01 / 21
\(\lambda\)
\(\lambda\)

Characteristic
Polynomial

Learn how to extract eigenvalues from a matrix by computing \(\det(A - \lambda I) = 0\), the cornerstone equation of linear algebra.

Linear Algebra

What Is an Eigenvalue?

Given a square matrix \(A\), a scalar \(\lambda\) is an eigenvalue if there exists a nonzero vector \(\mathbf{v}\) such that:

The vector \(\mathbf{v}\) is called an eigenvector. The matrix \(A\) stretches or flips \(\mathbf{v}\) by the factor \(\lambda\), without changing its direction.

Geometric Intuition

\(\mathbf{v}\)
\(A\mathbf{v} = \lambda\mathbf{v}\)
Same direction, scaled by \(\lambda\)

Why Do Eigenvalues Matter?

Stability
In dynamical systems, eigenvalues tell you whether solutions grow, decay, or oscillate.
Diagonalisation
A matrix with \(n\) independent eigenvectors can be decomposed as \(A = PDP^{-1}\).
Data Science
PCA, Google PageRank, and spectral clustering all depend on eigenvalues.

The characteristic polynomial is the tool we use to find these eigenvalues.

The Characteristic Polynomial

Starting from \(A\mathbf{v} = \lambda\mathbf{v}\), we rearrange to \((A - \lambda I)\mathbf{v} = \mathbf{0}\). For a nonzero \(\mathbf{v}\) to exist, the matrix \(A - \lambda I\) must be singular:

The polynomial \(p(\lambda)\) is called the characteristic polynomial of \(A\). Its roots are the eigenvalues of \(A\).

Core Theorem

1
Degree: For an \(n \times n\) matrix, \(p(\lambda)\) has degree \(n\).
2
Root count: There are at most \(n\) eigenvalues (counted with multiplicity).
3
Leading term: The leading coefficient is \((-1)^n\), so \(p(\lambda) = (-1)^n \lambda^n + \cdots\)

Trace and Determinant Connection

The coefficients of the characteristic polynomial encode important information about \(A\):

Sum of Eigenvalues
\(\lambda_1 + \lambda_2 + \cdots + \lambda_n = \text{tr}(A)\)
Product of Eigenvalues
\(\lambda_1 \cdot \lambda_2 \cdots \lambda_n = \det(A)\)

Algebraic vs Geometric Multiplicity

Algebraic Multiplicity Geometric Multiplicity
Definition Multiplicity of \(\lambda\) as a root of \(p(\lambda)\) Dimension of \(\text{null}(A - \lambda I)\)
Notation \(\text{AM}(\lambda)\) \(\text{GM}(\lambda)\)
Always true \(1 \le \text{GM}(\lambda) \le \text{AM}(\lambda)\)
Diagonalisable? \(\text{GM}(\lambda) = \text{AM}(\lambda)\) for every \(\lambda\)

Step-by-Step Method

1
Form \(A - \lambda I\): Subtract \(\lambda\) from every diagonal entry of \(A\).
2
Compute the determinant: Evaluate \(\det(A - \lambda I)\) to get a polynomial in \(\lambda\).
3
Set equal to zero: Solve \(p(\lambda) = 0\) by factoring, the quadratic formula, or other root-finding techniques.
4
Find eigenvectors: For each eigenvalue \(\lambda_i\), solve \((A - \lambda_i I)\mathbf{v} = \mathbf{0}\).

Example 1: \(2 \times 2\) Matrix

Find the eigenvalues of \(A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}\).

Verifying Example 1

Let us verify using the trace and determinant shortcut:

Trace Check
\(\text{tr}(A) = 4 + 3 = 7\)
\(\lambda_1 + \lambda_2 = 5 + 2 = 7\) ✓
Determinant Check
\(\det(A) = 4 \cdot 3 - 1 \cdot 2 = 10\)
\(\lambda_1 \cdot \lambda_2 = 5 \cdot 2 = 10\) ✓

Example 2: \(3 \times 3\) Matrix

Find the characteristic polynomial of \(B = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}\).

\(\lambda = 2\)
Algebraic multiplicity: 2
Geometric multiplicity: 1
\(\lambda = 3\)
Algebraic multiplicity: 1
Geometric multiplicity: 1

Example 3: Repeated Eigenvalue

Find the eigenvalues of \(C = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}\).

Contrast this with Example 2 where \(\lambda = 2\) had \(\text{GM} = 1 < \text{AM} = 2\).

Visualising \(p(\lambda)\)

For Example 1, \(p(\lambda) = \lambda^2 - 7\lambda + 10\). Its roots are the eigenvalues:

\(\lambda\)
\(p(\lambda)\)
 2 4 5 6
\(\lambda = 2\)
\(\lambda = 5\)

Common Mistakes

✕ Wrong sign convention
Using \(\det(\lambda I - A)\) instead of \(\det(A - \lambda I)\). They differ by a factor of \((-1)^n\) but yield the same roots. Be consistent with your textbook.
✕ Forgetting the identity matrix
Writing \(A - \lambda\) instead of \(A - \lambda I\). You must subtract \(\lambda\) from the diagonal only, not from every entry.
✕ Confusing AM and GM
Assuming a repeated root always means the matrix is not diagonalisable. Check the geometric multiplicity too: if \(\text{GM} = \text{AM}\), it is still diagonalisable.

Summary Table

Property \(2 \times 2\) \(3 \times 3\) \(n \times n\)
Polynomial degree 2 3 \(n\)
Max eigenvalues 2 3 \(n\)
Sum of eigenvalues \(\text{tr}(A)\) \(\text{tr}(A)\) \(\text{tr}(A)\)
Product of eigenvalues \(\det(A)\) \(\det(A)\) \(\det(A)\)
Quick formula \(\lambda^2 - \text{tr}(A)\lambda + \det(A)\) Cofactor expansion General methods

Question 1 of 5

What is the characteristic polynomial of \(A = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix}\)?

Question 2 of 5

If the eigenvalues of a \(3 \times 3\) matrix \(B\) are \(1, 2, 4\), what is \(\det(B)\)?

Question 3 of 5

The matrix \(C = \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix}\) has eigenvalue \(\lambda = 5\) with algebraic multiplicity 2. What is its geometric multiplicity?

Question 4 of 5

Which expression correctly defines the characteristic polynomial of a square matrix \(A\)?

Question 5 of 5

A \(4 \times 4\) matrix has characteristic polynomial \(p(\lambda) = \lambda^4 - 10\lambda^3 + \cdots\). What is the trace of the matrix?