Geometric multiplicity is a fundamental concept in linear algebra, specifically in the context of eigenvalues and eigenvectors of a matrix. When dealing with square matrices, each eigenvalue has an associated geometric multiplicity, which is defined as the dimension of the eigenspace corresponding to that eigenvalue. In simpler terms, it represents the number of linearly independent eigenvectors associated with a given eigenvalue.

Understanding geometric multiplicity is crucial for insights into matrix diagonalization and stability analysis in various applications, such as system dynamics and quantum mechanics.

What is Geometric Multiplicity?

Geometric multiplicity is a concept in linear algebra related to eigenvalues and eigenvectors of a square matrix. It refers to the number of linearly independent eigenvectors associated with a given eigenvalue. In other words, it is the dimension of the eigenspace corresponding to that eigenvalue.

To understand this better, let’s break it down:

• Eigenvalues and Eigenvectors
  • An eigenvalue λ of a matrix A is a scalar such that there exists a non-zero vector v (called an eigenvector) satisfying the equation Av = λv.
• Eigenspace
  • The eigenspace corresponding to an eigenvalue λ is the set of all vectors v that satisfy the equation Av = λv. This is essentially the null space of the matrix A − λI, where I is the identity matrix.
• Geometric Multiplicity
  • The geometric multiplicity of an eigenvalue λ is the dimension of its eigenspace, which is the same as the number of linearly independent eigenvectors associated with λ.
  • It is always less than or equal to the algebraic multiplicity of the eigenvalue, which is the number of times λ appears as a root of the characteristic polynomial of A.

Steps to Determine Geometric Multiplicity

To calculate the geometric multiplicity of an eigenvalue λ:

Step 1: Find the Eigenvalue: Solve the characteristic equation det⁡(A − λI) = 0 to find λ.

Step 2: Form the Matrix (A − λI): Subtract λ from A.

Step 3: Solve the System: Solve the homogeneous system (A − λI)x = 0.

Step 4: Determine the Dimension: The number of linearly independent solutions x gives the geometric multiplicity.

Example Problems on Geometric Multiplicity

Let’s discuss the solved examples of calculating geometric multiplicity for 2 by 2 and 3 by 3 matrix:

Example with a 2×2 Matrix

Consider the matrix A: [Tex]A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}[/Tex]

Step 1: Find the eigenvalues

The characteristic equation is obtained from det(A – λI) = 0:

[Tex]\Rightarrow \text{det} \begin{pmatrix} 2 – \lambda & 1 \\ 0 & 2 – \lambda \end{pmatrix} = (2 – \lambda)^2 = 0[/Tex]

So, the eigenvalue is: λ = 2

Step 2: Find the eigenspace

To find the eigenspace for λ = 2, solve (A – 2I)v = 0:

[Tex]\Rightarrow (A – 2I) = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}[/Tex]

The equation (A – 2I)v = 0 becomes:

[Tex]\Rightarrow \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}[/Tex]

This simplifies to y = 0 . Thus, any vector in the eigenspace is of the form [Tex]\begin{pmatrix} x \\ 0 \end{pmatrix}[/Tex].

So, the eigenspace is spanned by [Tex]\begin{pmatrix} 1 \\ 0 \end{pmatrix}[/Tex], and the geometric multiplicity of λ = 2 is 1.

Example with a 3×3 Matrix

Consider the matrix B: [Tex]B = \begin{pmatrix} 5 & 4 & 2 \\ 0 & 1 & -1 \\ 0 & 0 & 3 \end{pmatrix}[/Tex]

Step 1: Find the eigenvalues

The characteristic equation is obtained from det(B – λI) = 0:

[Tex]\Rightarrow \text{det} \begin{pmatrix} 5 – \lambda & 4 & 2 \\ 0 & 1 – \lambda & -1 \\ 0 & 0 & 3 – \lambda \end{pmatrix} = (5 – \lambda)(1 – \lambda)(3 – \lambda) = 0[/Tex]

So, the eigenvalues are:

λ = 5, λ = 1, and λ =3

Step 2: Find the eigenspace for each eigenvalue

For λ = 5:

[Tex](B – 5I) = \begin{pmatrix} 0 & 4 & 2 \\ 0 & -4 & -1 \\ 0 & 0 & -2 \end{pmatrix}[/Tex]

The equation (B – 5I)v = 0 implies:

[Tex]\Rightarrow \begin{pmatrix} 0 & 4 & 2 \\ 0 & -4 & -1 \\ 0 & 0 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}[/Tex]

From the first row, we get [Tex]4y + 2z = 0 \implies 2y + z = 0 \implies z = -2y[/Tex].

So, the eigenspace for λ = 5 is spanned by [Tex]\begin{pmatrix} 0 \\ 1 \\ -2 \end{pmatrix}[/Tex], and the geometric multiplicity of λ = 5 is 1.

For λ = 1:

[Tex](B – 1I) = \begin{pmatrix} 4 & 4 & 2 \\ 0 & 0 & -1 \\ 0 & 0 & 2 \end{pmatrix}[/Tex]

The equation (B – 1I)v = 0 implies:

[Tex]\Rightarrow \begin{pmatrix} 4 & 4 & 2 \\ 0 & 0 & -1 \\ 0 & 0 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}[/Tex]

From the second row, we get -z = 0 ⇒ z = 0.

From the first row, 4x + 4y = 0 ⇒ x = -y.

So, the eigenspace for λ = 1 is spanned by [Tex] \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}[/Tex], and the geometric multiplicity of λ = 1 is 1.

For λ = 3:

[Tex](B – 3I) = \begin{pmatrix} 2 & 4 & 2 \\ 0 & -2 & -1 \\ 0 & 0 & 0 \end{pmatrix}[/Tex]

The equation (B – 3I)v = 0 implies:

[Tex]\Rightarrow \begin{pmatrix} 2 & 4 & 2 \\ 0 & -2 & -1 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}[/Tex]

From the second row, -2y – z = 0 ⇒ z = -2y.

From the first row, 2x + 4y + 2(-2y) = 0 ⇒ 2x = 0 ⇒ x = 0.

So, the eigenspace for λ = 3 is spanned by [Tex]\begin{pmatrix} 0 \\ 1 \\ -2 \end{pmatrix}[/Tex], and the geometric multiplicity of λ = 3 is 1.

Read More,

• Eigenvalues
• Square Matrix
• Linear Algebra
• Linear Algebra Symbols
• Identity Matrix
• Inverse of a Matrix

Practice Problems on Geometric Multiplicity

Problem 1: Find the geometric multiplicity of the eigenvalues for the matrix:

[Tex]\begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}[/Tex]

Problem 1: Determine the geometric multiplicity of each eigenvalue for the matrix:

[Tex]\begin{pmatrix} 4 & 1 & 2 \\ 0 & 3 & -1 \\ 0 & 0 & 1 \end{pmatrix}[/Tex]

Problem 3: For the matrix below, calculate the geometric multiplicity of its eigenvalues:

[Tex]\begin{pmatrix} 1 & 2 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 3 \end{pmatrix}[/Tex]

Problem 4: Find the geometric multiplicity of the eigenvalues for the matrix:

[Tex]\begin{pmatrix} 5 & 4 & 2 \\ 0 & 1 & -1 \\ 0 & 0 & 3 \end{pmatrix}[/Tex]

Problem 5: Determine the geometric multiplicity of each eigenvalue for the matrix:

[Tex]\begin{pmatrix} 3 & 2 & 4 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \end{pmatrix}[/Tex]

FAQs on Geometric Multiplicity

What is geometric multiplicity?

Geometric multiplicity of an eigenvalue is the number of linearly independent eigenvectors associated with that eigenvalue. It is the dimension of the eigenspace corresponding to the eigenvalue.

How is geometric multiplicity related to algebraic multiplicity?

Algebraic multiplicity is the number of times an eigenvalue appears as a root of the characteristic polynomial. Geometric multiplicity is always less than or equal to algebraic multiplicity.

Can the geometric multiplicity of an eigenvalue be zero?

No, the geometric multiplicity of an eigenvalue is always at least one because there is always at least one eigenvector corresponding to an eigenvalue.

What does it mean if the geometric multiplicity is less than the algebraic multiplicity?

If the geometric multiplicity of an eigenvalue is less than its algebraic multiplicity, it means that the matrix is not diagonalizable. There are not enough linearly independent eigenvectors to form a basis of eigenvectors.

What is the geometric multiplicity of a distinct eigenvalue?

For a distinct eigenvalue (an eigenvalue that appears only once in the characteristic polynomial), the geometric multiplicity is always 1 because there is exactly one linearly independent eigenvector.