Systems of Linear Differential Equations

Orientation

When no quantity can be tracked alone

Two tanks of brine exchange liquid. The salt in the first drains into the second, and the second drains back. Neither amount can be worked out on its own: the rate at which the first changes depends on how much is in the second, whose rate depends on the first.

Writing $x_{1}$ and $x_{2}$ for the two amounts gives a coupled system,

x_{1}^{'} = a_{11} x_{1} + a_{12} x_{2}, x_{2}^{'} = a_{21} x_{1} + a_{22} x_{2},

which in matrix form is simply $x^{'} = A x$ . Nothing in the previous units solves this. Separation needs one unknown; the characteristic equation needs one equation.

The move that works is to ask whether some direction can be tracked alone. Try $x = e^{λ t} v$ for a fixed vector $v$ : a solution that keeps its direction and only changes in length. Substituting gives $λ e^{λ t} v = e^{λ t} A v$ , and cancelling the exponential leaves

A v = λ v .

That is the eigenvalue equation, arrived at from a differential equation rather than from geometry. The directions in which a coupled system behaves like a single scalar equation are exactly the eigenvectors of its matrix, and the rate along each is its eigenvalue.

So the whole method is: find the eigenpairs, solve the one-dimensional problem in each, and add. The coupling was never removed. It was aligned with directions where it acts as one number.

Two further things follow. The eigenvalues answer the question usually worth asking, which is not what $x (t)$ is but where it ends up: negative real parts drain everything to the origin, an imaginary part means rotation. And every equation from the previous unit is secretly a system, since setting $x_{1} = y$ , $x_{2} = y^{'}$ turns $y^{″} + p y^{'} + q y = 0$ into a $2 \times 2$ system with the same characteristic polynomial. A repeated root there is a defective companion matrix here.

Why this matters

Why eigenvectors, and not some other change of variable

The obstacle in $x^{'} = A x$ is the off-diagonal entries. If $A$ were diagonal,

A = (\begin{matrix} λ_{1} & 0 \\ 0 & λ_{2} \end{matrix}),

the system would read $x_{1}^{'} = λ_{1} x_{1}$ and $x_{2}^{'} = λ_{2} x_{2}$ : two separate scalar equations, each solved in one line by an exponential. The difficulty is entirely that the equations talk to each other.

So the natural question is whether a change of variable can make $A$ diagonal. That question has a name and an answer already in the corpus: $A$ is diagonalisable exactly when it has a full set of independent eigenvectors, and the diagonal entries are then the eigenvalues. Writing $x = P y$ with $P$ the matrix of eigenvectors gives

P y^{'} = A P y ⟹ y^{'} = P^{- 1} A P y = D y,

which is decoupled. Solve each component separately as $y_{i} = c_{i} e^{λ_{i} t}$ , then transform back with $x = P y$ , which is precisely $\sum c_{i} e^{λ_{i} t} v_{i}$ , the general solution stated in the definition.

That is why eigenvectors and no other change of variable. They are the directions along which the matrix acts by scaling alone, and scaling is the one behaviour a scalar exponential can express.

What this predicts. If the method depends on $A$ being diagonalisable, it must fail exactly when $A$ is not. The eigenvalue unit already identifies that obstruction: a repeated eigenvalue with too few eigenvectors. The prediction is correct, and the repair is the $t v$ term, the same shortfall and the same kind of patch as the repeated root that needed $x e^{r x}$ .

Why this is worth the trouble. The eigenvalues are usually the answer, not just a route to a formula. A system is stable when every real part is negative, and that is read off without solving anything. In circuits, mixing, populations and control loops, the question is almost always whether the system settles, oscillates or runs away, and the eigenvalues decide it.

Figure

A saddle and a centre, with eigendirections where they exist

the eigenvalues decide the shape of the flow

the eigenvalues decide the shape of the flow

Two phase portraits drawn side by side in separate coordinate frames, each with its own marked origin and its own pair of axis stubs labelled $x_{1}$ and $x_{2}$ ; the horizontal separation between them is page layout and carries no coordinate meaning. The left frame shows the system $x^{'} = A x$ with $A = (\begin{matrix} 1 & 2 \\ 2 & 1 \end{matrix})$ : a grid of grey arrows gives the direction of flow, and two straight lines through the origin mark the eigendirections. Along $(1, 1)$ , eigenvalue $3$ , the arrows point away from the origin; along $(1, - 1)$ , eigenvalue $- 1$ , they point toward it. Trajectories approach the outward direction, which is the signature of a saddle. The right frame shows $x^{'} = B x$ with $B = (\begin{matrix} 0 & - 2 \\ 2 & 0 \end{matrix})$ , whose equilibrium is likewise at its own origin. Its eigenvalues are $\pm 2 i$ , purely imaginary, so no real direction is preserved and no straight-line solution exists; two concentric closed circles show the flow circulating around the equilibrium, which is a centre.

The lesson's two matrices, each drawn as the flow it defines. The two panels are separate coordinate frames: each has its own origin, marked $0$ , and both systems have their equilibrium there.

Real eigenvalues, $A = (\begin{matrix} 1 & 2 \\ 2 & 1 \end{matrix})$ . Eigenvalues $3$ and $- 1$ , with eigenvectors $(1, 1)$ and $(1, - 1)$ drawn. Along $(1, 1)$ the flow runs outward, along $(1, - 1)$ inward; every other trajectory is pulled toward the unstable direction. That is a saddle, and the eigenvectors are the only straight-line solutions.

Complex eigenvalues, $B = (\begin{matrix} 0 & - 2 \\ 2 & 0 \end{matrix})$ . Eigenvalues $\pm 2 i$ , no real eigenvector, so no direction is preserved. The flow circulates: a centre, closed orbits around the origin.

These are two of the patterns eigenvalues produce, and both are readable before any solution is written down: real eigenvalues of opposite sign give a saddle, a pure imaginary pair gives a centre.

They are not the whole classification. Complex eigenvalues with nonzero real part spiral rather than close; repeated eigenvalues behave differently according to whether the matrix is diagonalizable or defective; a zero eigenvalue leaves a whole line of equilibria. What is general is that the eigenvalues decide.

Definition

Multiplicities, conjugate pairs, and counting solutions

Why $v$ must be nonzero. The trial $x = e^{λ t} v$ with $v = 0$ is the zero function, which solves every linear system and carries no information. Requiring $v \neq 0$ is what makes $A v = λ v$ a genuine constraint on $λ$ , and it is the same requirement the eigenvalue unit imposes, for the same reason.

Counting, and the two multiplicities. An $n \times n$ system needs $n$ independent solutions, because an initial vector has $n$ components and each imposes one condition. Eigenvalues supply solutions one per independent eigenvector, not one per eigenvalue, and the two counts can differ:

algebraic multiplicity, how many times $λ$ repeats as a root of $det (A - λ I)$ ;
geometric multiplicity, $\dim \ker (A - λ I)$ , the number of independent eigenvectors it actually supplies.

The geometric never exceeds the algebraic. When they are equal for every eigenvalue, the eigenvectors span and the method finishes; when the geometric falls short, the matrix is defective and the shortfall must be repaired.

Why a conjugate pair yields two real solutions, not four. A real matrix with a complex eigenvalue $α + β i$ also has $α - β i$ , with conjugate eigenvector. The two complex solutions $e^{(α \pm β i) t} v^{\pm}$ are not independent over the reals in the naive way: taking real and imaginary parts of either one produces two real solutions, and the conjugate adds nothing new. So one pair of complex eigenvalues contributes exactly two to the count of $n$ , which is what makes the arithmetic come out right for a $2 \times 2$ system with no real eigenvalues at all.

Concretely, $e^{(α + β i) t} v = e^{α t} (\cos β t + i \sin β t) (p + i q)$ separates into

e^{α t} (p \cos β t - q \sin β t) and e^{α t} (p \sin β t + q \cos β t),

both real, both solutions.

Why the eigenvalues decide the long run. Each term carries $e^{λ_{i} t}$ , whose magnitude is $e^{(Re λ_{i}) t}$ . The imaginary part contributes only rotation, since $| e^{i β t} | = 1$ . So a negative real part shrinks its term, a positive one grows it, and zero holds it steady. A single eigenvalue with positive real part is enough to make almost every trajectory escape, because that term eventually dominates the ones that fade; the exceptions are the initial vectors whose coefficient on that eigenvector happens to be zero, which form a lower-dimensional set.

That is why stability is read from the eigenvalues alone, and why no solving is needed to answer it.

Theorem

Independent eigenvectors give a basis of solutions

Theorem. Let $A$ be a constant $n \times n$ real matrix. If $v_{1}, \dots, v_{n}$ are linearly independent eigenvectors of $A$ with eigenvalues $λ_{1}, \dots, λ_{n}$ , then

x = c_{1} e^{λ_{1} t} v_{1} + \dots + c_{n} e^{λ_{n} t} v_{n}

is the general solution of $x^{'} = A x$ : every solution has this form, and the constants are uniquely determined by $x (t_{0})$ .

Each term is a solution. For $x = e^{λ t} v$ with $A v = λ v$ ,

x^{'} = λ e^{λ t} v = e^{λ t} (λ v) = e^{λ t} A v = A x .

The solution set is a vector space of dimension $n$ . Linearity of differentiation makes any combination of solutions a solution, exactly as in the scalar second-order case. Existence and uniqueness for linear systems gives one solution per initial vector $x (t_{0}) \in R^{n}$ , so the space has dimension $n$ .

The constants are uniquely determined. At $t_{0} = 0$ the requirement $x (0) = x_{0}$ reads

c_{1} v_{1} + \dots + c_{n} v_{n} = x_{0},

a linear system whose matrix has the eigenvectors as columns. Independence makes it invertible, so the $c_{i}$ exist and are unique, which is also why independence, not merely having $n$ eigenvalues, is the hypothesis that matters.

Verification on three systems. Each claimed solution was substituted and $x^{'} - A x$ evaluated numerically:

System	Eigenvalues	Solution checked	Max residual
$A = (\begin{matrix} 1 & 2 \\ 2 & 1 \end{matrix})$	$3, - 1$	$2 e^{3 t} (\binom{1}{1}) + e^{- t} (\binom{1}{- 1})$	$1.1 \times 10^{- 8}$
$B = (\begin{matrix} 0 & - 2 \\ 2 & 0 \end{matrix})$	$\pm 2 i$	$(\cos 2 t, \sin 2 t)$	$2.6 \times 10^{- 11}$
$C = (\begin{matrix} 3 & 1 \\ 0 & 3 \end{matrix})$	$3$ (twice)	$e^{3 t} (1 + 2 t, 2)$	$1.6 \times 10^{- 8}$

The residuals are finite-difference error. For $A$ the eigenvalues were confirmed against the invariants, trace $2 = 3 + (- 1)$ and determinant $- 3 = 3 \times (- 1)$ , and each eigenvector satisfies $A v - λ v = 0$ to $8 \times 10^{- 16}$ .

When the hypothesis fails. The third row is the case the theorem does not cover: $C$ has the single eigenvalue 3 with $rank (C - 3 I) = 1$ , so its eigenspace is one-dimensional and there is no basis of eigenvectors. The solution listed there is not of the form $\sum c_{i} e^{λ_{i} t} v_{i}$ , it carries a factor of $t$ , which is why the defective case needs separate treatment.

Corollary (stability). If every eigenvalue has negative real part, every solution tends to $0$ as $t \to \infty$ , since each term's magnitude is $| c_{i} | e^{(Re λ_{i}) t} ‖ v_{i} ‖ \to 0$ . If some eigenvalue has positive real part, solutions whose coefficient on that eigenvector is nonzero are unbounded. This holds for defective matrices too: a $t$ factor changes the rate of approach but not whether $e^{λ t} t \to 0$ , which it does whenever $Re λ < 0$ .

Derivation

The complex pair made real, and the defective repair

Complex eigenvalues to real solutions. Take $B = (\begin{matrix} 0 & - 2 \\ 2 & 0 \end{matrix})$ . The characteristic polynomial is

det (B - λ I) = λ^{2} + 4,

so $λ = \pm 2 i$ , confirmed against the invariants, trace $0 = 2 i + (- 2 i)$ and determinant $4 = (2 i) (- 2 i)$ .

For $λ = 2 i$ , solving $(B - 2 i I) v = 0$ gives $- 2 i v_{1} - 2 v_{2} = 0$ , so $v_{2} = - i v_{1}$ and $v = (\binom{1}{- i})$ . Then

e^{2 i t} (\binom{1}{- i}) = (\cos 2 t + i \sin 2 t) (\binom{1}{- i}) = (\binom{\cos 2 t + i \sin 2 t}{\sin 2 t - i \cos 2 t}) .

Separating real and imaginary parts gives two real solutions:

(\binom{\cos 2 t}{\sin 2 t}) and (\binom{\sin 2 t}{- \cos 2 t}) .

Verification. Substituting the first, $x^{'} - B x$ has magnitude below $2.6 \times 10^{- 11}$ at $t = 0, 0.5, 1$ . Its norm is $\sqrt{\cos^{2} 2 t + \sin^{2} 2 t} = 1$ at every $t$ , confirmed as $1.0$ at $t = 0, 1, 2$ . A closed orbit, because the real part is zero. A nonzero real part would multiply this by $e^{α t}$ and turn the circle into a spiral.

The conjugate eigenvalue $- 2 i$ produces the same two real solutions, which is why a conjugate pair contributes two and not four.

The defective case. Take $C = (\begin{matrix} 3 & 1 \\ 0 & 3 \end{matrix})$ . Being triangular, its characteristic polynomial is $(3 - λ)^{2}$ , so $λ = 3$ has algebraic multiplicity 2. But

C - 3 I = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})

has rank 1, so its kernel is one-dimensional: every eigenvector is a multiple of $v = (\binom{1}{0})$ . The geometric multiplicity is 1, short of the algebraic 2. The matrix is defective, and $c_{1} e^{3 t} v + c_{2} e^{3 t} v$ collapses to a single constant, exactly as $A e^{r x} + B e^{r x}$ did in the scalar case.

Finding the missing solution. Guided by the scalar repair, try $x = e^{λ t} (w + t v)$ . Differentiating,

x^{'} = e^{λ t} (λ w + v + λ t v),

while

A x = e^{λ t} (A w + t A v) = e^{λ t} (A w + λ t v),

using $A v = λ v$ . The $t v$ terms match, and equating what remains gives

λ w + v = A w ⟺ (A - λ I) w = v .

So $w$ is not an eigenvector but a generalised eigenvector: it is sent to $v$ rather than to zero. The equation is solvable precisely because $v$ lies in the image of $A - λ I$ , which is what being defective arranges.

Worked instance. With $λ = 3$ and $v = (\binom{1}{0})$ , solve $(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}) w = (\binom{1}{0})$ : the first row gives $w_{2} = 1$ , and $w_{1}$ is free, so take $w = (\binom{0}{1})$ . The general solution is

x = c_{1} e^{3 t} (\binom{1}{0}) + c_{2} e^{3 t} [(\binom{0}{1}) + t (\binom{1}{0})] .

Verification. With $c_{1} = 1$ , $c_{2} = 2$ this is $e^{3 t} (1 + 2 t, 2)$ , and $x^{'} - C x$ stays below $1.6 \times 10^{- 8}$ at $t = 0, 0.5, 1$ .

The scalar case was this case. Writing $y^{″} - 5 y^{'} + 6 y = 0$ as a system with $x_{1} = y$ , $x_{2} = y^{'}$ gives the companion matrix $(\begin{matrix} 0 & 1 \\ - 6 & 5 \end{matrix})$ , whose eigenvalues compute to exactly $2$ and $3$ . The roots of $λ^{2} - 5 λ + 6$ , the same characteristic polynomial. Substituting the previous unit's solution $y = 3 e^{2 t} - 2 e^{3 t}$ as $x = (y, y^{'})$ satisfies the system with residuals below $2.1 \times 10^{- 8}$ . A repeated root there is a defective companion matrix here, and $x e^{r x}$ is the $t v$ term.

Procedure

Solving a constant-coefficient linear system

Input. A system $x^{'} = A x$ with constant $A$ , optionally with an initial vector $x (t_{0})$ .

Step 1 — Write the system in matrix form. Collect the coefficients so that row $i$ of $A$ holds the coefficients appearing in $x_{i}^{'}$ . Missing terms are zeros.

Step 2 — Find the eigenvalues. Solve

det (A - λ I) = 0 .

For a $2 \times 2$ matrix this is $λ^{2} - (tr A) λ + det A = 0$ . Check them against the invariants before going on: the eigenvalues must sum to the trace and multiply to the determinant. This catches most arithmetic slips immediately.

Step 3 — Find an eigenvector for each eigenvalue. Solve $(A - λ I) v = 0$ . The matrix $A - λ I$ is singular by construction, so its rows are dependent and one equation suffices; any nonzero solution will do, since scaling $v$ only rescales the constant. Verify $A v = λ v$ . One multiplication settles whether the vector is right.

Step 4 — At a repeated eigenvalue, test for defectiveness. Compute $rank (A - λ I)$ . For a $2 \times 2$ matrix with a doubled eigenvalue:

rank 0. Every vector is an eigenvector; $A = λ I$ and any basis works;
rank 1, only one independent eigenvector; the matrix is defective, so go to step 5.

Never infer the count of eigenvectors from the multiplicity.

Step 5 — Supply the missing solution if defective. Solve $(A - λ I) w = v$ for a generalised eigenvector $w$ , and take the second solution to be $e^{λ t} (w + t v)$ .

Step 6 — Convert any complex pair to real form. For $λ = α + β i$ with eigenvector $v = p + i q$ , expand $e^{λ t} v$ by Euler's identity and take real and imaginary parts:

e^{α t} (p \cos β t - q \sin β t), e^{α t} (p \sin β t + q \cos β t) .

Do not also process the conjugate eigenvalue; it yields the same two solutions.

Step 7 — Assemble the general solution as a combination of the $n$ independent solutions found, with one constant each.

Step 8 — Apply the initial vector. Substituting $t = t_{0}$ gives a linear system in the constants; solve it as a system, not one constant at a time.

Step 9 — Verify, and read the behaviour. Substitute back into $x^{'} = A x$ and check the initial vector. Then state the long-run behaviour from the eigenvalues: all real parts negative means decay to $0$ ; any positive real part means escape; imaginary parts mean rotation.

Where it goes wrong.

Assuming a repeated eigenvalue gives two eigenvectors. It sometimes does and sometimes does not, and only the rank of $A - λ I$ says which.
Counting a conjugate pair twice. Two complex eigenvalues yield two real solutions, not four; a $2 \times 2$ system is then complete.
Solving for the constants one at a time. The initial condition couples them; $c_{1}$ generally cannot be found without $c_{2}$ .
Reading stability from the entries of $A$ . The diagonal entries are not the eigenvalues unless $A$ is triangular. A matrix with negative diagonal entries can still have a positive eigenvalue.
Applying the method to a non-constant $A$ . If the entries depend on $t$ , the trial $e^{λ t} v$ does not reduce to an eigenvalue equation.

Worked example

An initial value problem, worked in full

Problem. Solve

x_{1}^{'} = x_{1} + 2 x_{2}, x_{2}^{'} = 2 x_{1} + x_{2}, x (0) = (\binom{3}{1}) .

Step 1 — Matrix form.

x^{'} = A x, A = (\begin{matrix} 1 & 2 \\ 2 & 1 \end{matrix}) .

Step 2 — Eigenvalues. With $tr A = 2$ and $det A = 1 - 4 = - 3$ :

λ^{2} - 2 λ - 3 = 0 ⟹ (λ - 3) (λ + 1) = 0,

so $λ_{1} = 3$ and $λ_{2} = - 1$ .

Check against the invariants. Sum $3 + (- 1) = 2 = tr A$ ; product $3 \times (- 1) = - 3 = det A$ . Substituting confirms the polynomial vanishes: $9 - 6 - 3 = 0$ and $1 + 2 - 3 = 0$ .

Step 3 — Eigenvectors.

For $λ_{1} = 3$ : $A - 3 I = (\begin{matrix} - 2 & 2 \\ 2 & - 2 \end{matrix})$ , whose first row gives $- 2 v_{1} + 2 v_{2} = 0$ , so $v_{1} = v_{2}$ and

v_{1} = (\binom{1}{1}) .

For $λ_{2} = - 1$ : $A + I = (\begin{matrix} 2 & 2 \\ 2 & 2 \end{matrix})$ , giving $v_{1} + v_{2} = 0$ and

v_{2} = (\binom{1}{- 1}) .

Verify. $A v_{1} = (\binom{1 + 2}{2 + 1}) = (\binom{3}{3}) = 3 v_{1}$ and $A v_{2} = (\binom{1 - 2}{2 - 1}) = (\binom{- 1}{1}) = - v_{2}$ . Numerically both residuals $A v - λ v$ are below $8 \times 10^{- 16}$ .

Step 4 — Not repeated, so no defectiveness test is needed. The two eigenvectors are independent, since neither is a multiple of the other.

Step 7 — General solution.

x = c_{1} e^{3 t} (\binom{1}{1}) + c_{2} e^{- t} (\binom{1}{- 1}) .

Step 8 — Apply the initial vector. At $t = 0$ both exponentials are 1:

c_{1} (\binom{1}{1}) + c_{2} (\binom{1}{- 1}) = (\binom{3}{1}) ⟹ {\begin{cases} c_{1} + c_{2} = 3 \\ c_{1} - c_{2} = 1 . \end{cases}

Adding gives $2 c_{1} = 4$ , so $c_{1} = 2$ and $c_{2} = 1$ , confirmed by solving the system numerically. Both constants come from the pair together; neither could be found alone.

x = 2 e^{3 t} (\binom{1}{1}) + e^{- t} (\binom{1}{- 1})

Componentwise, $x_{1} = 2 e^{3 t} + e^{- t}$ and $x_{2} = 2 e^{3 t} - e^{- t}$ .

Step 9 — Verify and read.

Initial vector. $x (0) = 2 (\binom{1}{1}) + (\binom{1}{- 1}) = (\binom{3}{1})$ , confirmed numerically as $(3, 1)$ .

The system. Evaluating $x^{'} - A x$ gives residuals of $1.9 \times 10^{- 10}$ , $9.0 \times 10^{- 10}$ and $1.1 \times 10^{- 8}$ at $t = 0, 0.5, 1$ , finite-difference error.

Behaviour. The eigenvalue 3 is positive, so the solution grows without bound: $x (1) = (40.538953, 39.803194)$ and $x (2) = (806.992922, 806.722252)$ . The two components converge toward each other, because the $e^{- t}$ term that distinguishes them dies away while the $e^{3 t}$ term along $(\binom{1}{1})$ takes over. At $t = 2$ they differ by $0.27$ out of roughly $807$ .

The eigenvalue with the largest real part dominates the long run, and the trajectory aligns with its eigenvector, here $(\binom{1}{1})$ , the line $x_{1} = x_{2}$ .

Example

A system with no real eigenvalues

The worked example had two real eigenvalues, so both solutions were exponentials along fixed directions. Here is the other kind of $2 \times 2$ system: one with no real eigenvalues at all.

Problem. Solve $x^{'} = B x$ with

B = (\begin{matrix} 0 & - 2 \\ 2 & 0 \end{matrix}),

that is $x_{1}^{'} = - 2 x_{2}$ and $x_{2}^{'} = 2 x_{1}$ .

Eigenvalues. $tr B = 0$ and $det B = 0 - (- 4) = 4$ , so the characteristic equation is

λ^{2} - 0 \cdot λ + 4 = λ^{2} + 4 = 0 ⟹ λ = \pm 2 i .

Check against the invariants. Sum $2 i + (- 2 i) = 0 = tr B$ ; product $(2 i) (- 2 i) = - 4 i^{2} = 4 = det B$ .

No real eigenvalue exists, which says something geometric: no direction is preserved. Every vector is turned by this matrix, so no solution can keep a fixed direction and merely scale, and the real-exponential solutions of the worked example have no counterpart here.

Eigenvector for $λ = 2 i$ . Solving $(B - 2 i I) v = 0$ :

(\begin{matrix} - 2 i & - 2 \\ 2 & - 2 i \end{matrix}) (\binom{v_{1}}{v_{2}}) = (\binom{0}{0}) .

The first row gives $- 2 i v_{1} - 2 v_{2} = 0$ , so $v_{2} = - i v_{1}$ . Taking $v_{1} = 1$ :

v = (\binom{1}{- i}) .

The second row is $2 v_{1} - 2 i v_{2} = 2 - 2 i (- i) = 2 - 2 = 0$ , consistent, as it must be, since $B - 2 i I$ is singular.

Real solutions. Expand $e^{2 i t} v$ by Euler's identity:

e^{2 i t} (\binom{1}{- i}) = (\cos 2 t + i \sin 2 t) (\binom{1}{- i}) = (\binom{\cos 2 t + i \sin 2 t}{\sin 2 t - i \cos 2 t}),

using $- i (\cos 2 t + i \sin 2 t) = \sin 2 t - i \cos 2 t$ . Real and imaginary parts give two real solutions, and the general solution is

x = c_{1} (\binom{\cos 2 t}{\sin 2 t}) + c_{2} (\binom{\sin 2 t}{- \cos 2 t}) .

The conjugate eigenvalue $- 2 i$ is not processed separately: it yields these same two solutions, so the pair contributes two and the $2 \times 2$ system is complete.

Verification. For the first solution, $x = (\cos 2 t, \sin 2 t)$ gives $x^{'} = (- 2 \sin 2 t, 2 \cos 2 t)$ , while

B x = (\binom{- 2 \sin 2 t}{2 \cos 2 t}),

identical. Numerically $‖ x^{'} - B x ‖$ stays below $2.6 \times 10^{- 11}$ at $t = 0, 0.5, 1$ .

What the trajectories look like. The norm is

‖ x ‖ = \sqrt{\cos^{2} 2 t + \sin^{2} 2 t} = 1

for every $t$ , confirmed as exactly $1.0$ at $t = 0, 1, 2$ . The trajectory is a circle of constant radius, traversed with angular frequency 2 and period $π$ .

That constancy is the real part being zero. The general complex case contributes $e^{α t}$ times these trigonometric terms, so:

$α = Re λ$	Trajectory
$< 0$	spirals inward to the origin
$= 0$	closed orbit, constant radius
$> 0$	spirals outward without bound

Here $α = 0$ exactly, which is the knife-edge case: the system neither gains nor loses, and circles forever. Perturbing $B$ slightly, adding $- 0.1$ to each diagonal entry, say, would move the eigenvalues to $- 0.1 \pm 2 i$ and turn every circle into an inward spiral.

Comparison with the worked example.

	$A = (\begin{matrix} 1 & 2 \\ 2 & 1 \end{matrix})$	$B = (\begin{matrix} 0 & - 2 \\ 2 & 0 \end{matrix})$
Eigenvalues	$3, - 1$ (real, distinct)	$\pm 2 i$ (pure imaginary)
Invariant directions	two lines, $(\binom{1}{1})$ and $(\binom{1}{- 1})$	none
Solutions	real exponentials	sines and cosines
Long run	escapes along $(\binom{1}{1})$	circles forever

Both were solved by the same three steps, characteristic equation, eigenvectors, combine, and the difference in behaviour was settled by the eigenvalues before any solution was written down.

Contrast

Two matrices with the same repeated eigenvalue

A repeated eigenvalue does not by itself determine the shape of the solution. These two matrices both have the single eigenvalue 3, doubled, and their solutions differ.

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Case 1 — not defective.

A_{1} = (\begin{matrix} 3 & 0 \\ 0 & 3 \end{matrix}) = 3 I .

Characteristic polynomial $(3 - λ)^{2}$ , so $λ = 3$ with algebraic multiplicity 2.

A_{1} - 3 I = (\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}), rank = 0 .

The kernel is all of $R^{2}$ , so every nonzero vector is an eigenvector and the geometric multiplicity is 2. Taking the standard basis,

x = c_{1} e^{3 t} (\binom{1}{0}) + c_{2} e^{3 t} (\binom{0}{1}) = e^{3 t} (\binom{c_{1}}{c_{2}}) .

No $t$ factor appears. Both components simply scale by $e^{3 t}$ , and any initial vector is met by reading off its components.

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Case 2 — defective.

A_{2} = (\begin{matrix} 3 & 1 \\ 0 & 3 \end{matrix}) .

Being triangular, the characteristic polynomial is again $(3 - λ)^{2}$ . The same eigenvalue with the same multiplicity. But

A_{2} - 3 I = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}), rank = 1,

confirmed numerically. The kernel is one-dimensional, spanned by $v = (\binom{1}{0})$ : the geometric multiplicity is 1, and the matrix is defective.

Here $c_{1} e^{3 t} v + c_{2} e^{3 t} v$ collapses to $(c_{1} + c_{2}) e^{3 t} v$ , one constant with two names, and a family confined to the line $x_{2} = 0$ , which cannot meet an initial vector such as $(\binom{1}{2})$ at all. The second solution must come from a generalised eigenvector: solving $(A_{2} - 3 I) w = v$ gives $w = (\binom{0}{1})$ , so

x = c_{1} e^{3 t} (\binom{1}{0}) + c_{2} e^{3 t} [(\binom{0}{1}) + t (\binom{1}{0})] .

With $c_{1} = 1$ , $c_{2} = 2$ this is $e^{3 t} (1 + 2 t, 2)$ , whose residual against $x^{'} = A_{2} x$ stays below $1.6 \times 10^{- 8}$ .

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What differs, and what does not.

	$A_{1} = 3 I$	$A_{2} = (\begin{matrix} 3 & 1 \\ 0 & 3 \end{matrix})$
Eigenvalue	3, twice	3, twice
Algebraic multiplicity	2	2
$rank (A - 3 I)$	0	1
Geometric multiplicity	2	1
Defective	no	yes
Solution	$e^{3 t} (\binom{c_{1}}{c_{2}})$	$e^{3 t} [c_{1} v + c_{2} (w + t v)]$
$t$ factor	none	present

The characteristic polynomial is identical; everything that distinguishes the two lives in the rank of $A - λ I$ , which the polynomial cannot see. That is why step 4 of the procedure is a computation and not an inference.

The stability reading survives either way. Both have eigenvalue 3 with positive real part, so both grow without bound, $t$ factor or not. And had the eigenvalue been $- 3$ , both would decay, since $t e^{- 3 t} \to 0$ , a polynomial factor cannot defeat a decaying exponential. Defectiveness changes the shape of the approach, never its direction.

The same distinction, one unit earlier. For scalar equations this is the difference between distinct roots and a repeated root needing $x e^{r x}$ . The companion matrix of $y^{″} - 6 y^{'} + 9 y = 0$ is defective, and its $t v$ term is that $x e^{3 x}$ .

Application

Two connected tanks, solved by eigenvalues

Two tanks are connected by pipes running both ways. Brine flows out of the first into the second and back, and the salt in each is stirred uniformly. Writing $x_{1}$ and $x_{2}$ for the kilograms of salt in each tank, conservation gives a system: what leaves one arrives in the other, at a rate proportional to the concentration there.

A particular arrangement yields

x_{1}^{'} = - \frac{1}{2} x_{1} + \frac{1}{4} x_{2}, x_{2}^{'} = \frac{1}{2} x_{1} - \frac{1}{2} x_{2},

that is $x^{'} = M x$ with $M = (\begin{matrix} - 0.5 & 0.25 \\ 0.5 & - 0.5 \end{matrix})$ .

Why this cannot be done one tank at a time. The first equation involves $x_{2}$ , so integrating it requires knowing $x_{2}$ first, but the second equation involves $x_{1}$ . Neither is solvable alone, which is exactly the situation the eigenvalue method was built for.

The eigenvalues. $tr M = - 1$ and $det M = 0.25 - 0.125 = 0.125$ , so

λ^{2} + λ + 0.125 = 0 ⟹ λ = \frac{- 1 \pm \sqrt{1 - 0.5}}{2} .

Computing: $λ_{1} = - 0.146447$ and $λ_{2} = - 0.853553$ . Their sum is $- 1 = tr M$ and their product is $0.125 = det M$ .

Both eigenvalues are real and negative, so:

No oscillation. There is no imaginary part, so the salt does not slosh back and forth past its equilibrium, but approaches it monotonically once the transient has settled.
Both tanks drain to zero. Every term carries $e^{λ t}$ with $λ < 0$ . Physically this says the arrangement has no salt source, so the system runs down. A model of this kind that produced a positive eigenvalue would be telling us we had written the conservation wrongly.
One rate eventually dominates. The term with $λ_{2} = - 0.853553$ dies roughly six times faster than the one with $λ_{1} = - 0.146447$ . After the fast transient, the whole system decays at the slow rate, with a half-life of $\ln 2 / 0.146447 = 4.733105$ time units.

That last number is the practically useful one, and it came from an eigenvalue rather than from a solution. Asked "how long until the tanks are nearly clean?", the answer is about five time units per halving, obtained without ever writing down $x (t)$ .

The slow eigenvector is the shape the system settles into. Long after the start, the state lies almost along the eigenvector for $λ_{1}$ , whatever it was initially. The system forgets its initial condition except for a scale factor, which is why two tanks started very differently end up with the same ratio of salt, draining together.

The same reading elsewhere. Computing eigenvalues, reading signs and imaginary parts, and extracting a rate is what makes this method carry across fields:

System	Coupled quantities	What the eigenvalues give
Mixing tanks	salt in each tank	decay rates, settling time
$L R C$ circuit	charge and current	ringing frequency, damping
Two-species population	predator and prey numbers	growth or collapse; a complex pair means population cycles
Structure under load	displacements	natural frequencies, resonance risk

In each case the question asked is about behaviour rather than formula, and the eigenvalues answer it directly. A complex pair in the population model, for instance, predicts cycles, predator and prey numbers chasing each other around, and the sign of the real part says whether those cycles grow, shrink, or persist.

A caution the procedure block also raises. The diagonal entries of $M$ are both $- 0.5$ , and neither is an eigenvalue. Stability cannot be read from the diagonal unless the matrix is triangular; it must be read from $det (M - λ I) = 0$ .

Systems of Linear Differential Equations

When no quantity can be tracked alone

Why eigenvectors, and not some other change of variable

A saddle and a centre, with eigendirections where they exist

Multiplicities, conjugate pairs, and counting solutions

Independent eigenvectors give a basis of solutions

The complex pair made real, and the defective repair

Solving a constant-coefficient linear system

An initial value problem, worked in full

A system with no real eigenvalues

Two matrices with the same repeated eigenvalue

Two connected tanks, solved by eigenvalues

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