# Geometric multiplicity of an eigenvalue

• Ted123
In summary: I really think you should review vector spaces and linear independence. Why do you think having a zero component in a vector has anything to do with...In summary, the geometric multiplicity of an eigenvalue \lambda is the dimension of the eigenspace, which is the number of elements in a basis for the eigenspace. If all eigenvectors are multiples of a single vector, then the eigenspace is one-dimensional and that single vector is a basis. If the eigenvectors have different forms and are not multiples of each other, then the eigenspace is multi-dimensional and the basis can be determined by finding linearly independent solutions that span the eigenspace. Having a zero component in an eigenvector does not affect
Ted123
Say we have an eigenvalue $\lambda$ and corresponding eigenvectors of the form $(x,x,2x)^T$.

What is the geometric multiplicity?

Ted123 said:
Say we have an eigenvalue $\lambda$ and corresponding eigenvectors of the form $(x,x,2x)^T$.

What is the geometric multiplicity?

Well, what's the definition of geometric multiplicity?

Dick said:
Well, what's the definition of geometric multiplicity?

If we have a matrix $A$ and eigenvalue $\lambda$ then by definition the geometric multiplicity of $\lambda$ is the dimension of $\text{Ker}(A-\lambda I)$ which is just the dimension of the eigenspace.

So if we have found an eigenvector, say $\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$ then what is the geometric multiplicity? i.e. what is the dimension of the eigenspace? Is it the number of non-zero elements of an eigenvector or the number of different non-zero elements?

Ted123 said:
If we have a matrix $A$ and eigenvalue $\lambda$ then by definition the geometric multiplicity of $\lambda$ is the dimension of $\text{Ker}(A-\lambda I)$ which is just the dimension of the eigenspace.

So if we have found an eigenvector, say $\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$ then what is the geometric multiplicity? i.e. what is the dimension of the eigenspace? Is it the number of non-zero elements of an eigenvector or the number of different non-zero elements?

It's neither of those. It's the number of elements in a basis for your eigenspace. You've figured out that all of the eigenvectors are multiples of a single vector. So what's the dimension?

Dick said:
It's neither of those. It's the number of elements in a basis for your eigenspace. You've figured out that all of the eigenvectors are multiples of a single vector. So what's the dimension?

How do I write a basis for the eigenspace? Won't the eigenspace for an eigenvalue always be multiples of an eigenvector?

Ted123 said:
How do I write a basis for the eigenspace? Won't the eigenspace for an eigenvalue always be multiples of an eigenvector?

You just wrote a basis for the eigenspace. It's [1,1,2]. It spans the eigenspace. You might want to review the definition of basis and dimension. If the eigenspace had been given by [x,y,2x]^T, it would be two dimensional. What would be a basis for that?

Dick said:
You just wrote a basis for the eigenspace. It's [1,1,2]. It spans the eigenspace. You might want to review the definition of basis and dimension. If the eigenspace had been given by [x,y,2x]^T, it would be two dimensional. What would be a basis for that?

Ack. I misread your post. I thought you'd written [1,1,2]. All of the vectors in [x,x,2x]^T are multiples of the single vector [1,1,2]^T. Yes?

Dick said:
You just wrote a basis for the eigenspace. It's [1,1,2]. It spans the eigenspace. You might want to review the definition of basis and dimension. If the eigenspace had been given by [x,y,2x]^T, it would be two dimensional. What would be a basis for that?

A question I've just done has an eigenvector $$\begin{bmatrix} 1 \\ 4 \\ 2 \end{bmatrix}$$ for an eigenvalue (from the equation $x=\frac{1}{4}y=\frac{1}{2}z$).

How do I know whether this is the only linearly independent eigenvector in a basis for the eigenspace? (I know the algebraic multiplicity is 2 and that the geometric multiplicity must be less than or equal to this).

Ted123 said:
A question I've just done has an eigenvector $$\begin{bmatrix} 1 \\ 4 \\ 2 \end{bmatrix}$$ for an eigenvalue (from the equation $x=\frac{1}{4}y=\frac{1}{2}z$).

How do I know whether this is the only linearly independent eigenvector in a basis for the eigenspace? (I know the algebraic multiplicity is 2 and that the geometric multiplicity must be less than or equal to this).

It's one dimensional again. All the eigenvectors have the form [x,4x,2x]^T. They are all multiples of a single vector [1,4,2]^T. So that single vector is a basis.

Dick said:
It's one dimensional again. All the eigenvectors have the form [x,4x,2x]^T. They are all multiples of a single vector [1,4,2]^T. So that single vector is a basis.

In another question all the eigenvector equations are multiples of $2x-y-2z=0$

In this case the geometric multiplicity is 2 - how do you know there are 2 linearly independent eigenvectors?

$$\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$$ and $$\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}$$ are 2 linearly independent eigenvectors but isn't$$\begin{bmatrix}0 \\ 2 \\ -1 \end{bmatrix}$$ a 3rd?

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Ted123 said:
In another question all the eigenvector equations are multiples of $2x-y-2z=0$

In this case the geometric multiplicity is 2 - how do you know there are 2 linearly independent eigenvectors?

Solve for z in terms of x and y. So z=x-y/2. That means if you know x and y then z is determined. So it's a two parameter solution, it's a plane. It's two dimensional. Now you tell me a basis for it. Two linearly independent solutions that span the plane.

Dick said:
Solve for z in terms of x and y. So z=x-y/2. That means if you know x and y then z is determined. So it's a two parameter solution, it's a plane. It's two dimensional. Now you tell me a basis for it. Two linearly independent solutions that span the plane.

$$\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$$ and $$\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}$$ are 2 linearly independent eigenvectors but isn't$$\begin{bmatrix}0 \\ 2 \\ -1 \end{bmatrix}$$ a 3rd?

Ted123 said:
$$\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$$ and $$\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}$$ are 2 linearly independent eigenvectors but isn't$$\begin{bmatrix}0 \\ 2 \\ -1 \end{bmatrix}$$ a 3rd?

[0,2,-1]=(1)*[1,2,0]+(-1)*[1,0,1]. No, the third vector isn't linearly independent of the first two.

Dick said:
[0,2,-1]=(1)*[1,2,0]+(-1)*[1,0,1]. No, the third vector isn't linearly independent of the first two.

So it is!

If you have a zero in an eigenvector, say$$\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$$ then how does this affect things? Does it reduce the dimension by 1?

Ted123 said:
So it is!

If you have a zero in an eigenvector, say$$\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$$ then how does this affect things? Does it reduce the dimension by 1?

I really think you should review vector spaces and linear independence. Why do you think having a zero component in a vector has anything to do with it?

## 1. What is the definition of geometric multiplicity of an eigenvalue?

The geometric multiplicity of an eigenvalue is defined as the number of linearly independent eigenvectors associated with that eigenvalue.

## 2. How is the geometric multiplicity related to the algebraic multiplicity of an eigenvalue?

The geometric multiplicity is always less than or equal to the algebraic multiplicity of an eigenvalue. If the geometric multiplicity is equal to the algebraic multiplicity, then the eigenvalue is considered to be non-defective.

## 3. What information does the geometric multiplicity provide about a matrix?

The geometric multiplicity of an eigenvalue provides information about the dimension of the eigenspace associated with that eigenvalue. It also gives insight into the behavior and properties of the matrix, such as whether it is diagonalizable.

## 4. How is the geometric multiplicity calculated?

The geometric multiplicity can be calculated by finding the nullity of the matrix A - λI, where A is the original matrix and λ is the eigenvalue in question.

## 5. Can the geometric multiplicity of an eigenvalue be greater than the dimension of the matrix?

No, the geometric multiplicity of an eigenvalue cannot be greater than the dimension of the matrix. The maximum possible geometric multiplicity for an eigenvalue is equal to the dimension of the matrix.

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