Complex Conjugate Pair of Eigenvalues¶

Proposition¶

Let $A$ be a real $n\times n$ matrix and let $\lambda=a+b\mathrm{i}$ with $a,b\in\mathbb{R}$$\mathbb{R}$, $b\ne0$$0$, be an eigenvalue of $A$ with eigenvector $\boldsymbol{p}\in\mathbb{C}^n$${\mathbb{C}}^n$. Then the complex conjugate $\bar\lambda=a-b\mathrm{i}$ is also an eigenvalue of $A$, with eigenvector $\bar{\boldsymbol{p}}$ (entrywise conjugate), and $\lambda$ and $\bar\lambda$ have the same algebraic multiplicity.

Proof. Conjugating $A\boldsymbol{p}=\lambda\boldsymbol{p}$ and using $\bar{A}=A$ gives $A\bar{\boldsymbol{p}}=\bar\lambda\bar{\boldsymbol{p}}$$\bar{\lambda }\bar{\mathbf{p}}$, and $\bar{\boldsymbol{p}}\ne\boldsymbol{0}$$0$. The multiplicity statement holds because the characteristic polynomial has real coefficients, so its factorization is invariant under conjugation.

Interpretation¶

The characteristic polynomial $\det(A-\lambda I)$ of a real matrix has real coefficients, so this is the complex conjugate root theorem from Math II applied to eigenvalues. The hypothesis that $A$ is real is essential: the $1\times1$ complex matrix $(\mathrm{i})$ has eigenvalue $\mathrm{i}$ but not $-\mathrm{i}$.

The figure uses the numerical matrix from the examples, $A=\begin{pmatrix}1&-2\\1&3\end{pmatrix}$$\left(\begin{array}{cc}1& -2\\ 1& 3\end{array}\right)$. On the left its eigenvalues $\lambda=2+\mathrm{i}$ and $\bar\lambda=2-\mathrm{i}$ are reflections of each other in the real axis, with the same modulus $\rho=\sqrt{5}$ and opposite angles $\pm\theta$. On the right are the real and imaginary parts $\boldsymbol{u}=(-1,1)^{\top}$ and $\boldsymbol{w}=(1,0)^{\top}$ of the eigenvector $\boldsymbol{p}$: the image $A\boldsymbol{u}=(-3,2)^{\top}=2\boldsymbol{u}-\boldsymbol{w}$ stays in the plane of $\boldsymbol{u}$ and $\boldsymbol{w}$ rather than on the line through $\boldsymbol{u}$.

A nonreal eigenvalue has no real eigenvector, so it does not describe a stretched line. Instead the pair acts on a real plane: writing $\boldsymbol{p}=\boldsymbol{u}+\mathrm{i}\boldsymbol{w}$ with real $\boldsymbol{u},\boldsymbol{w}$, the plane $\operatorname{span}\{\boldsymbol{u},\boldsymbol{w}\}$ is mapped into itself, and in the coordinates given by $\boldsymbol{u},\boldsymbol{w}$ the coordinate block is $\begin{pmatrix}a&b\\-b&a\end{pmatrix}=\rho R_{-\theta}$$\rho R_{-\theta }$, with $\lambda=\rho \mathrm{e}^{\mathrm{i}\theta}$ in polar form. Thus the coordinate vector rotates by $-\theta$ and scales by $\rho$ each step. The basis $\boldsymbol{u},\boldsymbol{w}$ need not be orthonormal, so this is a similarity description: lengths and angles in the original Euclidean plane need not change by those same rules.

The pair matters for linear difference equations: for $\boldsymbol{x}_{t+1}=A\boldsymbol{x}_t$ it produces oscillating terms $\rho^t\cos(\theta t)$ and $\rho^t\sin(\theta t)$, which decay exactly when $\rho=\lvert\lambda\rvert<1$$1$.

Key Results¶

• Nonreal eigenvalues of a real matrix come in pairs, so their number (with multiplicity) is even; a real $n\times n$ matrix with $n$ odd has at least one real eigenvalue.
• If $\boldsymbol{p}=\boldsymbol{u}+\mathrm{i}\boldsymbol{w}$ is an eigenvector for $a+b\mathrm{i}$$b\mathrm{i}$, $b\neq0$$0$, then $\boldsymbol{u},\boldsymbol{w}$ are linearly independent and $A\boldsymbol{u}=a\boldsymbol{u}-b\boldsymbol{w}$$b\mathbf{w}$, $A\boldsymbol{w}=b\boldsymbol{u}+a\boldsymbol{w}$$a\mathbf{w}$, i.e. $A\,[\boldsymbol{u}\ \boldsymbol{w}]=[\boldsymbol{u}\ \boldsymbol{w}]\begin{pmatrix}a&b\\-b&a\end{pmatrix}$$[\mathbf{u}\mathbf{w}]\left(\begin{array}{cc}a& b\\ -b& a\end{array}\right)$.
• The pair contributes $\lambda\bar\lambda=a^2+b^2=\lvert\lambda\rvert^2>0$ to $\det A$ and $2a$ to $\operatorname{tr}A$.
• A $2\times2$ real matrix has a nonreal pair if and only if $(\operatorname{tr}A)^2-4\det A<0$$0$.
• By De Moivre's formula, $\lambda^t=\rho^t\big(\cos(\theta t)+\mathrm{i}\sin(\theta t)\big)$$\mathrm{i}\sin (\theta t))$, so $\lambda^t\to0$ if and only if $\rho<1$$1$.

Examples¶

A numerical matrix. For $A=\begin{pmatrix}1&-2\\1&3\end{pmatrix}$$\left(\begin{array}{cc}1& -2\\ 1& 3\end{array}\right)$, $\operatorname{tr}A=4$ and $\det A=5$$5$, so $\lambda^2-4\lambda+5=0$ and $\lambda=2\pm\mathrm{i}$$\mathrm{i}$. From the second row of $A-(2+\mathrm{i})I$$\mathrm{i})I$, namely $p_1+(1-\mathrm{i})p_2=0$$0$, an eigenvector is $\boldsymbol{p}=(-1+\mathrm{i},\,1)^{\top}$$\mathrm{i},1{)}^{\top }$; hence $(-1-\mathrm{i},\,1)^{\top}$ belongs to $2-\mathrm{i}$$\mathrm{i}$. Here $\boldsymbol{u}=(-1,1)^{\top}$$(-1,1{)}^{\top }$, $\boldsymbol{w}=(1,0)^{\top}$$(1,0{)}^{\top }$, and indeed $A\boldsymbol{u}=(-3,2)^{\top}=2\boldsymbol{u}-\boldsymbol{w}$$\mathbf{w}$. Both eigenvalues have modulus $\sqrt{5}$.

Rotation. $R_\theta=\begin{pmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{pmatrix}$ has characteristic polynomial $\lambda^2-2\cos\theta\,\lambda+1$ and eigenvalues $\mathrm{e}^{\pm\mathrm{i}\theta}$, with eigenvectors $(1,-\mathrm{i})^{\top}$ and $(1,\mathrm{i})^{\top}$. For $\theta$ not a multiple of $\pi$ they are nonreal and have modulus $1$.

Odd dimension. A real $3\times3$ matrix cannot have eigenvalues $\mathrm{i},1,2$: the conjugate $-\mathrm{i}$ would be a fourth eigenvalue.

Prerequisites¶

• Eigenvalue and Eigenvector
• Characteristic Polynomial
• Complex Conjugate
• Complex Conjugate Root Theorem

Related Concepts¶

• Second-Order Linear Difference Equation
• Stability of a System of Linear Difference Equations
• Matrix Representation of a Linear Map
• De Moivre's Formula

Sources¶

• Math V: Advanced Linear Algebra · Week 1

Aliases: conjugate pair of eigenvalues, complex conjugate eigenvalues, complex eigenvalues of a real matrix, conjugate eigenvalues Keywords: a + bi and a - bi, conjugate eigenvector, real canonical form, rotation-scaling block Symbols: complex numbers, element of, lambda, pi, real numbers, rho, square root, theta

Last updated: 27 October 2026 11:52