The algebraic multiplicity of an eigenvalue is the number of times it appears as a root of the characteristic polynomial (i.e., the polynomial whose roots are the eigenvalues of a matrix).

The geometric multiplicity of an eigenvalue is the dimension of the linear space of its associated eigenvectors (i.e., its eigenspace).

In this lecture we provide rigorous definitions of the two concepts of algebraic and geometric multiplicity and we prove some useful facts about them.

Let us start with a definition.

Definition Let be a matrix. Denote by the possibly repeated eigenvalues of , which solve the characteristic equationWe say that an eigenvalue has algebraic multiplicity if and only if there are no more and no less than solutions of the characteristic equation equal to .

Let us see some examples.

Example Consider the matrix The characteristic polynomial isThe roots of the polynomial, that is, the solutions of areThus, has two distinct eigenvalues. Their algebraic multiplicities are because they are not repeated.

Example Define the matrix Its characteristic polynomial isThe roots of the polynomial, that is, the solutions of areThus, has one repeated eigenvalue whose algebraic multiplicity is

Recall that each eigenvalue is associated to a linear space of eigenvectors, called eigenspace.

Definition Let be a matrix. Let be one of the eigenvalues of and denote its associated eigenspace by . The dimension of is called the geometric multiplicity of the eigenvalue .

Let's now make some examples.

Definition Consider the matrix The characteristic polynomial isThe roots of the polynomial areThe eigenvectors associated to are the vectors that solve the equationorThe last equation implies thatTherefore, the eigenspace of is the linear space that contains all vectors of the formwhere can be any scalar. Thus, the eigenspace of is generated by a single vectorTherefore, it has dimension . As a consequence, the geometric multiplicity of is .

Example Consider the matrix The characteristic polynomial isand its roots areThus, there is a repeated eigenvalue () with algebraic multiplicity equal to 2. Its associated eigenvectors solve the equationorThe equation is satisfied for and any value of . As a consequence, the eigenspace of is the linear space that contains all vectors of the formwhere can be any scalar. Since the eigenspace of is generated by a single vectorit has dimension . As a consequence, the geometric multiplicity of is 1, less than its algebraic multiplicity, which is equal to 2.

Example Define the matrix The characteristic polynomial isand its roots areThus, there is a repeated eigenvalue () with algebraic multiplicity equal to 2. Its associated eigenvectors solve the equationorThe equation is satisfied for any value of and . As a consequence, the eigenspace of is the linear space that contains all vectors of the formwhere and are scalars that can be arbitrarily chosen. Thus, the eigenspace of is generated by the two linearly independent vectorsHence, it has dimension . As a consequence, the geometric multiplicity of is 2, equal to its algebraic multiplicity.

A takeaway message from the previous examples is that the algebraic and geometric multiplicity of an eigenvalue do not necessarily coincide.

The following proposition states an important property of multiplicities.

Proposition Let be a matrix. Let be one of the eigenvalues of . Then, the geometric multiplicity of is less than or equal to its algebraic multiplicity.

Proof

Suppose that the geometric multiplicity of is equal to , so that there are linearly independent eigenvectors associated to . Arbitrarily choose vectors , all having dimension and such that the column vectors are linearly independent. Define the matrix For any , denote by the vector that solveswhich is guaranteed to exist because is full-rank (its columns are linearly independent). Define the matrixand denote by its upper block and by its lower block:Denote by the identity matrix. For any scalar , we have thatSince is full-rank and, as a consequence its determinant is non-zero, we can writewhere in step we have used a result about the determinant of block-matrices. The eigenvalues of solve the characteristic equation or, equivalently, the equationThis equation has a root that is repeated at least times. Therefore, the algebraic multiplicity of is at least equal to its geometric multiplicity . It can be larger if is also a root of

When the geometric multiplicity of a repeated eigenvalue is strictly less than
its algebraic multiplicity, then that eigenvalue is said to be
**defective**.

An eigenvalue that is not repeated has an associated eigenvector which is different from zero. Therefore, the dimension of its eigenspace is equal to 1, its geometric multiplicity is equal to 1 and equals its algebraic multiplicity. Thus, an eigenvalue that is not repeated is also non-defective.

Below you can find some exercises with explained solutions.

Find whether the matrixhas any defective eigenvalues.

Solution

The characteristic polynomial isand its roots areThus, there are no repeated eigenvalues and, as a consequence, no defective eigenvalues.

Define

Determine whether possesses any defective eigenvalues.

Solution

The characteristic polynomial iswhere in step we have used the Laplace expansion along the third row. The roots of the polynomial areThus, there is a repeated eigenvalue () with algebraic multiplicity equal to 2. Its associated eigenvectors solve the equationorThe equation is satisfied for any value of and . As a consequence, the eigenspace of is the linear space that contains all vectors of the formwhere the scalar can be arbitrarily chosen. Therefore, the eigenspace of is generated by a single vectorThus, it has dimension , the geometric multiplicity of is 1, its algebraic multiplicity is 2 and it is defective.

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