---
title: Repeated Eigenvalues, Fundamental Matrices, and Nonhomogeneous Systems
module: Systems of First-Order Linear Equations
moduleNumber: 6
lessonNumber: 3
order: 603
summary: >
  When a repeated eigenvalue supplies too few eigenvectors, a generalized
  eigenvector supplies the missing solution as ξte^{ρt} + ηe^{ρt}, giving an
  improper node. A fundamental set packaged as a matrix Φ(t) yields the matrix
  exponential e^{At}, the propagator mapping initial states to later ones.
  Variation of parameters solves the nonhomogeneous system x' = Ax + g(t).
topics: [Systems of First-Order Linear Equations]
sources:
  - book: Boyce
    ref: "Ch. 7 §7.7 Fundamental Matrices; §7.8 Repeated Eigenvalues"
  - book: Boyce
    ref: "§7.9 Nonhomogeneous Linear Systems"
  - book: Simmons
    ref: "Ch. 10 §56 Homogeneous Linear Systems with Constant Coefficients"
draft: false
---

Two gaps remain in the constant-coefficient theory. A repeated eigenvalue can
fail to supply enough eigenvectors, leaving the
[eigenpair method](/differential-equations/systems/constant-coefficient-systems-phase-portraits)
short of solutions; and a forcing term $\mathbf{g}(t)$ makes the system
nonhomogeneous. The connecting idea for both is the
**fundamental matrix**: package a fundamental set as the columns of one matrix,
and the whole solution theory compresses into matrix algebra. In the
constant-coefficient case that matrix turns out to be a matrix exponential.

## Fundamental matrices

Arrange a fundamental set of solutions side by side as the columns of a matrix.

> **Definition (Fundamental matrix).** If
> $\mathbf{x}^{(1)}(t), \dots, \mathbf{x}^{(n)}(t)$ form a fundamental set of
> solutions of $\mathbf{x}' = P(t)\mathbf{x}$, the matrix
> $$
> \Psi(t) = \big[\ \mathbf{x}^{(1)}(t)\ \ \mathbf{x}^{(2)}(t)\ \cdots\ \mathbf{x}^{(n)}(t)\ \big]
> $$
> whose columns are those solutions is a **fundamental matrix** for the system.
> It is nonsingular (its columns are independent) and satisfies the matrix
> equation $\Psi' = P(t)\Psi$.

The general solution collapses to one matrix-vector product. Writing
$\mathbf{x} = c_1\mathbf{x}^{(1)} + \cdots + c_n\mathbf{x}^{(n)}$ as
$\Psi(t)$ acting on the constant vector $\mathbf{c} = (c_1, \dots, c_n)^\top$ gives

$$
\mathbf{x} = \Psi(t)\,\mathbf{c}.
$$

For an initial value problem $\mathbf{x}(t_0) = \mathbf{x}^0$, the constant
vector must satisfy $\Psi(t_0)\mathbf{c} = \mathbf{x}^0$, so
$\mathbf{c} = \Psi^{-1}(t_0)\mathbf{x}^0$ and

$$
\mathbf{x}(t) = \Psi(t)\,\Psi^{-1}(t_0)\,\mathbf{x}^0.
$$

$$
% caption: A fundamental matrix (written M here, the Psi of the text) stacks
% independent solution vectors as columns; acting on a constant vector c it
% gives the general solution, and on an initial state it propagates it forward.
\begin{tikzpicture}[>=stealth, font=\footnotesize,
  col/.style={draw, minimum width=8mm, minimum height=18mm}]
  \definecolor{acc}{HTML}{4A6FA5}
  \node[font=\small] at (-1.3,0) {$M(t) =$};
  \node[col] (c1) at (0,0) {};
  \node[col] (c2) at (0.85,0) {};
  \node[font=\small] at (1.6,0) {...};
  \node[col] (cn) at (2.35,0) {};
  \node[font=\scriptsize, rotate=90] at (0,0) {$\mathbf{x}^{(1)}$};
  \node[font=\scriptsize, rotate=90] at (0.85,0) {$\mathbf{x}^{(2)}$};
  \node[font=\scriptsize, rotate=90] at (2.35,0) {$\mathbf{x}^{(n)}$};
  \draw[->, thick] (3.3,0) -- (4.5,0);
  \node[align=center, font=\scriptsize] at (6.0,0.05) {general solution\\$\mathbf{x} = M(t)\,\mathbf{c}$};
  \draw[->, thick] (7.6,0) -- (8.8,0);
  \node[align=center, font=\scriptsize] at (10.4,0.05) {IVP: solve\\$M(t_0)\,\mathbf{c} = \mathbf{x}^0$};
\end{tikzpicture}
$$

Among all fundamental matrices, one is distinguished by a convenient initial
value. Reserve the symbol $\Phi(t)$ for the fundamental matrix with

$$
\Phi(t_0) = I.
$$

Its columns are the solutions with initial vectors the standard basis vectors
$\mathbf{e}^{(1)}, \dots, \mathbf{e}^{(n)}$. With $\Phi$ the IVP solution is as
simple as it can be, since $\Phi^{-1}(t_0) = I$:

$$
\mathbf{x}(t) = \Phi(t)\,\mathbf{x}^0.
$$

$\Phi(t)$ is a transformation that carries the initial state $\mathbf{x}^0$ into
the state $\mathbf{x}(t)$ at any later time — a **propagator**. Once $\Phi$ is
known, every initial condition is solved by a single matrix multiplication.

## The matrix exponential

The scalar problem $x' = ax$, $x(0) = x_0$, has solution $x = x_0 e^{at}$. The
system $\mathbf{x}' = A\mathbf{x}$, $\mathbf{x}(0) = \mathbf{x}^0$, has solution
$\mathbf{x} = \Phi(t)\mathbf{x}^0$. The parallel is exact once $\Phi(t)$ is
identified as an exponential of the matrix $A$.

Define $\exp(At)$ by the same power series that defines the scalar exponential,
with $A$ in place of $a$ and $I$ in place of $1$:

$$
\exp(At) = I + At + \frac{A^2 t^2}{2!} + \cdots + \frac{A^n t^n}{n!} + \cdots
= I + \sum_{n=1}^{\infty} \frac{A^n t^n}{n!}.
$$

Every entry of this matrix series converges for all $t$. Differentiating term by
term factors out one $A$:

$$
\frac{\d}{\d t}\exp(At) = A\exp(At), \qquad \exp(A \cdot 0) = I.
$$

So $\exp(At)$ solves the same matrix initial value problem $\Phi' = A\Phi$,
$\Phi(0) = I$ that defines the special fundamental matrix. By uniqueness they are
the same object:

> **Theorem (Matrix exponential as propagator).** For a constant matrix $A$, the
> fundamental matrix with $\Phi(0) = I$ is
> $$
> \Phi(t) = \exp(At),
> $$
> and the solution of $\mathbf{x}' = A\mathbf{x}$, $\mathbf{x}(0) = \mathbf{x}^0$
> is $\mathbf{x}(t) = \exp(At)\,\mathbf{x}^0$.

The exponential shares the algebra of the scalar exponential, including the
propagator law $\exp(At)\exp(As) = \exp(A(t+s))$ and the inverse
$\exp(At)^{-1} = \exp(-At)$. When $A$ has a full set of eigenvectors the series
is not the practical route: diagonalizing $A = TDT^{-1}$ gives
$\exp(At) = T\exp(Dt)T^{-1}$ with $\exp(Dt) = \operatorname{diag}(e^{r_1 t},
\dots, e^{r_n t})$, so the eigenpairs already compute it. The series is needed
only when eigenvectors run short.

> **Worked example.** The matrix $A = \begin{pmatrix} -2 & 1 \\ 1 & -2 \end{pmatrix}$
> is diagonalizable, with eigenpairs $r_1 = -1$, $\boldsymbol{\xi}^{(1)} = (1, 1)^\top$
> and $r_2 = -3$, $\boldsymbol{\xi}^{(2)} = (1, -1)^\top$. Collecting the
> eigenvectors as $T = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$, with
> $T^{-1} = \tfrac12\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ and
> $D = \operatorname{diag}(-1, -3)$,
>
> $$
> \exp(At) = T\exp(Dt)T^{-1}
> = T\begin{pmatrix} e^{-t} & 0 \\ 0 & e^{-3t} \end{pmatrix}T^{-1}
> = \frac{1}{2}\begin{pmatrix} e^{-t} + e^{-3t} & e^{-t} - e^{-3t} \\
>   e^{-t} - e^{-3t} & e^{-t} + e^{-3t} \end{pmatrix}.
> $$
>
> At $t = 0$ every entry collapses to $\exp(A \cdot 0) = I$, as it must.

## Repeated eigenvalues

An eigenvalue $\rho$ of algebraic multiplicity $m$ may have geometric
multiplicity $q < m$: fewer than $m$ independent eigenvectors, and so fewer than
$m$ solutions of the pure form $\boldsymbol{\xi}e^{\rho t}$. The missing
solutions must be found in a new form, and the pattern copies the
[repeated-root case](/differential-equations/second-order-linear/complex-and-repeated-roots)
for a single second-order equation, where a double root $r$ gave the second
solution $te^{rt}$.

> **Worked example.** For $A = \begin{pmatrix} 1 & -1 \\ 1 & 3 \end{pmatrix}$ the
> characteristic equation is $(1-r)(3-r) + 1 = r^2 - 4r + 4 = (r-2)^2 = 0$,[^boyce-78]
> so $r = 2$ is a double eigenvalue. Solving $(A - 2I)\boldsymbol{\xi} = \mathbf{0}$
> gives the single condition $\xi_1 + \xi_2 = 0$, so there is only one independent
> eigenvector, $\boldsymbol{\xi} = (1, -1)^\top$. One solution is
>
> $$
> \mathbf{x}^{(1)}(t) = \begin{pmatrix} 1 \\ -1 \end{pmatrix} e^{2t}.
> $$
>
> A naive second guess $\mathbf{x} = \boldsymbol{\xi}te^{2t}$ fails: substituting
> forces $\boldsymbol{\xi} = \mathbf{0}$. Allowing an extra constant vector
> alongside the $te^{2t}$ term,
>
> $$
> \mathbf{x} = \boldsymbol{\xi}\,t e^{2t} + \boldsymbol{\eta}\,e^{2t},
> $$
>
> substituting into $\mathbf{x}' = A\mathbf{x}$ and matching the coefficients of
> $te^{2t}$ and $e^{2t}$ separately gives two conditions:
>
> $$
> (A - 2I)\boldsymbol{\xi} = \mathbf{0}, \qquad
> (A - 2I)\boldsymbol{\eta} = \boldsymbol{\xi}.
> $$
>
> The first says $\boldsymbol{\xi}$ is the eigenvector. The second determines
> $\boldsymbol{\eta}$: although $A - 2I$ is singular, the right-hand side
> $\boldsymbol{\xi}$ lies in its column space, so the equation is solvable. Here
> $-\eta_1 - \eta_2 = 1$, and a convenient choice is
> $\boldsymbol{\eta} = (0, -1)^\top$, giving the second solution
>
> $$
> \mathbf{x}^{(2)}(t) = \begin{pmatrix} 1 \\ -1 \end{pmatrix} t e^{2t}
> + \begin{pmatrix} 0 \\ -1 \end{pmatrix} e^{2t}.
> $$
>
> Their Wronskian is $-e^{4t} \neq 0$, so $\{\mathbf{x}^{(1)}, \mathbf{x}^{(2)}\}$
> is a fundamental set and the general solution is
> $\mathbf{x} = c_1\mathbf{x}^{(1)} + c_2\mathbf{x}^{(2)}$.

> **Definition (Generalized eigenvector).** For a double eigenvalue $\rho$ with a
> single eigenvector $\boldsymbol{\xi}$ (so $(A - \rho I)\boldsymbol{\xi} =
> \mathbf{0}$), a vector $\boldsymbol{\eta}$ solving
> $(A - \rho I)\boldsymbol{\eta} = \boldsymbol{\xi}$ is a **generalized
> eigenvector**. It satisfies $(A - \rho I)^2\boldsymbol{\eta} = \mathbf{0}$, and
> the second solution is
> $\mathbf{x}^{(2)}(t) = \boldsymbol{\xi}te^{\rho t} + \boldsymbol{\eta}e^{\rho t}$.

$$
% caption: A defective double eigenvalue r chains the eigenvector v to a
% generalized eigenvector w (the xi and eta of the text); the second solution
% carries the extra t e^{rt} term.
\begin{tikzpicture}[>=stealth, font=\footnotesize,
  b/.style={draw, minimum width=34mm, minimum height=11mm, align=center, font=\scriptsize}]
  \definecolor{acc}{HTML}{4A6FA5}
  \node[b] (e1) at (0,0) {$A\mathbf{v} = r\mathbf{v}$};
  \node[b] (e2) at (5.2,0) {$A\mathbf{w} = r\mathbf{w} + \mathbf{v}$};
  \node[b] (s1) at (0,-1.9) {$\mathbf{x}^{(1)} = \mathbf{v}\,e^{rt}$};
  \node[b, fill=acc!10] (s2) at (5.2,-1.9) {$\mathbf{x}^{(2)} = \mathbf{v}\,te^{rt} + \mathbf{w}\,e^{rt}$};
  \draw[->, thick] (e1) -- (e2) node[midway, above, font=\scriptsize, text=black]{feeds};
  \draw[->, thick] (e1) -- (s1);
  \draw[->, thick] (e2) -- (s2);
  \node[font=\scriptsize, text=black, anchor=west] at (-1.7,0.95) {eigenvector};
  \node[font=\scriptsize, text=black, anchor=west] at (3.5,0.95) {generalized eigenvector};
\end{tikzpicture}
$$

The phase portrait for a defective double eigenvalue is an **improper node**.
Every term carries the factor $e^{\rho t}$, and every trajectory approaches the
origin tangent to the single eigendirection (for $\rho < 0$) or departs along it
(for $\rho > 0$), but no trajectory settles onto a straight line — the $te^{\rho
t}$ term keeps bending it.

$$
% caption: Improper node from a defective double eigenvalue: every trajectory
% is tangent to the one eigendirection at the origin, bent by the t e^{rho t}
% term.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, ->] (-3,0) -- (3,0) node[right, black]{$x_1$};
  \draw[black, ->] (0,-2.6) -- (0,2.6) node[above, black]{$x_2$};
  % single eigenline x2 = -x1
  \draw[acc, thick] (-2.4,2.4) -- (2.4,-2.4);
  % inflow arrows along eigenline
  \draw[acc, thick, ->] (1.7,-1.7) -- (1.3,-1.3);
  \draw[acc, thick, ->] (-1.7,1.7) -- (-1.3,1.3);
  % curved trajectories tangent to eigenline at origin
  \draw[black, ->] (2.6,1.4) .. controls (1.4,0.2) and (0.9,-0.4) .. (0.25,-0.28);
  \draw[black, ->] (-2.6,-1.4) .. controls (-1.4,-0.2) and (-0.9,0.4) .. (-0.25,0.28);
  \draw[black, ->] (0.6,2.5) .. controls (0.9,0.9) and (0.4,0.1) .. (-0.1,0.12);
  \draw[black, ->] (-0.6,-2.5) .. controls (-0.9,-0.9) and (-0.4,-0.1) .. (0.1,-0.12);
  \fill (0,0) circle (1.6pt);
  \node[acc, anchor=west, font=\scriptsize] at (1.3,-2.4) {$x_1 + x_2 = 0$};
  \node[anchor=north, font=\scriptsize, text=black] at (0,-2.75) {improper node};
\end{tikzpicture}
$$

This defective example also shows the matrix exponential explicitly. Assembling
$\mathbf{x}^{(1)}, \mathbf{x}^{(2)}$ as columns gives a fundamental matrix, and
normalizing to $\Phi(0) = I$ by $\Phi(t) = \Psi(t)\Psi^{-1}(0)$ yields the
propagator

$$
\exp(At) = \Phi(t) = e^{2t}\begin{pmatrix} 1 - t & -t \\ t & 1 + t \end{pmatrix},
$$

the polynomial-in-$t$ entries being exactly what the eigenvector shortage
produces — the series $I + At + \tfrac12 A^2 t^2 + \cdots$ does not collapse to a
single exponential per direction, as it would for a diagonalizable matrix.

For higher multiplicities the chain lengthens: a triple defective eigenvalue
adds a $t^2 e^{\rho t}/2$ term with a second generalized eigenvector solving
$(A - \rho I)\boldsymbol{\zeta} = \boldsymbol{\eta}$. Packaged as a similarity
transformation, the eigenvector and its generalized companions turn $A$ into its
**Jordan form** $J = T^{-1}AT$: the eigenvalue on the diagonal, a $1$ just above
it in the column of each missing eigenvector, and zeros elsewhere. The Jordan
form is the reason the $te^{\rho t}$ pattern appears — $\exp(Jt)$ generates
exactly those powers of $t$.

## Nonhomogeneous systems

Restore a forcing term and consider

$$
\mathbf{x}' = P(t)\,\mathbf{x} + \mathbf{g}(t).
$$

The structure copies the single-equation case exactly: the general solution is
the general solution of the homogeneous system plus any one particular solution.

> **Theorem (Structure of the general solution).** If
> $\mathbf{x}^{(c)} = c_1\mathbf{x}^{(1)} + \cdots + c_n\mathbf{x}^{(n)}$ is the
> general solution of $\mathbf{x}' = P(t)\mathbf{x}$ and $\mathbf{v}(t)$ is any
> particular solution of $\mathbf{x}' = P(t)\mathbf{x} + \mathbf{g}(t)$, then
> every solution of the nonhomogeneous system is
> $\mathbf{x} = \mathbf{x}^{(c)} + \mathbf{v}$.

Three methods find the particular solution $\mathbf{v}$, in increasing order of
generality.

- **Diagonalization.** When $A$ is constant and diagonalizable, $\mathbf{x} =
  T\mathbf{y}$ turns the system into $n$ uncoupled scalar equations $y_j' = r_j
  y_j + h_j(t)$ with $\mathbf{h} = T^{-1}\mathbf{g}$, each solved by the
  first-order [integrating factor](/differential-equations/first-order/linear-first-order-integrating-factors).
- **Undetermined coefficients.** When $A$ is constant and $\mathbf{g}$ is
  polynomial, exponential, or sinusoidal, guess a matching form. The one twist
  versus the scalar case: if $\mathbf{g}$ contains $\mathbf{u}e^{\lambda t}$ with
  $\lambda$ an eigenvalue, the trial must include both $\mathbf{a}te^{\lambda t}$
  and $\mathbf{b}e^{\lambda t}$, not $\mathbf{a}te^{\lambda t}$ alone.
- **Variation of parameters.** The general method, needing only a fundamental
  matrix — no restriction on $P(t)$ or $\mathbf{g}(t)$.

> **Worked example.** Solve
> $\mathbf{x}' = \begin{pmatrix} -2 & 1 \\ 1 & -2 \end{pmatrix}\mathbf{x} + \begin{pmatrix} 2e^{-t} \\ 3t \end{pmatrix}$
> by undetermined coefficients.[^boyce-79] The coefficient matrix has eigenvalues
> $-1$ and $-3$; since $\mathbf{g}$ mixes an exponential $e^{-t}$ (whose exponent
> $-1$ is an eigenvalue) with a linear term, the trial solution is
>
> $$
> \mathbf{v} = \mathbf{a}\,te^{-t} + \mathbf{b}\,e^{-t} + \mathbf{c}\,t + \mathbf{d}.
> $$
>
> Substituting and matching the coefficients of $te^{-t}$, $e^{-t}$, $t$, and the
> constant term in turn gives
>
> $$
> (A + I)\mathbf{a} = \mathbf{0}, \quad
> (A + I)\mathbf{b} = \mathbf{a} - \begin{pmatrix} 2 \\ 0 \end{pmatrix}, \quad
> A\mathbf{c} = \begin{pmatrix} 0 \\ -3 \end{pmatrix}, \quad
> A\mathbf{d} = \mathbf{c}.
> $$
>
> The first makes $\mathbf{a} = (1, 1)^\top$ an eigenvector for $r = -1$; the
> second is then solvable and gives $\mathbf{b} = (0, -1)^\top$; the last two give
> $\mathbf{c} = (1, 2)^\top$ and $\mathbf{d} = -\tfrac13(4, 5)^\top$. The particular
> solution is
>
> $$
> \mathbf{v}(t) = \begin{pmatrix} 1 \\ 1 \end{pmatrix}te^{-t}
> + \begin{pmatrix} 0 \\ -1 \end{pmatrix}e^{-t}
> + \begin{pmatrix} 1 \\ 2 \end{pmatrix}t
> - \frac{1}{3}\begin{pmatrix} 4 \\ 5 \end{pmatrix}.
> $$

### Variation of parameters

Start from the homogeneous general solution $\Psi(t)\mathbf{c}$ and promote the
constant vector to an unknown function, $\mathbf{x} = \Psi(t)\mathbf{u}(t)$.
Differentiating and using $\Psi' = P\Psi$,

$$
\Psi'\mathbf{u} + \Psi\mathbf{u}' = P\Psi\mathbf{u} + \mathbf{g}
\;\Longrightarrow\;
\Psi(t)\,\mathbf{u}'(t) = \mathbf{g}(t).
$$

The homogeneous part cancels, leaving a purely algebraic condition on
$\mathbf{u}'$. Since $\Psi$ is invertible, $\mathbf{u}'(t) = \Psi^{-1}(t)
\mathbf{g}(t)$, and integrating gives the general solution.

> **Theorem (Variation of parameters).** With $\Psi(t)$ a fundamental matrix for
> $\mathbf{x}' = P(t)\mathbf{x}$, the general solution of
> $\mathbf{x}' = P(t)\mathbf{x} + \mathbf{g}(t)$ is
> $$
> \mathbf{x}(t) = \Psi(t)\,\mathbf{c}
> + \Psi(t)\int_{t_1}^{t} \Psi^{-1}(s)\,\mathbf{g}(s)\,\d s.
> $$
> The first term is the homogeneous solution; the second is a particular
> solution. For the initial value problem $\mathbf{x}(t_0) = \mathbf{x}^0$, take
> $t_1 = t_0$ and $\mathbf{c} = \Psi^{-1}(t_0)\mathbf{x}^0$.

> **Worked example.** Solve the same system
> $\mathbf{x}' = \begin{pmatrix} -2 & 1 \\ 1 & -2 \end{pmatrix}\mathbf{x} + \begin{pmatrix} 2e^{-t} \\ 3t \end{pmatrix}$
> by variation of parameters.[^boyce-79] The eigenvectors $(1, -1)^\top$ and
> $(1, 1)^\top$ (for $-3$ and $-1$) give a fundamental matrix and its inverse
>
> $$
> \Psi(t) = \begin{pmatrix} e^{-3t} & e^{-t} \\ -e^{-3t} & e^{-t} \end{pmatrix},
> \qquad
> \Psi^{-1}(t) = \frac{1}{2}\begin{pmatrix} e^{3t} & -e^{3t} \\ e^{t} & e^{t} \end{pmatrix}.
> $$
>
> Then $\mathbf{u}'(t) = \Psi^{-1}(t)\mathbf{g}(t)$ has components
>
> $$
> u_1' = e^{2t} - \tfrac32 t e^{3t}, \qquad u_2' = 1 + \tfrac32 t e^{t}.
> $$
>
> Integrating (the additive constants become the homogeneous part),
>
> $$
> u_1 = \tfrac12 e^{2t} - \tfrac12 t e^{3t} + \tfrac16 e^{3t}, \qquad
> u_2 = t + \tfrac32 t e^{t} - \tfrac32 e^{t}.
> $$
>
> Assembling $\mathbf{x} = \Psi\mathbf{u}$ and absorbing a homogeneous $e^{-t}$
> multiple into $c_2$ gives the general solution
>
> $$
> \mathbf{x} = c_1\begin{pmatrix} 1 \\ -1 \end{pmatrix}e^{-3t}
> + c_2\begin{pmatrix} 1 \\ 1 \end{pmatrix}e^{-t}
> + \begin{pmatrix} 1 \\ 1 \end{pmatrix}t e^{-t}
> + \begin{pmatrix} 0 \\ -1 \end{pmatrix}e^{-t}
> + \begin{pmatrix} 1 \\ 2 \end{pmatrix}t
> - \frac{1}{3}\begin{pmatrix} 4 \\ 5 \end{pmatrix},
> $$
>
> the first two terms the homogeneous solution and the rest a particular solution,
> matching the undetermined-coefficients result above.

$$
% caption: Variation of parameters with M the fundamental matrix (Psi in the
% text): substituting x = M u collapses the equation to u' = M^{-1} g, then
% integration and one matrix product rebuild x.
\begin{tikzpicture}[>=stealth, font=\footnotesize,
  b/.style={draw, minimum width=32mm, minimum height=12mm, align=center, font=\scriptsize}]
  \definecolor{acc}{HTML}{4A6FA5}
  \node[b] (a) at (0,0) {$\mathbf{x} = M(t)\,\mathbf{u}(t)$};
  \node[b] (b) at (4.8,0) {$M\,\frac{d\mathbf{u}}{dt} = \mathbf{g}$};
  \node[b] (c) at (9.6,0) {$\frac{d\mathbf{u}}{dt}$ known};
  \node[b] (d) at (9.6,-2.0) {$\mathbf{u}(t)$ by integration};
  \node[b] (e) at (2.4,-2.0) {$\mathbf{x} = M(t)\,\mathbf{u}(t)$};
  \draw[->, thick] (a) -- (b) node[midway, above, font=\scriptsize, text=black]{substitute};
  \draw[->, thick] (b) -- (c) node[midway, above, font=\scriptsize, text=black]{cancel};
  \draw[->, thick] (c) -- (d) node[midway, right, font=\scriptsize, text=black]{integrate};
  \draw[->, thick] (d) -- (e) node[midway, above, font=\scriptsize, text=black]{rebuild};
\end{tikzpicture}
$$

## Summary

The fundamental matrix organizes the constant-coefficient theory.

- **Fundamental matrix.** Columns are a fundamental set; $\mathbf{x} =
  \Psi(t)\mathbf{c}$ is the general solution, and the special $\Phi(t)$ with
  $\Phi(0) = I$ propagates initial states via $\mathbf{x} = \Phi(t)\mathbf{x}^0$.
- **Matrix exponential.** For constant $A$, $\Phi(t) = \exp(At)$, defined by the
  same series as the scalar exponential and computed from eigenpairs when they
  are complete.
- **Defective eigenvalues.** Missing eigenvectors are replaced by
  generalized eigenvectors solving $(A - \rho I)\boldsymbol{\eta} =
  \boldsymbol{\xi}$; the extra solution carries $te^{\rho t}$ and the portrait is
  an improper node.
- **Nonhomogeneous systems.** General $=$ homogeneous $+$ particular; variation
  of parameters, $\mathbf{u}' = \Psi^{-1}\mathbf{g}$, finds the particular
  solution with only a fundamental matrix in hand.

[^boyce-78]: **Boyce**, §7.8, Examples 1–2: the defective double eigenvalue of $\begin{pmatrix} 1 & -1 \\ 1 & 3 \end{pmatrix}$ and the generalized-eigenvector second solution.
[^boyce-79]: **Boyce**, §7.9, Examples 1–3: the system $\mathbf{x}' = \begin{pmatrix} -2 & 1 \\ 1 & -2 \end{pmatrix}\mathbf{x} + \begin{pmatrix} 2e^{-t} \\ 3t \end{pmatrix}$ solved by diagonalization, undetermined coefficients, and variation of parameters, all yielding the same general solution.