6.2 The Characteristic Equation
Introduction
In the previous post, we learned what eigenvalues and eigenvectors are. The equation $A\vec{v} = \lambda\vec{v}$ tells us that certain special vectors only get scaled by a transformation. But given an actual matrix, how do we find these values? We certainly cannot guess eigenvectors one by one and check.
Fortunately, linear algebra gives us a systematic procedure. The key insight is that rearranging $A\vec{v} = \lambda\vec{v}$ brings the determinant into play. Setting that determinant equal to zero produces the characteristic equation, and solving it yields the eigenvalues. In this post, we will derive the characteristic equation step by step and work through a complete example.
Deriving the Characteristic Equation
The Core Idea: When Does a Homogeneous System Have Nontrivial Solutions?
Start from the eigenvalue definition $A\vec{v} = \lambda\vec{v}$ and rearrange:
\[A\vec{v} - \lambda\vec{v} = \vec{0}\] \[A\vec{v} - \lambda I\vec{v} = \vec{0}\] \[(A - \lambda I)\vec{v} = \vec{0}\]This is a homogeneous system with coefficient matrix $(A - \lambda I)$. We need a nontrivial solution ($\vec{v} \neq \vec{0}$). If $(A - \lambda I)$ were invertible, the only solution would be $\vec{v} = (A-\lambda I)^{-1}\vec{0} = \vec{0}$, which is useless. So for eigenvectors to exist, $(A - \lambda I)$ must be singular (non-invertible), which happens exactly when its determinant is zero:
\[\det(A - \lambda I) = 0\]This is the characteristic equation.
The Characteristic Polynomial
Expanding $\det(A - \lambda I)$ as a polynomial in $\lambda$ gives the characteristic polynomial. For an $n \times n$ matrix, this polynomial has degree $n$, so (counting multiplicities and allowing complex roots) there are exactly $n$ eigenvalues.
A 2x2 Example: Finding Eigenvalues
The Matrix
\[A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}\]Step 1: Form $(A - \lambda I)$
\[A - \lambda I = \begin{pmatrix} 4-\lambda & 1 \\ 2 & 3-\lambda \end{pmatrix}\]Step 2: Set Up the Characteristic Equation
\[\det(A - \lambda I) = (4-\lambda)(3-\lambda) - (1)(2) = 0\] \[= 12 - 4\lambda - 3\lambda + \lambda^2 - 2 = 0\] \[= \lambda^2 - 7\lambda + 10 = 0\]Step 3: Factor the Characteristic Polynomial
\[(\lambda - 5)(\lambda - 2) = 0\] \[\therefore \quad \lambda_1 = 5, \quad \lambda_2 = 2\]Finding the Eigenvectors for Each Eigenvalue
Eigenvectors for $\lambda_1 = 5$
Solve $(A - 5I)\vec{v} = \vec{0}$:
\[A - 5I = \begin{pmatrix} 4-5 & 1 \\ 2 & 3-5 \end{pmatrix} = \begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix}\]Row reducing reveals that both rows are proportional (dependent):
\[\begin{pmatrix} -1 & 1 \\ 0 & 0 \end{pmatrix} \rightarrow -v_1 + v_2 = 0 \rightarrow v_1 = v_2\]Setting the free variable $v_2 = t$: $\vec{v} = t\begin{pmatrix}1\1\end{pmatrix}$
Eigenvector: $\vec{v}_1 = \begin{pmatrix}1\1\end{pmatrix}$
Verification: $A\vec{v}_1 = \begin{pmatrix}4&1\2&3\end{pmatrix}\begin{pmatrix}1\1\end{pmatrix} = \begin{pmatrix}5\5\end{pmatrix} = 5\begin{pmatrix}1\1\end{pmatrix}$ ✓
Eigenvectors for $\lambda_2 = 2$
Solve $(A - 2I)\vec{v} = \vec{0}$:
\[A - 2I = \begin{pmatrix} 4-2 & 1 \\ 2 & 3-2 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix}\]Row reducing:
\[\begin{pmatrix} 2 & 1 \\ 0 & 0 \end{pmatrix} \rightarrow 2v_1 + v_2 = 0 \rightarrow v_2 = -2v_1\]Setting the free variable $v_1 = t$: $\vec{v} = t\begin{pmatrix}1\-2\end{pmatrix}$
Eigenvector: $\vec{v}_2 = \begin{pmatrix}1\-2\end{pmatrix}$
Verification: $A\vec{v}_2 = \begin{pmatrix}4&1\2&3\end{pmatrix}\begin{pmatrix}1\-2\end{pmatrix} = \begin{pmatrix}2\-4\end{pmatrix} = 2\begin{pmatrix}1\-2\end{pmatrix}$ ✓
Repeated Eigenvalues and Special Cases
Algebraic and Geometric Multiplicity
When an eigenvalue $\lambda$ appears $k$ times as a root of the characteristic polynomial, its algebraic multiplicity is $k$. The number of linearly independent eigenvectors for that eigenvalue is called its geometric multiplicity, and it satisfies:
\[1 \leq \text{geometric multiplicity} \leq \text{algebraic multiplicity}\]General Formula for 2x2 Matrices
For any $2 \times 2$ matrix $A = \begin{pmatrix}a&b\c&d\end{pmatrix}$:
\[\det(A - \lambda I) = \lambda^2 - (a+d)\lambda + (ad-bc) = 0\] \[= \lambda^2 - \text{tr}(A)\lambda + \det(A) = 0\]Here $\text{tr}(A) = a + d$ is the trace and $\det(A) = ad - bc$ is the determinant. This gives two elegant relationships:
\[\lambda_1 + \lambda_2 = \text{tr}(A), \qquad \lambda_1 \cdot \lambda_2 = \det(A)\]Key Takeaways
| Concept | Description / Formula |
|---|---|
| Derivation | $A\vec{v}=\lambda\vec{v} \Rightarrow (A-\lambda I)\vec{v}=\vec{0}$ |
| Characteristic equation | $\det(A - \lambda I) = 0$ |
| Degree of characteristic polynomial | $n$ for an $n \times n$ matrix |
| Number of eigenvalues | $n$ over the complex numbers (counting multiplicity) |
| 2x2 formula | $\lambda^2 - \text{tr}(A)\lambda + \det(A) = 0$ |
| Sum of eigenvalues | $\lambda_1 + \cdots + \lambda_n = \text{tr}(A)$ |
| Product of eigenvalues | $\lambda_1 \cdots \lambda_n = \det(A)$ |
| Finding eigenvectors | For each $\lambda$, solve $(A-\lambda I)\vec{v}=\vec{0}$ |
Next up: Diagonalization – using eigenvalues and eigenvectors to compute matrix powers efficiently.