Systems of first-order linear ODEs — matrix method: Question bank

Before the traps, pin down the notation so nothing sneaks in undefined:

Reminders of the rest of the vocabulary you must use precisely:

eigenpair $(λ, v)$ : a number $λ$ and a nonzero vector $v$ with $A v = λ v$ — see Eigenvalues and Eigenvectors.
algebraic multiplicity: how many times $λ$ is a root of $det (A - λ I) = 0$ .
geometric multiplicity: how many independent eigenvectors $λ$ actually has.
defective: geometric $<$ algebraic multiplicity (too few eigenvectors) — see Diagonalization.

The next figure is your visual dictionary for the trajectory shapes referenced throughout — keep it in view.

Figure — Systems of first-order linear ODEs — matrix method

True or false — justify

True or false: a system

x^{'} = A x

always has exactly

n

eigenvalues, so it always has

n

independent solutions.

False. It has

n

eigenvalues counted with multiplicity, but independent solutions come from independent eigenvectors (plus generalized eigenvectors). A defective matrix has fewer eigenvectors, so you must build the missing modes from generalized eigenvectors solving

(A - λ I) w = v

True or false: if every eigenvalue of

A

has negative real part, every solution decays to

0

t \to \infty

True. Each mode carries

e^{λ t}

, and

∣ e^{λ t} ∣ = e^{(Re λ) t}

; if every real part is negative, every mode shrinks, so the origin is asymptotically stable — see Phase Portraits and Stability.

True or false: multiplying an eigenvector by

- 5

gives a different solution of the system.

False in the sense of "new information."

- 5 v

is still an eigenvector for the same

λ

; the solution

c e^{λ t} v

just absorbs the

- 5

into the constant

c

. It spans the same solution line.

True or false: two distinct eigenvalues always give two linearly independent solutions.

True. Suppose

c_{1} v_{1} + c_{2} v_{2} = 0

; apply

A

to get

c_{1} λ_{1} v_{1} + c_{2} λ_{2} v_{2} = 0

, subtract

λ_{1}

times the first to get

c_{2} (λ_{2} - λ_{1}) v_{2} = 0

. Since

λ_{2} \neq = λ_{1}

and

v_{2} \neq = 0

c_{2} = 0

, then

c_{1} = 0

— so the eigenvectors, and hence the two modes, are independent.

True or false: a

2 \times 2

real matrix with complex eigenvalues has real eigenvectors.

False. Complex eigenvalues force complex eigenvectors

v = p + i q

; the real solutions are then rebuilt from the real part

p

and imaginary part

q

via Euler's Formula.

True or false: the general solution

x = c_{1} e^{λ_{1} t} + c_{2} e^{λ_{2} t}

is correct as long as you found the right

λ

's.

False. That expression is a scalar — each term must carry its eigenvector:

x = c_{1} e^{λ_{1} t} v_{1} + c_{2} e^{λ_{2} t} v_{2}

. Without

v_{i}

there is no direction and the answer isn't a vector.

True or false: if

λ = 0

is an eigenvalue, that mode contributes a solution that stays put.

True. The mode is

e^{0 \cdot t} v = v

, a constant vector. So there is a whole line of equilibria along

v

; nothing moves along it.

True or false: swapping to

det (λ I - A) = 0

instead of

det (A - λ I) = 0

changes the eigenvalues.

False. The two polynomials differ only by a factor

(- 1)^{n}

, so they have identical roots. Just stay consistent so the equation reads cleanly.

Spot the error

A student writes: "

A = (3013)

has eigenvalue

3

twice, so the solution is

c_{1} e^{3 t} v_{1} + c_{2} e^{3 t} v_{2}

with two eigenvectors."

Error: this matrix is defective — it has only one eigenvector

(0 1)

. The second mode must use a generalized eigenvector

w

with

(A - 3 I) w = v

, giving

e^{3 t} (t v + w)

with a genuine

t

factor.

A student solves complex

λ = α \pm i β

and keeps four solutions: two from

α + i β

and two from

α - i β

Error: double-counting. One conjugate pair yields exactly two independent real solutions (the real and imaginary parts of a single branch). The conjugate branch adds nothing new.

A student concludes: "

λ = 2 i

so trajectories grow, because

i

makes things blow up."

Error: growth/decay is controlled by the real part

α = Re (λ)

, here

0

. The imaginary part only sets rotation frequency, so trajectories are closed circles (a center), not blow-up.

A student cancels

e^{λ t}

from

λ e^{λ t} v = e^{λ t} A v

but worries it might be zero for some

t

No error, and the worry is unfounded:

e^{λ t} \neq = 0

for every real (and complex)

t

, so dividing it out is always legal. That's exactly why the exponential ansatz collapses to the algebraic eigenvalue equation.

A student writes the generalized-eigenvector solution as

e^{3 t} (t w + v)

Error: the roles are swapped. The correct form is

e^{λ t} (t v + w)

— the eigenvector

v

multiplies

t

, because it's

v

that

(A - λ I)

annihilates.

A student says "the origin is always a solution, so it's stable."

Error:

x \equiv 0

is always an equilibrium, but stability depends on the eigenvalues. A positive real part makes nearby trajectories fly away — the origin is then unstable.

Why questions

Why do we guess

x = e^{λ t} v

rather than some other function?

Because in a fixed direction

v

the equation must reduce to a scalar equation

u^{'} = λ u

, whose only solution is an exponential. The exponential is the one function that reproduces itself under differentiation.

Why does a repeated eigenvalue sometimes still give two eigenvectors and not require generalized ones?

Because algebraic multiplicity and geometric multiplicity can be equal — e.g.

A = 3 I

has

λ = 3

twice but a full 2-dimensional eigenspace. Only when eigenvectors run short (

A

is defective) do you need the

t v + w

trick.

Why does the

t

factor appear in the defective case?

Substituting

e^{λ t} (t v + w)

and using

(A - λ I) w = v

makes both sides of

x^{'} = A x

balance. The

t

supplies the extra degree of freedom the missing eigenvector couldn't.

Why does a complex eigenvalue

α \pm i β

produce spirals rather than straight-line motion?

Because

e^{λ t} = e^{α t} (cos β t \pm i sin β t)

by Euler's Formula:

α

scales radius (grow/decay) while

β

rotates. Straight-line motion needs a real eigenvector, which complex

λ

lacks.

Why can any solution be written as a superposition of the eigenvector solutions?

Because the system is linear, so sums of solutions are solutions, and the

n

independent eigenvector modes span an

n

-parameter family — matching the

n

free constants a first-order

n

-dimensional system must have.

Why does the sign of the real part alone decide long-run behaviour, ignoring the eigenvectors?

Eigenvectors only set the directions of the modes; the factor

e^{(Re λ) t}

decides whether each mode's amplitude grows or shrinks. So stability is read straight off

Re (λ)

— see Phase Portraits and Stability.

Edge cases

Refer back to the figure above for each portrait shape named here.

What is the solution when

A

is already diagonal,

A = diag (λ_{1}, λ_{2})

The variables are already decoupled, so

x_{1} = c_{1} e^{λ_{1} t}

x_{2} = c_{2} e^{λ_{2} t}

independently. The standard basis vectors are the eigenvectors — no mixing to undo.

What happens if

A = 0

(the zero matrix)?

Then

x^{'} = 0

, so every solution is a constant

x (t) = x_{0}

. Every direction is an eigenvector with

λ = 0

; the whole plane is equilibria.

What if

A

has one eigenvalue

λ = 0

and one

λ < 0

The

λ = 0

mode gives a fixed direction (a line of equilibria) while the

λ < 0

mode decays onto it. Trajectories slide toward that line — a degenerate / non-isolated stable case.

What is the phase portrait when

λ = α \pm i β

with

α > 0

An outward spiral (unstable spiral / focus): each turn is scaled by

e^{α t} > 1

, so the radius grows while

β

keeps it rotating.

What is the phase portrait when

λ = α \pm i β

with

α < 0

An inward spiral (stable spiral / focus): each turn is scaled by

e^{α t} < 1

, so the radius shrinks toward the origin while

β

keeps it rotating — trajectories spiral in.

What if the two real eigenvalues have opposite signs,

λ_{1} > 0 > λ_{2}

A saddle: trajectories are pushed out along

v_{1}

and pulled in along

v_{2}

. Only points starting exactly on the

v_{2}

line reach the origin; all others escape.

What if

β = 0

inside a supposedly complex pair — i.e. the discriminant collapses?

Then the eigenvalues are actually real and equal (

λ = α

); there is no rotation. You've hit the repeated-eigenvalue case and must check whether

A

is defective.

Recall One-line summary of the traps

Count eigenvectors not eigenvalues; keep the eigenvector attached to each exponential term; read growth from $Re (λ)$ and rotation from $Im (λ)$ ; and always check the degenerate boundaries ( $λ = 0$ , repeated, defective, $α = 0$ ).