Complex matrices

Lecture 30

Minjae Park

Auburn University
MATH 2660 - Spring 2026

March 30, 2026

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Recap

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Solving linear ODEs

• Higher-order linear ODEs can be rewritten as a system of linear ODEs: \[ Y'(t)=AY(t). \]

• The solution is given by \[ Y(t)=e^{At}Y(0). \]

• If \(A\) is diagonalizable (\(A=PDP^{-1}\)), then \[ e^{At}=Pe^{Dt}P^{-1}, \] which makes computation much easier because \(e^{Dt}\) is just exponentials of diagonal entries.

• In particular, if \(\vec{u}_1,\dots,\vec{u}_n\) are eigenvectors with eigenvalues \(\lambda_1,\dots,\lambda_n\), then the general solution is \[ Y(t)=c_1 e^{\lambda_1 t}\vec{u}_1 + \cdots + c_n e^{\lambda_n t}\vec{u}_n. \]

• If an initial condition \(Y(0)\) is given, the constants \(c_1,\dots,c_n\) can be determined by solving a linear system.

Example

• Consider the second-order ODE \[ y'' + 3y' + 2y = 0, \] with \(y(0)=1\), \(y'(0)=0\).

• Convert to a first-order system. Let \[ Y(t)=\langle y(t),y'(t) \rangle. \] Then \[ Y'(t)=\langle y',y'' \rangle=\langle y',-3y'-2y \rangle =\begin{pmatrix} 0 & 1\\ -2 & -3 \end{pmatrix}Y(t). \]

• So \[ A=\begin{pmatrix} 0 & 1\\ -2 & -3 \end{pmatrix}. \]

• Compute eigenvalues: \[ \det(A-\lambda I)= \begin{vmatrix} -\lambda & 1\\ -2 & -3-\lambda \end{vmatrix} =\lambda^2+3\lambda+2 =(\lambda+1)(\lambda+2). \]

• Thus eigenvalues are \(\lambda_1=-1\), \(\lambda_2=-2\) (distinct real eigenvalues).

• Find eigenvectors:

  • For \(\lambda=-1\): \[ (A+I)\vec{u}=0 \Rightarrow \begin{pmatrix} 1 & 1\\ -2 & -2 \end{pmatrix}\vec{u}=0 \Rightarrow \vec{u}_1=\langle 1,-1 \rangle. \]
  • For \(\lambda=-2\): \[ (A+2I)\vec{u}=0 \Rightarrow \begin{pmatrix} 2 & 1\\ -2 & -1 \end{pmatrix}\vec{u}=0 \Rightarrow \vec{u}_2=\langle 1,-2 \rangle. \]

• General solution: \[ Y(t)=c_1 e^{-t}\langle 1,-1 \rangle + c_2 e^{-2t}\langle 1,-2 \rangle. \]

• Extract \(y(t)\) (first component): \[ y(t)=c_1 e^{-t} + c_2 e^{-2t}. \]

• Use initial conditions: \[ y(0)=c_1+c_2=1, \] \[ y'(t)=-c_1 e^{-t}-2c_2 e^{-2t} \Rightarrow y'(0)=-c_1-2c_2=0. \]

• Solve: \[ \begin{cases} c_1+c_2=1\\ -c_1-2c_2=0 \end{cases} \Rightarrow c_2=-1,\; c_1=2. \]

• Final solution: \[ y(t)=2e^{-t}-e^{-2t}. \]

Complex matrices

Preliminary: complex numbers

• Complex numbers are of the form \(z=a+bi\) where \(i^2=-1\) is the imaginary unit.

• We denote by \(\operatorname{Re}(z)=a\) the real part of \(z\) and by \(\operatorname{Im}(z)=b\) the imaginary part of \(z\).

• Let \[ z_1=a_1+b_1 i,\quad z_2=a_2+b_2 i. \]

• Addition: \[ z_1+z_2=(a_1+a_2)+(b_1+b_2)i. \]

• Multiplication: \[ z_1 z_2=(a_1+b_1 i)(a_2+b_2 i) \] \[ = a_1a_2 - b_1b_2 + (a_1b_2 + a_2b_1)i, \] using \(i^2=-1\).

• Complex conjugate:

  • For \(z=a+bi\), define \[ \overline{z}=a-bi. \]

• Properties: \[ z\overline{z}=a^2+b^2 \quad (\text{always real and nonnegative}), \] \[ \operatorname{Re}(z)=\frac{z+\overline{z}}{2}, \quad \operatorname{Im}(z)=\frac{z-\overline{z}}{2i}. \]

• Example: \[ (1+2i)(3-i)=3 - i + 6i -2i^2 = 5+5i. \]

Revisit an example

• We have discussed an example (spring motion, Hooke’s law):
  • By Newton’s law, \(F=ma\) and acceleration is \(y''(t)\).
  • For a spring, the restoring force is proportional to displacement: \[ F=-ky(t) \]
  • So: \[ m y''(t) = -k y(t) \]
  • This gives the ODE: \[ y''(t)+\frac{k}{m}y(t)=0. \]
• Solution: \[ y(t)=C_1\cos(\omega t)+C_2\sin(\omega t), \quad \omega=\sqrt{k/m}. \]

Example (continued)

• Let \(Y(t)=\langle y(t),y'(t) \rangle\). Then \[ Y'(t)=\langle y',y'' \rangle=\langle y',-\frac{k}{m}y \rangle =\begin{pmatrix} 0 & 1\\ -\frac{k}{m} & 0 \end{pmatrix} Y(t). \]

• So \[ A=\begin{pmatrix} 0 & 1\\ -\frac{k}{m} & 0 \end{pmatrix}. \]

• If \(A\) were diagonalizable over \(\mathbb R\), the solution would be a combination of real exponentials \(e^{\lambda t}\).

• Compute the characteristic polynomial: \[ \det(A-\lambda I)= \begin{vmatrix} -\lambda & 1\\ -\frac{k}{m} & -\lambda \end{vmatrix} =\lambda^2+\frac{k}{m}. \]

• The eigenvalues are: \[ \lambda=\pm i\sqrt{\frac{k}{m}}. \]

• These are complex numbers, so \(A\) is not diagonalizable over \(\mathbb R\).

• Question: can we still diagonalize \(A\) using complex numbers?

Complex eigenvalues

• For an \(n\times n\) matrix \(A\), if there is a nonzero vector with complex entries \(\vec{u}\in\mathbb C^n\) such that \[ A\vec{u} = \lambda \vec{u}, \] then \(\vec{u}\) is a complex eigenvector and \(\lambda\) is a complex eigenvalue.

• Complex eigenvalues are found from the same characteristic polynomial as before.

• Complex eigenvectors are obtained by solving \((A-\lambda I)\vec{u}=0\) over \(\mathbb C\).

• Geometric intuition:

  • Real eigenvectors correspond to directions that are preserved.
  • For complex eigenvalues, there is no real direction that is preserved.
  • Instead, the transformation behaves like a rotation (and possibly scaling) in the plane.

Example

• Let \[ A=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix} \] (rotation by \(90^\circ\)).

• Characteristic equation: \[ \det(A-\lambda I)=\lambda^2+1=0 \Rightarrow \lambda=\pm i. \]

• Eigenvectors:

  • For \(\lambda=i\): \[ (A-iI)\vec{u}=0 \Rightarrow \vec{u}=\langle 1,-i \rangle. \]
  • For \(\lambda=-i\): \[ (A+iI)\vec{u}=0 \Rightarrow \vec{u}=\langle 1,i \rangle. \]

• Interpretation:

  • This matrix rotates every real vector, so no real direction is fixed.
  • But in the complex space, it preserves complex lines.

Example (continued)

• For the spring system: \[ A=\begin{pmatrix} 0 & 1\\ -\omega^2 & 0 \end{pmatrix}, \quad \omega=\sqrt{\frac{k}{m}}. \]

• Eigenvalues: \[ \lambda=\pm i\omega. \]

• Solve \((A-\lambda I)\vec{u}=0\):

  • For \(\lambda=i\omega\), one eigenvector is \[ \vec{u}_1=\langle 1,i\omega \rangle. \]
  • For \(\lambda=-i\omega\), one eigenvector is \[ \vec{u}_2=\langle 1,-i\omega \rangle. \]

• Let \[ P=[\vec{u}_1 \ \vec{u}_2], \quad D=\begin{pmatrix} i\omega & 0\\ 0 & -i\omega \end{pmatrix}. \]

• Then \[ A=PDP^{-1}, \] which is complex diagonalization.

Matrix exponential and complex diagonalization

• Recall that \(e^x\) is defined for complex numbers \(x\in\mathbb C\) by the Taylor series: \[ e^x=1+x+\frac{x^2}{2!}+\cdots \]

• Therefore, we can define \(e^A\) for complex matrices in the same way.

• If \(A=PDP^{-1}\) (even over \(\mathbb C\)), then \[ e^{At}=Pe^{Dt}P^{-1}. \]

• For example, if \(D=\operatorname{diag}(i\omega, -i\omega)\) as before: \[ e^{Dt}=\begin{pmatrix} e^{i\omega t} & 0\\ 0 & e^{-i\omega t} \end{pmatrix}. \]

Example (continued)

• Recall \(Y(t)=\langle y,y' \rangle\) and \(Y'=AY\).

• The solution is: \[ Y(t)=e^{At}Y(0)=Pe^{Dt}P^{-1}Y(0). \]

• Let \(\vec{c}=P^{-1}Y(0)\), then \[ Y(t)=c_1 e^{i\omega t}\vec{u}_1 + c_2 e^{-i\omega t}\vec{u}_2. \]

• This expression involves complex exponentials, but we know the physical solution is real.

• How do we recover sine and cosine?

Euler’s formula

Let \(\theta\in \mathbb R\). Then \[ e^{i\theta}= \cos \theta + i \sin \theta. \]

• This follows from Taylor series expansions: \[ e^{i\theta}=1+i\theta+\frac{(i\theta)^2}{2!}+\frac{(i\theta)^3}{3!}+\cdots \]

• Recall the Taylor series: \[ \cos\theta = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots \] \[ \sin\theta = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots \]

• Substitute powers of \(i\) into \(e^{i\theta}\): \[ e^{i\theta} =1+i\theta-\frac{\theta^2}{2!}-i\frac{\theta^3}{3!} +\frac{\theta^4}{4!}+i\frac{\theta^5}{5!}-\cdots \]

• Group real and imaginary parts: \[ e^{i\theta} =\left(1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\cdots\right) +i\left(\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\cdots\right) \]

• Therefore: \[ e^{i\theta}=\cos\theta+i\sin\theta. \]

• Useful identities: \[ e^{i\theta}+e^{-i\theta}=2\cos\theta \] \[ e^{i\theta}-e^{-i\theta}=2i\sin\theta \]

• Real and imaginary parts: \[ \operatorname{Re}(e^{i\theta})=\cos\theta, \quad \operatorname{Im}(e^{i\theta})=\sin\theta. \]

Example (continued)

• From \[ Y(t)=c_1 e^{i\omega t}\vec{u}_1 + c_2 e^{-i\omega t}\vec{u}_2, \] where \(c_1,c_2\in\mathbb C\) and \(\vec{u}_1,\vec{u}_2\in\mathbb C^2\).

• Write the constants in terms of real and imaginary parts: \[ c_1=a_1+i b_1,\quad c_2=a_2+i b_2, \quad a_i,b_i\in\mathbb R. \]

• Using Euler’s formula: \[ e^{i\omega t}=\cos(\omega t)+i\sin(\omega t),\quad e^{-i\omega t}=\cos(\omega t)-i\sin(\omega t). \]

• Substitute into \(Y(t)\): \[ Y(t)= (a_1+i b_1)(\cos\omega t+i\sin\omega t)\vec{u}_1 \] \[ \qquad +(a_2+i b_2)(\cos\omega t-i\sin\omega t)\vec{u}_2. \]

• Expand each product: \[ (a_1+i b_1)(\cos\omega t+i\sin\omega t) \] \[ = a_1\cos\omega t - b_1\sin\omega t + i(a_1\sin\omega t + b_1\cos\omega t), \] \[ (a_2+i b_2)(\cos\omega t-i\sin\omega t) \] \[ = a_2\cos\omega t + b_2\sin\omega t + i(-a_2\sin\omega t + b_2\cos\omega t). \]

• Therefore, \[ Y(t)=\text{(real part)} + i\,\text{(imaginary part)}, \] where the real part is: \[ \cos(\omega t)\,(a_1\vec{u}_1+a_2\vec{u}_2) \] \[ + \sin(\omega t)\,(-b_1\vec{u}_1 + b_2\vec{u}_2). \]

• Since the original ODE has real coefficients, we take the real part: \[ Y(t)=\vec{a}\,\cos(\omega t)+\vec{b}\,\sin(\omega t), \] for some real vectors \(\vec{a},\vec{b}\).

• Taking the first component gives: \[ y(t)=C_1\cos(\omega t)+C_2\sin(\omega t). \]

• Key idea:

  • Complex exponentials produce sine/cosine after expansion
  • Taking real parts extracts the real-valued solution

Summary

• Let \(A\) be an \(n\times n\) matrix with complex entries. We can define the matrix exponential \(e^{A}\) using the same Taylor series as in the real case: \[ e^{A}=I+A+\frac{A^2}{2!}+\cdots \]

• The system of linear ODEs \[ Y'(t)=AY(t) \] has solution \[ Y(t)=e^{At}Y(0), \] which holds even when \(A\) has complex entries.

• The characteristic equation \[ \det(A-\lambda I)=0 \] always has \(n\) complex solutions (counting multiplicity). These are called the complex eigenvalues of \(A\).

• For each eigenvalue \(\lambda_i\in\mathbb C\), a (nonzero) vector \(\vec{u}_i\in\mathbb C^n\) satisfying \[ A\vec{u}_i=\lambda_i\vec{u}_i \] is called a complex eigenvector.

• If we can find \(n\) linearly independent eigenvectors \(\vec{u}_1,\dots,\vec{u}_n\), they form a basis of \(\mathbb C^n\) (a complex eigenbasis).

• In this case, \(A\) is diagonalizable over \(\mathbb C\), and we can write \[ A=PDP^{-1}, \quad P=[\vec{u}_1 \ \cdots \ \vec{u}_n], \quad D=\operatorname{diag}(\lambda_1,\dots,\lambda_n). \]

• Then the matrix exponential simplifies to \[ e^{At}=Pe^{Dt}P^{-1}, \] where \[ e^{Dt}=\operatorname{diag}\big(e^{\lambda_1 t},\dots,e^{\lambda_n t}\big). \]

• If \(A\) is a real matrix, complex eigenvalues appear in conjugate pairs, and using Euler’s formula, \[ e^{i\theta}=\cos\theta+i\sin\theta, \] the solution can be rewritten entirely in terms of real-valued functions (sines and cosines) instead of complex exponentials.