\begin{align*} C &= \mattwoc{\lambda_1}{0}{0}{\lambda_2}, D &= \mattwo{\sigma}{-\tau}{\tau}{\sigma}, E &= \mattwoc{\lambda_1}{1}{0}{\lambda_1}, \end{align*}$

that is, we need to verify that $p_C(C) = 0 \qquad p_D(D)=0 \qquad p_E(E)=0.$

Using the fact that $p_A(\lambda )=\lambda ^2-\trace (A)\lambda +\det (A)$ , we see that

$\begin{eqnarray*} p_C(\lambda) & = & (\lambda-\lambda_1)(\lambda-\lambda_2) \\ p_D(\lambda) & = & \lambda^2 - 2\sigma \lambda + (\sigma^2+\tau^2) \\ p_E(\lambda) & = & (\lambda-\lambda_1)^2. \end{eqnarray*}$

It now follows that $\begin{align*} p_C(C) &= (C-\lambda_1I_2)(C-\lambda_2I_2) \\ &=\mattwoc{0}{0}{0}{\lambda_2-\lambda_1}\mattwoc{\lambda_1-\lambda_2}{0}{0}{0} \\ &=0, \end{align*}$

and

$\begin{align*} p_D(D) &= \mattwoc{\sigma^2-\tau^2}{-2\sigma\tau}{2\sigma\tau}{\sigma^2-\tau^2} -\\ &\quad 2\sigma \mattwo{\sigma}{-\tau}{\tau}{\sigma} + \\ &\quad (\sigma^2+\tau^2)\mattwo{1}{0}{0}{1} = 0, \end{align*}$

and $p_E(E) = (E-\lambda _1I_2)^2 = \mattwo {0}{1}{0}{0}^2 = 0.$ $\blacksquare$

Verification of (??)

Let $C$ be a $2\times 2$ matrix with eigenvalues $\lambda _1\neq \lambda _2$ . Then the characteristic polynomial of $C$ is: $p_C(\lambda ) = (\lambda -\lambda _1)(\lambda -\lambda _2).$

We begin our verification of (??) by showing that

$\begin{equation} \label{E:Inpart} I_2 = a_1 (A - \lambda_2I_2) + a_2 (A - \lambda_1I_2), \end{equation}$ where $\begin{equation} \label{E:pfcoeff} a_1 = \frac{1}{\lambda_2-\lambda_1} \AND a_2 = \frac{1}{\lambda_1-\lambda_2}. \end{equation}$ Using partial fractions we can write $\begin{equation} \label{E:partfrac} \frac{1}{p_C(\lambda)} = \frac{a_1}{\lambda-\lambda_1} + \frac{a_2}{\lambda-\lambda_2}, \end{equation}$ where $a_j$ is as in (??). Multiplying (??) by $p_C(\lambda )$ yields $1 = a_1 (\lambda -\lambda _2) + a_2 (\lambda -\lambda _1).$ Now let $v_j$ be the eigenvector of $C$ corresponding to the eigenvalue $\lambda _j$ and compute $(a_1 (C - \lambda _2I_2) + a_2 (C - \lambda _1I_2))v_1 = a_1(C - \lambda _2I_2)v_1 = a_1(\lambda _1-\lambda _2)v_1 = v_1.$ Similarly, $(a_1 (C - \lambda _2I_2) + a_2 (C - \lambda _1I_2))v_2 = v_2.$ Since $v_1$ and $v_2$ form a basis, (??) holds by linearity.

We use the Cayley Hamilton theorem to show that

$\begin{equation} \label{E:almost} \begin{array}{rcl} (C - \lambda_2I_2)e^{tC} & = & e^{\lambda_1 t}(C - \lambda_2I_2)\\ (C - \lambda_1I_2)e^{tC} & = & e^{\lambda_2 t}(C - \lambda_1I_2) \end{array} \end{equation}$ Assuming that (??) is valid, we see that $e^{tC} = (a_1 (C - \lambda _2I_2) + a_2 (C - \lambda _1I_2))e^{tC}$ by (??) and that $e^{tC} = a_1 e^{\lambda _1 t}(C - \lambda _2I_2) + a_2 e^{\lambda _2 t}(C - \lambda _1I_2)$ by (??). Thus we have verified formula (??). To validate (??), calculate $e^{tC} = e^{tC - \lambda _1 tI_2 + \lambda _1 tI_2} = e^{\lambda _1t}e^{t(C - \lambda _1 I_2)}$ using Proposition ?? and the fact that $I_2$ commutes with every $2\times 2$ matrix. Second, compute

$\begin{eqnarray*} (C - \lambda_2I_2)e^{tC} & = & e^{\lambda_1 t}(C - \lambda_2I_2)e^{t(C - \lambda_1 I_2)} \\ & = & e^{\lambda_1 t}(C - \lambda_2I_2)(I_2 + t(C - \lambda_1 I_2) + \cdots)\\ & = & e^{\lambda_1 t}(C - \lambda_2I_2), \end{eqnarray*}$

since every other term in the infinite series contains a factor of $p_C(C) = (C - \lambda _2I_2)(C - \lambda _1 I_2)$ which vanishes by the Cayley Hamilton theorem. The second equation in (??) is proved similarly by interchanging the roles of $\lambda _1$ and $\lambda _2$ .

Verification of (??)

The verification of (??) is less complicated. Since $C$ is assumed to have a double eigenvalue $\lambda _1$ , it follows that $N = C - \lambda _1 I_2$ has zero as a double eigenvalue. Hence, the characteristic polynomial $p_N(\lambda ) = \lambda ^2$ and the Cayley Hamilton theorem implies that $N^2=0$ . Therefore, $e^{tC} = e^{t(C-\lambda _1 I_2)+\lambda _1 tI_2} = e^{\lambda _1 t}e^{tN} = e^{\lambda _1 t}(I_2+tN),$ as desired.

Exercises

Solve the initial value problem $\frac {dX}{dt} = \mattwo {0}{1}{-2}{3} X$ where $X(0)=(2,1)^t$ .

Find all solutions to the linear system of ODEs $\frac {dX}{dt} = \mattwo {-2}{4}{-1}{1} X.$

Solve the initial value problem $\frac {dX}{dt} = \mattwo {2}{1}{-2}{0}X$ where $X(0)=(1,1)^t$ .

Let $A$ be a $2\times 2$ matrix. Show that $A^2 = \trace (A)A-\det (A)I_2.$