Complex Eigenvalues: Conjugate Pairs & Rotation

When Complex Eigenvalues Appear

A real matrix with real entries can have complex eigenvalues. This occurs when the characteristic polynomial — a polynomial with real coefficients — has roots that are not real.

For a

2 \times 2

matrix, the discriminant of the characteristic polynomial determines the eigenvalue type:

Discriminant Classification 2x2

\Delta = \text{tr}(A)^2 - 4\det(A) \quad \begin{cases} \Delta > 0: & \text{two distinct real eigenvalues} \\ \Delta = 0: & \text{one repeated real eigenvalue} \\ \Delta < 0: & \text{complex conjugate pair} \end{cases}

Discriminant classification 2x2 →

Complex eigenvalues appear in the

\Delta < 0

case: the quadratic formula produces

\lambda = \frac{\text{tr}(A) \pm \sqrt{\text{negative}}}{2}

, which involves

\sqrt{-1} = i

.

The geometric interpretation is clear: no real direction in

\mathbb{R}^2

is mapped to a scalar multiple of itself. Every vector is rotated, not just stretched or compressed. The simplest example is rotation by

90°

, with matrix

\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}

and characteristic polynomial

\lambda^2 + 1 = 0

. The eigenvalues are

\pm i

— purely imaginary, reflecting pure rotation with no scaling.

Complex Eigenvalue Notation

Notation

Complex Eigenvalue Notation

A bar that has to be applied twice — once to a number and once to a whole vector — plus a

\pm

that is a statement about pairing rather than a shorthand for two options, and a normal form whose block is chosen so that no complex number has to be written down at all. All catalogued among the linear algebra symbols.

\lambda

, the spectrum and the characteristic polynomial come from eigenvalue notation; the overbar and

a + bi

from conjugate and complex number notation;

\Delta

as discriminant from quadratic notation;

P^{-1}AP

from matrix representation notation.

A\mathbf{v} = \lambda\mathbf{v} \;\Rightarrow\; A\bar{\mathbf{v}} = \bar{\lambda}\bar{\mathbf{v}}

conjugating both the eigenvalue and the eigenvector gives another eigenpair

The overbar works at two levels here: over

\lambda

it conjugates a single number, over

\mathbf{v}

it conjugates every entry of a vector. The same mark, applied to objects of different type — and the implication only holds because

A

is real, as Conjugate Pairs below proves.

CasesPhysics and much of numerical linear algebra prefer

A^{*}

or

A^{H}

for the conjugate transpose of a matrix, reserving the plain bar for entrywise conjugation without the transpose; the two differ, and on a real matrix the distinction disappears, which is why it goes unnoticed until complex entries arrive.

Do not confuseConjugating only the eigenvalue.

\bar{\lambda}

pairs with

\bar{\mathbf{v}}

, not with

\mathbf{v}

— the original eigenvector does not belong to the conjugate eigenvalue, and dropping the bar on the vector produces a statement that is simply false.

\lambda = a \pm bi

a pair of eigenvalues, not a choice between two

Elsewhere

\pm

offers alternatives; here it asserts that both values occur, together, as a package — a real matrix cannot have one without the other. The compression is honest because the pair always appears in full, which is what makes the single line a complete description of two eigenvalues.

CasesThe sign convention runs through the geometry:

b

is the rotation rate and its sign fixes the direction of turning, so writing

a \pm bi

records a rotation together with its mirror image;

b = 0

collapses the pair to a single real eigenvalue and the notation degenerates gracefully.

Do not confuseThe

\pm

of the quadratic formula, where the two signs are two candidate roots you may pick between. Here selecting one is not an option — the eigenvalues are both present in the matrix.

\begin{pmatrix} a & -b \\ b & a \end{pmatrix}

the real canonical block standing in for the pair a plus or minus b i

A

2 \times 2

block that represents a conjugate pair without writing any complex number — the whole point of Real Canonical Form below. The off-diagonal sign asymmetry is the notation:

-b

above the diagonal and

b

below is what encodes rotation.

CasesScaled out, the block reads

r\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{pmatrix}

with

r = |\lambda|

and

\theta = \arg\lambda

— the modulus and argument of the eigenvalue reappearing as a scaling factor and a rotation angle.

Do not confuseA diagonal matrix. The whole reason this form exists is that a real matrix with complex eigenvalues cannot be diagonalised over

\mathbb{R}

; the block is the closest real substitute, and treating it as diagonal loses the rotation it was built to carry.

All linear algebra symbols →↑ Eigenvalues and Eigenvectors

Conjugate Pairs

For a real matrix, complex eigenvalues always come in conjugate pairs. If

\lambda = a + bi

is an eigenvalue (with

b \neq 0

), then

\bar{\lambda} = a - bi

is also an eigenvalue. The corresponding eigenvectors are also conjugates:

Complex Conjugate Pairs

A \in \mathbb{R}^{n \times n}, \;\; A\mathbf{v} = \lambda\mathbf{v} \Rightarrow A\bar{\mathbf{v}} = \bar{\lambda}\bar{\mathbf{v}}

Complex conjugate pairs →

The proof uses the fact that the characteristic polynomial has real coefficients. If

p(\lambda) = 0

and every coefficient of

p

is real, then

p(\bar{\lambda}) = \overline{p(\lambda)} = \overline{0} = 0

. Complex conjugation passes through the polynomial because the coefficients are their own conjugates.

For the eigenvector identity: applying complex conjugation to

A\mathbf{v} = \lambda\mathbf{v}

and using that

A

has real entries gives

A\bar{\mathbf{v}} = \bar{\lambda}\bar{\mathbf{v}}

.

One consequence for odd-dimensional real matrices: at least one eigenvalue must be real. Complex roots pair up, consuming an even number of the

n

roots. When

n

is odd, at least one root is left unpaired, and an unpaired root of a real polynomial must be real.

The 2×2 Case in Detail

A

2 \times 2

real matrix with complex eigenvalues

\lambda = a \pm bi

(where

b \neq 0

) acts as a rotation composed with a scaling. The eigenvalues have modulus

r = \sqrt{a^2 + b^2}

and argument

\theta = \arctan(b/a)

.

When

r = 1

(equivalently,

\det(A) = a^2 + b^2 = 1

), the transformation is a pure rotation by angle

\theta

. When

r > 1

, it is a rotation with outward spiraling. When

r < 1

, it is a rotation with inward spiraling.

Worked Example

For

A = \begin{pmatrix} 3 & -2 \\ 1 & 1 \end{pmatrix}

:

\text{tr}(A) = 4

,

\det(A) = 5

, discriminant

= 16 - 20 = -4

. The eigenvalues are

\lambda = \frac{4 \pm \sqrt{-4}}{2} = 2 \pm i

The modulus is

r = \sqrt{4 + 1} = \sqrt{5}

and the argument is

\theta = \arctan(1/2)

. The transformation rotates by

\arctan(1/2) \approx 26.6°

while scaling distances by

\sqrt{5} \approx 2.24

. Since

r > 1

, repeated application spirals outward.

Download

PNGRaster · 2×PDFPrint quality

Eigen · complex pairs

Three regimes, set by the modulus

The argument $\theta$ fixes how fast the plane turns and never affects whether it grows. That is decided entirely by the modulus, and the three cases below are the only ones there are.

3regimes

Neither growing nor decaying1

On the unit circle§ 3

r = 1

geometric actionpure rotation by

\theta

, no scaling

det of the 2×2 block

\det A = 1

orbit of a pointa closed circle

Lengths are preserved exactly, so the matrix is orthogonal. Every orbit closes if

\theta

is a rational multiple of

2\pi

and fills the circle densely otherwise — the boundary case in every stability argument.

Growing or decaying2

Outside the unit circle§ 7

r > 1

geometric actionrotation with outward scaling by

r

det of the 2×2 block

\det A > 1

orbit of a pointa spiral outward

Each application turns by

\theta

and stretches by

r

, so iterates diverge while circling. In a discrete system this is instability — and the divergence is geometric, so it is fast.

Inside the unit circle§ 7

r < 1

geometric actionrotation with inward scaling by

r

det of the 2×2 block

\det A < 1

orbit of a pointa spiral inward to the origin

The stable case for iteration: every orbit converges to the origin, still turning as it goes. This is what

|\lambda| < 1

means geometrically, and why the condition is stated on the modulus rather than on the real part.

The determinant of the

2 \times 2

block is

a^2 + b^2 = r^2

, so the area scaling and the modulus are the same fact stated twice — which is why the determinant can be read as a stability marker without computing an eigenvalue at all.

Three regimes, set by the modulus·/linear-algebra/eigen/complexLearn Math Class

The three cases exhaust the possibilities because

r

is a positive real and can only be less than, equal to, or greater than one. The argument

\theta

plays no part in the classification — it sets the speed of rotation and nothing else, which is why two systems can spiral at completely different rates and be equally stable.

Every direction turned, none left fixed

No arrow survives this matrix pointing where it started: every direction is rotated. That is what a complex eigenvalue looks like from the real plane — not an exotic number but the absence of any invariant line. The rotation angle is the argument of the eigenvalue. Turn the matrix and watch the fixed directions vanish on the complex eigenvalue explorer.

The conjugate partner of this eigenvalue describes the same rotation turning the other way, which is why the pair always arrives together.

Complex Eigenvectors

The eigenvectors corresponding to complex eigenvalues have complex entries. To find them, solve

(A - \lambda I)\mathbf{v} = \mathbf{0}

using complex arithmetic.

Continuing the example with

A = \begin{pmatrix} 3 & -2 \\ 1 & 1 \end{pmatrix}

and

\lambda = 2 + i

:

A - (2 + i)I = \begin{pmatrix} 1 - i & -2 \\ 1 & -1 - i \end{pmatrix}

The second row gives

v_1 + (-1 - i)v_2 = 0

, so

v_1 = (1 + i)v_2

. Setting

v_2 = 1

:

\mathbf{v} = (1 + i, 1)^T

.

This eigenvector splits into real and imaginary parts:

\mathbf{v} = \underbrace{(1, 1)^T}_{\mathbf{u}} + i\underbrace{(1, 0)^T}_{\mathbf{w}}

. The real vectors

\mathbf{u}

and

\mathbf{w}

encode the rotation — they span the two-dimensional subspace on which the rotation-scaling acts. The conjugate eigenvector for

\bar{\lambda} = 2 - i

is

\bar{\mathbf{v}} = (1 - i, 1)^T

.

Real Canonical Form

A

2 \times 2

real matrix with eigenvalues

a \pm bi

is similar (over

\mathbb{R}

) to a rotation-scaling block:

Real Canonical Form 2x2

\lambda = a \pm bi \Rightarrow P^{-1}AP = \begin{pmatrix} a & -b \\ b & a \end{pmatrix} = r\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}

Real canonical form 2x2 →

where

r = \sqrt{a^2 + b^2}

and

\theta = \arctan(b/a)

. This is a rotation by

\theta

scaled by

r

.

The similarity is achieved by the real matrix

P = [\mathbf{u} \; \mathbf{w}]

, where

\mathbf{v} = \mathbf{u} + i\mathbf{w}

is the complex eigenvector. Then

P^{-1}AP = \begin{pmatrix} a & -b \\ b & a \end{pmatrix}

.

This is the real alternative to diagonalization: instead of a diagonal matrix with complex entries (which is valid over

\mathbb{C}

but not over

\mathbb{R}

), we get a real

2 \times 2

rotation-scaling block. The transformation is expressed in its simplest real form.

Complex Eigenvalues in Larger Matrices

For an

n \times n

real matrix, complex eigenvalues still come in conjugate pairs. Each pair contributes a

2 \times 2

rotation-scaling block to the real canonical form. Real eigenvalues contribute

1 \times 1

blocks (just the eigenvalue itself).

A

4 \times 4

matrix with eigenvalues

2 \pm 3i

and

-1 \pm i

has real canonical form

\begin{pmatrix} 2 & -3 & 0 & 0 \\ 3 & 2 & 0 & 0 \\ 0 & 0 & -1 & -1 \\ 0 & 0 & 1 & -1 \end{pmatrix}

A

3 \times 3

matrix with eigenvalues

5

and

1 \pm 2i

has real canonical form

\begin{pmatrix} 5 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 2 & 1 \end{pmatrix}

The real canonical form is the real analogue of diagonalization: it achieves the simplest possible real matrix that is similar to

A

, with the block structure directly reflecting the eigenvalue spectrum.

Dynamical Systems Interpretation

Complex eigenvalues produce oscillatory behavior in dynamical systems.

In the discrete system

\mathbf{x}_{n+1} = A\mathbf{x}_n

, an eigenvalue

\lambda = a + bi

with modulus

r = |\lambda| = \sqrt{a^2 + b^2}

generates a spiraling trajectory. When

r > 1

, the spiral grows outward — the system is unstable. When

r < 1

, the spiral decays inward — the system converges to the origin. When

r = 1

, the trajectory traces a closed curve — perpetual oscillation without growth or decay.

In the continuous system

\mathbf{x}' = A\mathbf{x}

, an eigenvalue

\lambda = a + bi

contributes a term

e^{at}(\cos bt, \sin bt)

to the solution. The real part

a

determines exponential growth (

a > 0

) or decay (

a < 0

). The imaginary part

b

determines the oscillation frequency. When

a = 0

, the oscillation is sustained. When

a < 0

, it is damped. When

a > 0

, it grows without bound.

Stability of a linear system reduces to eigenvalue analysis: the system is stable if and only if every eigenvalue has negative real part (continuous) or modulus less than

1

(discrete).

Regime	Discrete: x_n+1 = A x_n	Continuous: x′ = A x
Stable (converging)	\|λ\| < 1 → trajectory spirals inward to the origin	Re(λ) < 0 → damped oscillation decaying to zero
Marginally stable	\|λ\| = 1 → trajectory traces a closed (periodic) curve	Re(λ) = 0 → sustained oscillation at constant amplitude
Unstable (diverging)	\|λ\| > 1 → trajectory spirals outward without bound	Re(λ) > 0 → oscillation with exponentially growing amplitude

Spiralling in, then spiralling out

The rotation is the same in both; what differs is the modulus of the eigenvalue. Below one, repeated application pulls everything towards the origin; above one, it throws everything outward. The argument sets how fast it turns and the modulus sets whether it survives, which is the whole stability story for a discrete system. Tune both on the complex eigenvalue explorer.

Whether a system settles, oscillates forever or blows up is therefore read off a single number: the modulus.

The Fundamental Theorem of Algebra Connection

Over

\mathbb{C}

, every polynomial of degree

n

factors completely into

n

linear factors. This means every

n \times n

matrix — real or complex — has exactly

n

eigenvalues counted with algebraic multiplicity when the characteristic polynomial is factored over

\mathbb{C}

.

Over

\mathbb{R}

, some factors may be irreducible quadratics rather than linear, corresponding to conjugate pairs of complex eigenvalues. The characteristic polynomial of a real matrix factors into linear terms (real eigenvalues) and irreducible quadratic terms (complex conjugate pairs).

Working over

\mathbb{C}

simplifies the theory considerably. Every matrix is triangularizable over

\mathbb{C}

(Schur decomposition):

A = UTU^*

where

U

is unitary and

T

is upper triangular with eigenvalues on the diagonal. Diagonalizability depends only on whether the geometric multiplicities match the algebraic multiplicities, with no additional complications from irreducible quadratics.

The choice between working over

\mathbb{R}

and

\mathbb{C}

is a recurring theme. Real matrices are the natural objects of computation, but complex eigenvalues are the natural objects of spectral theory. Both perspectives are needed.

Summary: Complex Eigenvalues at a Glance

The behavior of complex eigenvalues for a real matrix is governed by a handful of formulas and statements that connect the discriminant, the modulus, the argument, and the real part to specific geometric and dynamical consequences. The table below collects each one alongside what it produces and what it means, providing a single reference for the page's key facts.

A complex eigenvalue is often met as a nuisance — the characteristic polynomial refuses to factor over the reals and the calculation appears to have failed. It has not. The pair is describing a rotation, and once that is granted the two numbers it carries are exactly the two things a rotation has: an angle and a scale. The grouping below is recognition first, then those two numbers, then what they predict.

Download

PNGRaster · 2×PDFPrint quality

Eigen · complex eigenvalues

Reading a conjugate pair

A complex pair is not a failure of the real case — it is a rotation, and its two parts say by how much and through what angle. The modulus and argument are the whole content, and everything below is one of them applied.

8readings

Recognising the pair3

1

When they appear§ 1

2 \times 2

real matrices

\operatorname{tr}(A)^2 - 4\det(A) < 0

The discriminant settles it before any solving. A negative value means no real direction is mapped to a multiple of itself — the transformation turns every vector, so no real eigenvector can exist.

2

Conjugate pairing§ 2

A

real — not for complex entries

\lambda = a + bi \Rightarrow \bar{\lambda} = a - bi

Complex roots of a real polynomial arrive in pairs, and the eigenvectors pair with them:

\mathbf{v}

and

\bar{\mathbf{v}}

. Nothing has to be computed twice — finding one half of the pair gives the other by conjugation.

3

Odd dimension§ 6

real matrices only

n

odd

\Rightarrow

at least one real

\lambda

Complex roots come in pairs, so an odd number of them cannot pair off entirely and at least one must be real. This is why a rotation in

\mathbb{R}^3

has an axis while a rotation in

\mathbb{R}^2

has none — the leftover real eigenvalue is the axis.

Reading the pair3

4

Modulus§ 3

the scaling factor

r = \sqrt{a^2 + b^2}

How much the transformation stretches the plane per application. Since

r^2 = \det

of the

2 \times 2

block, the determinant is the area scaling and the modulus is its square root — consistent with the geometric reading.

5

Argument§ 3

the rotation angle

\theta = \arctan(b/a)

How far the plane turns per application. A pair with

a = 0

is a quarter turn; the sign of

b

picks the direction, and the two conjugates give the two orientations of the same rotation.

6

Real canonical form§ 5

similar over

\mathbb{R}

, no complex arithmetic

\begin{bmatrix} a & -b \\ b & a \end{bmatrix} = r \cdot R(\theta)

The pair does not have to be handled in

\mathbb{C}

at all. Over

\mathbb{R}

the matrix is similar to a scaled rotation block, so the whole analysis stays real — which is what makes complex eigenvalues usable in applied work.

What the pair predicts2

7

Discrete stability§ 7

for

\mathbf{x}_{k+1} = A\mathbf{x}_k

|\lambda| < 1

for every eigenvalue

The modulus alone decides whether iteration spirals inward or outward; the argument only sets how fast it turns. Note

|\lambda| = 1

is the boundary case — a pure rotation that neither grows nor decays.

8

Continuous stability§ 7

for

\mathbf{x}' = A\mathbf{x}

\operatorname{Re}(\lambda) < 0

for every eigenvalue

A different criterion for a different system, and confusing the two is the usual error. Discrete systems compare the modulus against one; continuous systems compare the real part against zero — because the solution is

e^{\lambda t}

and its growth is governed by

\operatorname{Re}(\lambda)

.

The single idea underneath: a real matrix with eigenvalues

re^{\pm i\theta}

acts on the corresponding real plane as a rotation by

\theta

scaled by

r

. Recognition, canonical form and stability are all that statement read at different moments — which is why the modulus and argument are worth computing before anything else.

Reading a conjugate pair·/linear-algebra/eigen/complexLearn Math Class

The real canonical form is what keeps this practical. A conjugate pair can be handled entirely in

\mathbb{R}

: the matrix is similar to a block

\begin{bmatrix} a & -b \\ b & a \end{bmatrix}

, which is a rotation by

\theta

scaled by

r

, and no complex arithmetic is required at any point. The complex numbers are a route to the answer rather than a property of it.

The two stability criteria are worth separating carefully, because they look alike and are not. A discrete system

\mathbf{x}_{k+1} = A\mathbf{x}_k

is stable when every

|\lambda| < 1

— the modulus decides, since each step multiplies by it. A continuous system

\mathbf{x}' = A\mathbf{x}

is stable when every

\operatorname{Re}(\lambda) < 0

— the real part decides, since the solution behaves as

e^{\lambda t}

. Applying the wrong test gives a confidently wrong answer.

Complex Eigenvalues FAQ

Does $\bar{\lambda}$ pair with $\mathbf{v}$ or with $\bar{\mathbf{v}}$ ?

+

With

\bar{\mathbf{v}}

. Conjugating the whole eigenvalue equation conjugates the vector as well, so the conjugate eigenvalue belongs to the conjugate eigenvector. The original

\mathbf{v}

does not satisfy the equation for

\bar{\lambda}

, and writing it that way produces a statement that is simply false rather than merely imprecise.Read more →

Is the $\pm$ in $a \pm bi$ like the one in the quadratic formula?

+

It looks identical but does a different job. In the quadratic formula the two signs offer two candidate roots and you pick whichever the problem needs. Here selecting one is not an option: both eigenvalues genuinely belong to the matrix at the same time, and the mark is recording a pair rather than a choice.Read more →

Is the real canonical block just a diagonal matrix?

+

No, and that is exactly why it exists. A real matrix with complex eigenvalues cannot be diagonalised over the reals at all, so the block is the closest real substitute. Written out, it is a scaled rotation, with the eigenvalue's modulus setting the scale and its argument the angle. Treating it as diagonal discards the rotation.Read more →

Complex Eigenvalues

When No Real Direction Survives

When Complex Eigenvalues Appear

Complex Eigenvalue Notation

Complex Eigenvalue Notation

Conjugate Pairs

The 2×2 Case in Detail

Worked Example

Three regimes, set by the modulus

Complex Eigenvectors

Real Canonical Form

Complex Eigenvalues in Larger Matrices

Dynamical Systems Interpretation

The Fundamental Theorem of Algebra Connection

Summary: Complex Eigenvalues at a Glance

Reading a conjugate pair

Complex Eigenvalues FAQ

Does λˉ\bar{\lambda}λˉ pair with v\mathbf{v}v or with vˉ\bar{\mathbf{v}}vˉ?

Is the ±\pm± in a±bia \pm bia±bi like the one in the quadratic formula?

Is the real canonical block just a diagonal matrix?

Does $\bar{\lambda}$ pair with $\mathbf{v}$ or with $\bar{\mathbf{v}}$ ?

Is the $\pm$ in $a \pm bi$ like the one in the quadratic formula?