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Algebraic vs.
Geometric Multiplicity

Understand how eigenvalues can repeat in the characteristic polynomial and how many independent eigenvectors each one produces.

Linear Algebra

What Is an Eigenvalue?

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

The vector \( \mathbf{v} \) is the eigenvector. Eigenvalues tell us how a transformation stretches or compresses along special directions.

The Characteristic Polynomial

We find eigenvalues by solving:

This determinant produces a polynomial in \( \lambda \). Each root of the polynomial is an eigenvalue. The polynomial's degree equals the size of the matrix.

Two Kinds of Multiplicity

Algebraic Multiplicity
How many times \( \lambda \) appears as a root of the characteristic polynomial. Notation: \( \mathrm{AM}(\lambda) \).
Geometric Multiplicity
The dimension of the eigenspace \( \ker(A - \lambda I) \): the number of linearly independent eigenvectors. Notation: \( \mathrm{GM}(\lambda) \).

Visualising Eigenspaces

\( \mathrm{GM} = 1 \)
 (a line through origin) 0 v
\( \mathrm{GM} = 2 \)
 (a plane through origin) 0 v₁ v₂

When \( \mathrm{GM} = 1 \), the eigenspace is a line: all eigenvectors are scalar multiples of one vector. When \( \mathrm{GM} = 2 \), the eigenspace is a plane: two independent eigenvectors span it.

The Fundamental Inequality

1
\( \mathrm{GM} \geq 1 \): every eigenvalue has at least one eigenvector by definition.
2
\( \mathrm{GM} \leq \mathrm{AM} \): you can never have more independent eigenvectors than the root's multiplicity in the polynomial.
3
Equality is the best case: when \( \mathrm{GM} = \mathrm{AM} \) for all eigenvalues, the matrix is diagonalisable.

When \( \mathrm{GM} < \mathrm{AM} \): Defective Eigenvalues

If \( \mathrm{GM}(\lambda) < \mathrm{AM}(\lambda) \), the eigenvalue is called defective. The matrix does not have enough eigenvectors to form a basis.

Defective Matrix
\( \mathrm{AM}(\lambda) = 2 \)
\( \mathrm{GM}(\lambda) = 1 \)
 Not diagonalisable needs Jordan Normal Form Jordan block of size 2 Uses generalized eigenvectors instead

AM vs. GM: Side by Side

Algebraic (AM) Geometric (GM)
Definition Multiplicity as root of \( \det(A - \lambda I) \) \( \dim\!\ker(A - \lambda I) \)
Computed from Characteristic polynomial Null space of \( A - \lambda I \)
Counts Repeated roots Independent eigenvectors
Range \( 1 \leq \mathrm{AM} \leq n \) \( 1 \leq \mathrm{GM} \leq \mathrm{AM} \)
If equal for all \( \lambda \) Matrix is diagonalisable

Step-by-Step Method

1
Compute \( \det(A - \lambda I) \): expand the determinant to get the characteristic polynomial.
2
Find the roots: factor the polynomial. The power of each factor \( (\lambda - \lambda_i)^{k} \) gives \( \mathrm{AM}(\lambda_i) = k \).
3
For each eigenvalue, form \( A - \lambda_i I \): substitute the eigenvalue into the matrix.
4
Row-reduce \( A - \lambda_i I \): find the null space. The number of free variables equals \( \mathrm{GM}(\lambda_i) \).
5
Compare: if \( \mathrm{GM} = \mathrm{AM} \) for every eigenvalue, the matrix is diagonalisable.

Example 1: Diagonalisable Case

1
Characteristic polynomial: \( \det(A - \lambda I) = (5 - \lambda)(3 - \lambda) \).
2
Roots: \( \lambda_1 = 5 \) with \( \mathrm{AM} = 1 \), and \( \lambda_2 = 3 \) with \( \mathrm{AM} = 1 \).
3
Eigenspaces: Each eigenvalue has 1 free variable, so \( \mathrm{GM} = 1 \) for both.

Example 2: A Repeated Eigenvalue

1
Characteristic polynomial: \( (4 - \lambda)^2(7 - \lambda) \).
2
\( \lambda = 4 \): \( \mathrm{AM} = 2 \). Row-reduce \( A - 4I = \mathrm{diag}(0,0,3) \): 2 free variables, so \( \mathrm{GM} = 2 \).
3
\( \lambda = 7 \): \( \mathrm{AM} = 1 \), \( \mathrm{GM} = 1 \).

Example 3: A Defective Matrix

1
Characteristic polynomial: \( (2 - \lambda)^2 = 0 \), so \( \lambda = 2 \) with \( \mathrm{AM} = 2 \).
2
Eigenspace: \( A - 2I = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \). One free variable, so \( \mathrm{GM} = 1 \).
3
\( \mathrm{GM} = 1 < \mathrm{AM} = 2 \). The eigenvalue is defective.

Why Does This Matter?

Diagonalisation
Computing \( A^{100} \) or \( e^{At} \) is trivial for diagonal matrices. If AM = GM everywhere, we can diagonalise.
Differential Equations
Solutions to \( \mathbf{x}' = A\mathbf{x} \) depend on whether eigenvectors span the space. Defective matrices produce \( te^{\lambda t} \) terms.
Stability Analysis
In dynamical systems and control theory, defective eigenvalues cause polynomial growth that purely exponential models miss.

Common Mistakes

✕ Confusing AM with GM
AM comes from the polynomial. GM comes from the null space. They measure different things.
✕ Assuming repeated root = defective
A repeated eigenvalue can still have GM = AM (e.g., \( 4I \) has \( \lambda = 4 \) with AM = GM = 4).
✕ Forgetting \( \mathrm{GM} > 0 \)
Every eigenvalue has at least one eigenvector, so \( \mathrm{GM} \geq 1 \). \( \mathrm{GM} = 0 \) is impossible.

Quick-Reference Summary

Scenario AM GM Diagonalisable?
All distinct eigenvalues 1 each 1 each Always yes
Repeated, full eigenspace \( k \) \( k \) Yes
Repeated, defective \( k \) \( < k \) No
Symmetric / Hermitian \( k \) \( k \) Always yes

Question 1 of 5

If the characteristic polynomial of a \( 4 \times 4 \) matrix is \( (\lambda - 3)^3(\lambda + 1) \), what is the algebraic multiplicity of \( \lambda = 3 \)?

Question 2 of 5

For the matrix \( A = \begin{pmatrix} 5 & 1 \\ 0 & 5 \end{pmatrix} \), what is the geometric multiplicity of \( \lambda = 5 \)?

Question 3 of 5

A \( 3 \times 3 \) matrix has eigenvalues \( \lambda = 2 \) (\( \mathrm{AM} = 2 \)) and \( \lambda = 7 \) (\( \mathrm{AM} = 1 \)). If the matrix is diagonalisable, what must \( \mathrm{GM}(2) \) equal?

Question 4 of 5

Which of the following is always true for any eigenvalue \( \lambda \) of a square matrix?

Question 5 of 5

A real symmetric matrix has a repeated eigenvalue \( \lambda = 4 \) with \( \mathrm{AM} = 3 \). What is \( \mathrm{GM}(4) \)?