What is the dimension of the eigenspace

The dimension of the eigenspace is called the geometric multiplicity of λ. The algebraic multiplicity of an eigenvalue is the multiplicity of the root. The algebraic multiplicity of an eigenvalue is the multiplicity of the root. For example, the characteristic polynomial of   1 2 3 0 1 1 0 0 2   is (1 − λ)2(2 − λ).

How do you find the dimension of 1 eigenspace?

The dimension of the eigenspace is given by the dimension of the nullspace of A−8I=(1−11−1), which one can row reduce to (1−100), so the dimension is 1. Note that the number of pivots in this matrix counts the rank of A−8I.

What is the largest possible dimension for an eigenspace of A?

The solution given is that, for each each eigenspace, the smallest possible dimension is 1 and the largest is the multiplicity of the eigenvalue (the number of times the root of the characteristic polynomial is repeated).

What is the eigenspace of an eigenvalue?

The set of all eigenvectors of T corresponding to the same eigenvalue, together with the zero vector, is called an eigenspace, or the characteristic space of T associated with that eigenvalue. If a set of eigenvectors of T forms a basis of the domain of T, then this basis is called an eigenbasis.

How do you find Eigenspaces?

The eigenvalues are the roots of the characteristic polynomial, λ = 2 and λ = -3. To find the eigenspace associated with each, we set (A – λI)x = 0 and solve for x. This is a homogeneous system of linear equations, so we put A-λI in row echelon form. 1 ] , or equivalently of [ 1 2 ] .

How many Eigenspaces does a matrix have?

two eigenvalues
Since the characteristic polynomial of matrices is always a quadratic polynomial, it follows that matrices have precisely two eigenvalues — including multiplicity — and these can be described as follows.

Can an eigenspace have dimension 0?

It doesn’t imply that dimension 0 is possible. You know by definition that the dimension of an eigenspace is at least 1. So if the dimension is also at most 1 it means the dimension is exactly 1. It’s a classic way to show that something is equal to exactly some number.

What is the rank of the matrix?

The maximum number of its linearly independent columns (or rows ) of a matrix is called the rank of a matrix. The rank of a matrix cannot exceed the number of its rows or columns. So, there are no independent rows or columns. Hence the rank of a null matrix is zero.

Is an eigenspace a subspace of RN?

A vector x ∈ V is called an eigenvector of A (corre- sponding to λ) if Ax = λx. has a nontrivial solution. The set of all solutions of (3) is just the null space of matrix A − λI. So this set is a subspace of Rn and is called the eigenspace of A corresponding to λ.

How do you describe Eigenspaces?

Vibration analysis – Eigenspace describes the shapes of the vibration modes of an object for each eigenvalue or natural frequency, referred to in this context as an eigenfrequency.

Is eigenspace a vector space?

Finite Dimensional Vector Spaces (In fact, this is why the word “space” appears in the term “eigenspace.”) Let A be an n × n matrix, and let λ be an eigenvalue for A, having eigenspace Eλ.

Is eigenspace a zero vector?

We do not consider the zero vector to be an eigenvector: since A 0 = 0 = λ 0 for every scalar λ , the associated eigenvalue would be undefined.

If A is a 6x6 matrix with characteristic polynomial: x^2(x-1)(x-2)^3 what are the possible dimensions of the eigenspaces?

The Attempt at a Solution

The solution given is that, for each each eigenspace, the smallest possible dimension is 1 and the largest is the multiplicity of the eigenvalue (the number of times the root of the characteristic polynomial is repeated). So, for the eigenspace corresponding to the eigenvalue 2, the dimension is 1, 2, or 3.

I do not understand where this answer comes from. Please come someone help me understand this.

Answers and Replies

Since you are given a square matrix i.e. 6x6 matrix and the characteristic polynomial What are the eigenvalues? What are the multiplicities of each eigenvalue?

What can you say about the dimension for the eigenspace of the eigenvalues? it is "blank" than the multiplicity of the eigenvalue [tex]\lambda_{k}[/tex] for [tex]1 \leq k \leq p[/tex] where p is the number of distinct eigenvalues, which is basically what the book is telling you.

Last edited: Jul 12, 2008

Yes, I understand that. That was me axplaining the answer. I'm asking how they got to that answer. Sorry about the ambiguity.

If an eigenvalue has multiplicity n in the characteristic polynomial, why should its eigenspace have dimension between 1 and n? That's what I don't understand, why is that the case?

The eigenspace is the null space(kernel) of the matrix [tex]A- \lambda I[/tex], which is subspace of [tex]R^{n}[/tex]. By definition of a dimension of a non-zero subspace, it is the number of independent vectors in the basis for the eigenspace.

The multiplicity of the eigenvalue would then be the maximum number, n, of linearly independent vectors in the subspace. Since the eigenspace is non-zero then the dimension must be greater or equal to 1 and the maximum number of independent vectors in the basis is n. If n=3 when [tex]\lambda = 2[/tex], then the dimension can be one-dimensional, two-dimensional, or three-dimensional.

HallsofIvy

From knowing that the characteristic equation is x2(x-1)(x-2)3, you immediately know that the eigenvalues are 0, 1, and 2. Since the x-1 term is linear, you know that the dimension of the subspace of eigenvectors with eigenvalue 1 must be of dimension 1. Since x2 term is of degree two, you know that the subspace of eigenvectors with eigenvalue 1 has dimension either 1 or 2. Since (x-2)3 is of degree 3, you know that the subspace of eigenvectors with eigenvalue 2 has dimension 1, 2, or 3.

I'm feeling a bit thick. Please help me connect the dots. (x-2)^3 is cubic, and P_3 has dimension four? Why can't a cubic term have dimension 4? But it can have dimension less than 3?

I just don't make th connexion from knowing the degree of the term in the characteristic polynomial, and deducing the dimension of the corresponding eigenspace.

matt grime

Just think about the Jordan Normal form (or take it as a definition).

Sorry, I haven't done that (Jordan Normal form) yet.