Matrix Exponentials and Fundamental Matrices, differential equations lesson

With $t$ as the independent variable and a prime meaning $d/dt$ , the scalar problem $y' = ay$ with $y(0) = y_0$ has the solution $y = e^{at}y_0$ , a single number times the initial value. For a constant matrix $A$ , the system $X' = AX$ has a solution of the same shape, $X(t) = e^{tA}X_0$ , once the exponential of a matrix is defined. This lesson defines $e^{tA}$ , collects independent solutions into a fundamental matrix that carries every initial state forward, and uses Liouville’s formula to follow its determinant. It also shows that the scalar rule $e^{a + b} = e^ae^b$ can fail for matrices that do not commute, so a coefficient matrix that changes with time has to be handled in the right order.

Key terms

Matrix exponential

Definition For a constant $n \times n$ matrix $A$ with real or complex entries and a real number $t$ , the matrix exponential is

\begin{aligned} e^{tA} &= \sum_{k=0}^{\infty} \frac{t^k}{k!}A^k \\[0.4em] &= I + tA + \frac{t^2}{2!}A^2 + \cdots, \end{aligned}

where $A^0 = I$ and every power is a matrix product. The series converges for every $A$ and every $t$ , and

\begin{gathered} \frac{d}{dt}e^{tA} = Ae^{tA} = e^{tA}A, \\[0.4em] e^{0A} = I. \end{gathered}

Hence $X(t) = e^{(t - t_0)A}X_0$ solves $X' = AX$ with $X(t_0) = X_0$ for all real $t$ , and by uniqueness it is the only solution. The rule $e^{sA}e^{tA} = e^{(s + t)A}$ holds, so $e^{tA}$ is invertible with inverse $e^{-tA}$ . If $A = VDV^{-1}$ with $D$ diagonal, then $e^{tA} = Ve^{tD}V^{-1}$ , where $e^{tD}$ is the diagonal matrix with entries $e^{\lambda_j t}$ and $\lambda_1, \dots, \lambda_n$ are the diagonal entries of $D$ .

In plain words The matrix exponential is the scalar exponential series with powers of a matrix in place of powers of a number. It is one matrix that takes the state at time $0$ to the state at time $t$ , and it gives the same solutions that eigenvectors and Jordan chains produce, collected in a single formula.

Example Let $A$ have rows $(0, 1)$ and $(-4, 0)$ . Then $A^2 = -4I$ , so $A^{2m} = (-4)^mI$ and $A^{2m+1} = (-4)^mA$ . In the series the even powers add up to $\cos 2t$ times $I$ , and the odd powers add up to $\tfrac12\sin 2t$ times $A$ , which gives

e^{tA} = \begin{pmatrix} \cos 2t & \frac{\sin 2t}{2} \\[0.3em] -2\sin 2t & \cos 2t \end{pmatrix}.

At $t = 0$ this is $I$ , and differentiating each entry gives $A$ times the same matrix. The system is the first-order form of $y'' + 4y = 0$ with state $X = (y, y')^{T}$ .

Fundamental matrix

Definition Let $A(t)$ be an $n \times n$ matrix whose entries are continuous on an open interval $I$ , which includes a constant $A$ with $I = \mathbb{R}$ . A fundamental matrix of $X' = A(t)X$ on $I$ is an $n \times n$ matrix function $\Phi$ whose columns are $n$ linearly independent solutions on $I$ . Equivalently, $\Phi' = A(t)\Phi$ on $I$ and $\det\Phi(t_0) \ne 0$ at one point $t_0$ of $I$ , and then $\det\Phi(t) \ne 0$ at every point of $I$ . Every solution on $I$ is $X(t) = \Phi(t)c$ for exactly one constant vector $c$ . If $C$ is an invertible constant matrix, then $\Phi C$ is another fundamental matrix, and every fundamental matrix arises from $\Phi$ in this way.

In plain words A fundamental matrix stores a basis of solutions side by side. At each instant its columns form a basis of the state space, so every solution is a combination of the columns. The constants multiply from the right. A constant matrix on the left, as in $C\Phi$ , generally breaks the equation, because $(C\Phi)' = CA(t)\Phi$ , and this differs from $A(t)C\Phi$ unless $C$ and $A(t)$ commute.

Example For the matrix $A$ above, the columns $(\cos 2t, -2\sin 2t)^{T}$ and $(\sin 2t, 2\cos 2t)^{T}$ are solutions, which gives the matrix function

\Psi = \begin{pmatrix} \cos 2t & \sin 2t \\ -2\sin 2t & 2\cos 2t \end{pmatrix}.

Its determinant is $2\cos^2 2t + 2\sin^2 2t = 2$ for every $t$ , so $\Psi$ is a fundamental matrix on $\mathbb{R}$ . So is $e^{tA}$ , and the two are related by $\Psi(t) = e^{tA}\operatorname{diag}(1, 2)$ .

Principal fundamental matrix

Definition The principal fundamental matrix at a point $t_0$ of $I$ is the fundamental matrix that equals $I$ at $t = t_0$ . Starting from any fundamental matrix $\Phi$ it is

U(t, t_0) = \Phi(t)\,\Phi(t_0)^{-1},

and the result does not depend on which $\Phi$ is used. The unique solution of $X' = A(t)X$ with $X(t_0) = X_0$ is $X(t) = U(t, t_0)X_0$ on all of $I$ . The matrix $U(t, t_0)$ is also called the state transition matrix or the propagator from $t_0$ . It satisfies $U(t_0, t_0) = I$ and $U(t, s)\,U(s, t_0) = U(t, t_0)$ for all $t$ , $s$ and $t_0$ in $I$ . For a constant matrix, $U(t, t_0) = e^{(t - t_0)A}$ , so $e^{tA}$ is the principal fundamental matrix at $0$ .

In plain words The factor $\Phi(t_0)^{-1}$ turns the initial state into coordinates with respect to the columns of $\Phi(t_0)$ , and $\Phi(t)$ then rebuilds a state from the same coordinates at time $t$ . The product carries any state at time $t_0$ to the state at time $t$ , whichever basis of solutions was used to build it.

Example For $\Psi$ above, $\Psi(0) = \operatorname{diag}(1, 2)$ , so $\Psi(t)\Psi(0)^{-1} = \Psi(t)\operatorname{diag}(1, \tfrac12) = e^{tA}$ , the principal fundamental matrix at $0$ . The solution with $X(0) = (1, 2)^{T}$ is $X(t) = e^{tA}(1, 2)^{T}$ , whose components are

\begin{aligned} x_1(t) &= \cos 2t + \sin 2t, \\[0.4em] x_2(t) &= 2\cos 2t - 2\sin 2t. \end{aligned}

The derivative of the first component is the second component, and the derivative of the second is $-4$ times the first, as the rows of $A$ require. The product $\Psi(t)(1, 2)^{T}$ is also a solution, but it starts at $(1, 4)^{T}$ , not at the required state.

Liouville’s formula

Definition Let $A(t)$ be continuous on an open interval $I$ , and let $\Phi$ be any $n \times n$ matrix solution of $\Phi' = A(t)\Phi$ on $I$ . Then for all $t$ and $t_0$ in $I$ ,

\begin{aligned} \det\Phi(t) &= \det\Phi(t_0)\,e^{\mu(t)}, \\[0.4em] \mu(t) &= \int_{t_0}^{t} \operatorname{tr}A(\tau)\,d\tau, \end{aligned}

where $\operatorname{tr}A$ , the trace, is the sum of the diagonal entries. Hence $\det\Phi$ is either zero at every point of $I$ or zero at no point of $I$ . For a constant matrix, $\det e^{tA} = e^{t\operatorname{tr}A}$ . For the first-order form of $y'' + p(t)y' + q(t)y = 0$ the trace is $-p(t)$ , and the formula becomes Abel’s identity for the Wronskian.

In plain words The determinant of $\Phi(t)$ is the signed area, or volume in higher dimensions, spanned by its columns. The trace of $A(t)$ is the rate at which the system expands or contracts that volume, relative to its current size. So a basis of solutions can never collapse at any point of the interval.

Example The oscillator matrix $A$ above has trace $0$ , so $\det e^{tA} = 1$ for every $t$ , in agreement with $\cos^2 2t + \sin^2 2t = 1$ . The flow preserves area, even though its trajectories other than the origin are ellipses rather than circles. For a time-dependent case, $y_1 = t$ and $y_2 = t^2$ solve $t^2y'' - 2ty' + 2y = 0$ on $(0, \infty)$ . The first-order form and the matrix of these solutions with their derivatives are

\begin{gathered} A(t) = \begin{pmatrix} 0 & 1 \\ -2/t^2 & 2/t \end{pmatrix}, \\[0.4em] \Phi(t) = \begin{pmatrix} t & t^2 \\ 1 & 2t \end{pmatrix}. \end{gathered}

Directly, $\det\Phi(t) = 2t^2 - t^2 = t^2$ . Liouville’s formula from $t_0 = 1$ gives the same result, since $\det\Phi(1) = 1$ and the trace $2/t$ integrates to $2\ln t$ , so $\det\Phi(t) = e^{2\ln t} = t^2$ . The interval is $(0, \infty)$ because $A(t)$ is not defined at $t = 0$ .

Noncommuting coefficient matrices

Definition Two $n \times n$ matrices commute when $AB = BA$ , and their commutator is $AB - BA$ . If $AB = BA$ , then $e^{A + B} = e^Ae^B = e^Be^A$ . For matrices that do not commute, these three matrices can all differ. Let $A(t)$ be continuous on $I$ and commute with itself at different times, so that $A(t)A(s) = A(s)A(t)$ for all $t$ and $s$ in $I$ . Then the principal fundamental matrix is the exponential of the integral of $A$ ,

\begin{aligned} U(t, t_0) &= e^{B(t)}, \\[0.4em] B(t) &= \int_{t_0}^{t} A(\tau)\,d\tau. \end{aligned}

Without this condition the formula does not give a solution in general. When $A(t)$ is constant on successive intervals, equal to $A_1$ on $[t_0, t_1)$ and to $A_2$ on $[t_1, t_2]$ , the state is continuous at the switch, and with the step lengths $h_1 = t_1 - t_0$ and $h_2 = t_2 - t_1$

X(t_2) = e^{h_2A_2}e^{h_1A_1}X(t_0),

with the later factor on the left.

In plain words When the matrix changes with time, the order of events matters. The second stage acts on the state that the first stage produced, so its matrix stands to the left. Exponentiating the integral of $A(t)$ treats all times as if they acted at once. That is justified when the matrices at different times commute, but not in general.

Example Let $A_1$ have rows $(0, 1)$ and $(0, 0)$ and act for $0 \le t < 1$ , and let $A_2$ have rows $(0, 0)$ and $(1, 0)$ and act for $1 \le t \le 2$ . Each squares to zero, so $e^{A_1} = I + A_1$ and $e^{A_2} = I + A_2$ , and the two-stage map is

\begin{aligned} e^{A_2}e^{A_1} &= \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \\[0.4em] &= \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}. \end{aligned}

The integral of $A(t)$ over $[0, 2]$ is $S = A_1 + A_2$ , which has rows $(0, 1)$ and $(1, 0)$ and squares to $I$ . Its exponential is

e^{S} = \begin{pmatrix} \cosh 1 & \sinh 1 \\ \sinh 1 & \cosh 1 \end{pmatrix},

with entries about $1.543$ and $1.175$ , so it is not the ordered product. The reason is the commutator $A_1A_2 - A_2A_1 = \operatorname{diag}(1, -1)$ , which is not zero. Both matrices have determinant $1$ , as Liouville’s formula predicts for coefficient matrices with trace zero, so the determinant cannot tell them apart.

Check yourself

The functions $e^t(1, 1)^{T}$ and $e^{-t}(1, -1)^{T}$ solve $X' = AX$ , where $A$ has rows $(0, 1)$ and $(1, 0)$ . Build a fundamental matrix from them, find the principal fundamental matrix at $0$ , and compare it with the series for $e^{tA}$ .

Show the answer

The fundamental matrix and its value at $0$ are

\begin{gathered} \Phi(t) = \begin{pmatrix} e^t & e^{-t} \\ e^t & -e^{-t} \end{pmatrix}, \\[0.4em] \Phi(0) = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}. \end{gathered}

Its determinant is $-2$ for every $t$ . It is constant, as Liouville’s formula predicts for a matrix with trace $0$ , and it is not zero, so $\Phi$ is a fundamental matrix on $\mathbb{R}$ . The inverse is $\Phi(0)^{-1} = \tfrac12\Phi(0)$ . The product $\Phi(t)\Phi(0)^{-1}$ has diagonal entries $\tfrac12(e^t + e^{-t}) = \cosh t$ and off-diagonal entries $\tfrac12(e^t - e^{-t}) = \sinh t$ , so

U(t, 0) = \begin{pmatrix} \cosh t & \sinh t \\ \sinh t & \cosh t \end{pmatrix}.

In the series, $A^2 = I$ , so the even powers add up to $\cosh t$ times $I$ and the odd powers to $\sinh t$ times $A$ . That is the same matrix, as it must be, since both are the principal fundamental matrix at $0$ .

A matrix function $\Phi$ solves $\Phi' = A(t)\Phi$ on $(0, \infty)$ with

A(t) = \begin{pmatrix} 1/t & t^2 \\ \sin t & 1 \end{pmatrix},

and $\det\Phi(1) = 3$ . Find $\det\Phi(t)$ for $t > 0$ . The result tends to $0$ as $t \to 0^+$ . Does that contradict the statement that the determinant is either never zero or always zero?

Show the answer

The trace is $1/t + 1$ , and its integral from $1$ to $t$ is $\ln t + t - 1$ . Liouville’s formula gives

\det\Phi(t) = 3te^{t - 1}

for every $t > 0$ . This is positive on all of $(0, \infty)$ , so the columns stay independent there. There is no contradiction. Liouville’s formula and the dichotomy are statements about an interval on which $A(t)$ is continuous. The point $t = 0$ is not in $(0, \infty)$ , and $A(t)$ is not even defined there.

Write $X = (x_1, x_2)^{T}$ and let $A(t)$ have rows $(0, t)$ and $(0, 1)$ . Solve $X' = A(t)X$ with $X(0) = (0, 1)^{T}$ and find $X(1)$ . Then compute the shortcut $\exp\left(\int_0^1 A(\tau)\,d\tau\right)X(0)$ and explain the difference.

Show the answer

The second equation $x_2' = x_2$ with $x_2(0) = 1$ gives $x_2 = e^t$ . The first is then $x_1' = te^t$ with $x_1(0) = 0$ , so $x_1 = (t - 1)e^t + 1$ . Hence $X(1) = (1, e)^{T}$ . The integral of $A$ is the matrix $B$ with rows $(0, \tfrac12)$ and $(0, 1)$ . It satisfies $B^2 = B$ , so every power $B^k$ with $k \ge 1$ equals $B$ and

\begin{aligned} e^B &= I + (e - 1)B \\[0.4em] &= \begin{pmatrix} 1 & \tfrac12(e - 1) \\ 0 & e \end{pmatrix}. \end{aligned}

The shortcut gives $e^B(0, 1)^{T} = (\tfrac12(e - 1), e)^{T}$ , and its first component, about $0.859$ , is wrong. The coefficient matrices do not commute at different times, since

\begin{aligned} &A(t)A(s) - A(s)A(t) \\[0.4em] &\quad = \begin{pmatrix} 0 & t - s \\ 0 & 0 \end{pmatrix}, \end{aligned}

which is not zero for $t \ne s$ . So the exponential of the integral is not a solution here.

Matrix Exponentials and Fundamental Matrices

Key terms

Matrix exponential

Fundamental matrix

Principal fundamental matrix

Liouville’s formula

Noncommuting coefficient matrices

Check yourself

How this lesson connects

Builds on

Used later in