The foundation for understanding systems of differential equations
By the end of this section, you will be able to:
A system of first-order differential equations has the form:
where $\mathbf{x} = \begin{pmatrix} x_1(t) \\ x_2(t) \\ \vdots \\ x_n(t) \end{pmatrix}$ is the state vector and $A$ is an $n \times n$ coefficient matrix. Matrices are the natural language for writing and solving systems of differential equations. Understanding eigenvalues and eigenvectors is essential because:
Think of a matrix $A$ as a linear transformation that stretches, rotates, and scales space. Most vectors get complicated; but eigenvectors are special — they only get scaled by the eigenvalue. Finding eigenvectors means finding the "natural directions" of the transformation.
Add matrices of the same size element-by-element:
Example:
Multiply every element by a scalar $c$:
Example: $2 \begin{pmatrix} 1 & 3 \\ -2 & 5 \end{pmatrix} = \begin{pmatrix} 2 & 6 \\ -4 & 10 \end{pmatrix}$
The product $C = AB$ is computed as:
In other words, element $(i,j)$ of $AB$ is the dot product of row $i$ of $A$ with column $j$ of $B$.
Worked Example: Multiply two 2×2 matrices step by step.
Compute each element of $AB$:
Matrix multiplication is NOT commutative: $AB \neq BA$ in general! Always be careful about the order of multiplication. This is fundamentally different from arithmetic with real numbers.
A matrix $A$ multiplied by a vector $\mathbf{x}$ produces a new vector $A\mathbf{x}$:
Geometric Interpretation: The matrix $A$ transforms the vector $\mathbf{x}$. It can be understood as showing how $A$ acts on the basis vectors:
The columns of $A$ show where the standard basis vectors are mapped!
For a 2×2 matrix, the determinant is computed as:
Example:
For a 3×3 matrix, expand along the first row:
Each minor (2×2 determinant) is computed using the formula above. You can expand along any row or column.
A matrix $A$ is invertible (has an inverse $A^{-1}$) if and only if:
If invertible, the matrices satisfy:
For a 2×2 matrix, the inverse formula is:
where $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and $\det(A) = ad - bc$.
Example: Find the inverse of $A = \begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix}$.
First, $\det(A) = (2)(3) - (1)(1) = 6 - 1 = 5 \neq 0$, so the inverse exists.
This is the most important section. Eigenvalues and eigenvectors are the key to understanding all behavior of systems of differential equations.
A scalar $\lambda$ is an eigenvalue of an $n \times n$ matrix $A$ if there exists a nonzero vector $\mathbf{v}$ such that:
The nonzero vector $\mathbf{v}$ is called an eigenvector corresponding to eigenvalue $\lambda$.
When we apply the matrix $A$ to an eigenvector $\mathbf{v}$, the result is just a scaled version of $\mathbf{v}$. The eigenvalue $\lambda$ is the scaling factor:
In a differential system $\frac{d\mathbf{x}}{dt} = A\mathbf{x}$, if $\mathbf{x}(0)$ is an eigenvector $\mathbf{v}$ with eigenvalue $\lambda$, then the solution is simply $\mathbf{x}(t) = e^{\lambda t}\mathbf{v}$. The system evolves exponentially along the eigenvector direction! This is why eigenvectors are sometimes called the "natural modes" of the system.
Starting from $A\mathbf{v} = \lambda\mathbf{v}$, rearrange:
For a nonzero solution $\mathbf{v}$ to exist, the matrix $A - \lambda I$ must be singular (non-invertible), which means:
The solutions $\lambda$ to this equation are the eigenvalues of $A$. This equation is a polynomial of degree $n$ (for an $n \times n$ matrix), so it has exactly $n$ roots (counting multiplicity) in the complex numbers.
Depending on the characteristic equation, we can have three cases:
The characteristic equation has two different real roots $\lambda_1 \neq \lambda_2$. Each eigenvalue has its own eigenvector, and the system exhibits exponential growth or decay.
The characteristic equation has a repeated root $\lambda$ with multiplicity $m > 1$. Depending on the structure of $A$, there may be multiple linearly independent eigenvectors (diagonalizable case) or fewer (Jordan form case). In systems of DEs, this leads to polynomial factors multiplying exponentials.
The characteristic equation has complex roots $\lambda = \alpha \pm \beta i$ (complex conjugate pairs). These lead to oscillatory solutions with exponential envelope: $e^{\alpha t}\cos(\beta t)$ and $e^{\alpha t}\sin(\beta t)$.
If $\mathbf{v}$ is an eigenvector with eigenvalue $\lambda$, then so is $c\mathbf{v}$ for any nonzero constant $c$. So eigenvectors are not unique—they have arbitrary scaling. We typically normalize them or choose simple integer/rational values.
Click each example to reveal the complete step-by-step solution.
Step 1: Write $A - \lambda I$
Step 2: Compute $\det(A - \lambda I)$
Step 3: Solve $\det(A - \lambda I) = 0$
Eigenvalues: $\lambda_1 = 5$ and $\lambda_2 = 2$ (distinct real roots) ✓
Step 4a: Find eigenvector for $\lambda_1 = 5$
Solve $(A - 5I)\mathbf{v} = \mathbf{0}$:
From the first row: $-v_1 + v_2 = 0 \Rightarrow v_2 = v_1$
Eigenvector (choosing $v_1 = 1$): $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$
Verification: $A\mathbf{v}_1 = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 5 \end{pmatrix} = 5\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ ✓
Step 4b: Find eigenvector for $\lambda_2 = 2$
Solve $(A - 2I)\mathbf{v} = \mathbf{0}$:
From the first row: $2v_1 + v_2 = 0 \Rightarrow v_2 = -2v_1$
Eigenvector (choosing $v_1 = 1$): $\mathbf{v}_2 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$
Verification: $A\mathbf{v}_2 = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}\begin{pmatrix} 1 \\ -2 \end{pmatrix} = \begin{pmatrix} 2 \\ -4 \end{pmatrix} = 2\begin{pmatrix} 1 \\ -2 \end{pmatrix}$ ✓
Answer:
Eigenvalues: $\lambda_1 = 5$, $\lambda_2 = 2$
Eigenvectors: $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$, $\mathbf{v}_2 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$
Step 1: Write $A - \lambda I$
Step 2: Compute $\det(A - \lambda I)$
Wait! Let me recalculate: $(3-\lambda)(1-\lambda) = 3 - 3\lambda - \lambda + \lambda^2 = 3 - 4\lambda + \lambda^2$. Adding $-(-2)(1) = 2$: we get $\lambda^2 - 4\lambda + 5$.
Actually, let me verify this matrix should give a repeated eigenvalue. Let me use $A = \begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix}$ instead for this example, which has a repeated eigenvalue of $\lambda = 1$.
Revised: Find eigenvalues of $A = \begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix}$
Eigenvalue: $\lambda = 1$ (with multiplicity 2) ✓
Step 4: Find eigenvectors for $\lambda = 1$
Solve $(A - I)\mathbf{v} = \mathbf{0}$:
From the first row: $v_1 - v_2 = 0 \Rightarrow v_1 = v_2$
Eigenvector (choosing $v_1 = 1$): $\mathbf{v} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$
Note: Even though the eigenvalue has multiplicity 2, there is only one linearly independent eigenvector (geometric multiplicity 1). This matrix is not diagonalizable.
Answer:
Eigenvalue: $\lambda = 1$ (repeated, multiplicity 2)
Eigenvector: $\mathbf{v} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ (unique direction, up to scaling)
Step 1: Write $A - \lambda I$
Step 2: Compute $\det(A - \lambda I)$
Step 3: Solve the characteristic equation
Using the quadratic formula:
Eigenvalues: $\lambda_1 = 2 + i$ and $\lambda_2 = 2 - i$ (complex conjugate pair with $\alpha = 2$, $\beta = 1$) ✓
Step 4: Find eigenvector for $\lambda_1 = 2 + i$
Solve $(A - (2+i)I)\mathbf{v} = \mathbf{0}$:
From the first row: $(-1-i)v_1 - 2v_2 = 0 \Rightarrow v_2 = \frac{-1-i}{2}v_1$
Choosing $v_1 = 2$: $v_2 = -(1+i) = -1-i$
Eigenvector: $\mathbf{v}_1 = \begin{pmatrix} 2 \\ -1-i \end{pmatrix}$ (or any scalar multiple)
For the conjugate eigenvalue $\lambda_2 = 2 - i$:
The eigenvector is the complex conjugate: $\mathbf{v}_2 = \begin{pmatrix} 2 \\ -1+i \end{pmatrix}$
Answer:
Eigenvalues: $\lambda_1 = 2 + i$, $\lambda_2 = 2 - i$
Eigenvectors: $\mathbf{v}_1 = \begin{pmatrix} 2 \\ -1-i \end{pmatrix}$, $\mathbf{v}_2 = \begin{pmatrix} 2 \\ -1+i \end{pmatrix}$
For DEs, these lead to solutions of the form $e^{2t}\cos(t)$ and $e^{2t}\sin(t)$ (with appropriate coefficients).
Step 1: Write $A - \lambda I$
Step 2: Compute $\det(A - \lambda I)$
Step 3: Solve the characteristic equation
Eigenvalues: $\lambda_1 = 6$ and $\lambda_2 = -1$ (distinct real) ✓
Step 4a: Eigenvector for $\lambda_1 = 6$
From the first row: $-4v_1 + 3v_2 = 0 \Rightarrow 3v_2 = 4v_1 \Rightarrow v_2 = \frac{4}{3}v_1$
Choosing $v_1 = 3$: $\mathbf{v}_1 = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$
Step 4b: Eigenvector for $\lambda_2 = -1$
From the first row: $3v_1 + 3v_2 = 0 \Rightarrow v_1 = -v_2$
Choosing $v_1 = 1$: $\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$
Answer:
Eigenvalues: $\lambda_1 = 6$, $\lambda_2 = -1$
Eigenvectors: $\mathbf{v}_1 = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$, $\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$
Adjust the matrix entries below to visualize how the matrix transforms vectors. The red vectors are eigenvectors—notice how they only get scaled (not rotated) by the matrix transformation.
Eigenvalues and Eigenvectors:
Test your understanding with these practice problems. Click each problem to reveal the step-by-step solution.
Step 1: Characteristic equation
Step 2: Solve characteristic equation
Eigenvalues: $\lambda_1 = 4$, $\lambda_2 = -1$
Step 3: Find eigenvectors
For $\lambda_1 = 4$:
Eigenvector: $\mathbf{v}_1 = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$
For $\lambda_2 = -1$:
Eigenvector: $\mathbf{v}_2 = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$
Answer: $\boxed{\lambda_1 = 4, \mathbf{v}_1 = \begin{pmatrix} 2 \\ 1 \end{pmatrix}; \quad \lambda_2 = -1, \mathbf{v}_2 = \begin{pmatrix} 1 \\ 3 \end{pmatrix}}$
Step 1: Characteristic equation
Step 2: Solve
Eigenvalue: $\lambda = 2$ (repeated, multiplicity 2)
Step 3: Find eigenvector
From first row: $-2v_1 - 4v_2 = 0 \Rightarrow v_1 = -2v_2$
Eigenvector: $\mathbf{v} = \begin{pmatrix} -2 \\ 1 \end{pmatrix}$ (or $\begin{pmatrix} 2 \\ -1 \end{pmatrix}$)
Answer: $\boxed{\lambda = 2 \text{ (repeated)}, \quad \mathbf{v} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}}$
Step 1: Characteristic equation
Step 2: Solve using quadratic formula
Answer: $\boxed{\lambda = 2 \pm i\sqrt{5}}$ (complex conjugate pair with $\alpha = 2$, $\beta = \sqrt{5}$)