The discussion revolves around the concept of geometric multiplicity of an eigenvalue, particularly focusing on the relationship between eigenvectors and the dimension of the eigenspace. Participants explore definitions, properties, and implications of geometric multiplicity in the context of linear algebra.
The discussion is active, with participants questioning definitions and clarifying concepts related to eigenvectors and eigenspaces. Some guidance has been offered regarding the nature of bases and dimensions, and there is an ongoing exploration of examples to illustrate these concepts.
Participants reference specific eigenvectors and their forms, discussing how these relate to the dimensionality of the eigenspace. There is mention of algebraic multiplicity and its relationship to geometric multiplicity, as well as the implications of linear dependence among eigenvectors.
Ted123 said:Say we have an eigenvalue \lambda and corresponding eigenvectors of the form (x,x,2x)^T.
What is the geometric multiplicity?
Dick said:Well, what's the definition of geometric multiplicity?
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?
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?
Ted123 said:How do I write a basis for the eigenspace? Won't the eigenspace for an eigenvalue always be multiples of an eigenvector?
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?
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?
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).
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.
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?
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.
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?
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.
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?