- #1
- 446
- 0
Say we have an eigenvalue [itex]\lambda[/itex] and corresponding eigenvectors of the form [itex](x,x,2x)^T[/itex].
What is the geometric multiplicity?
What is the geometric multiplicity?
Well, what's the definition of geometric multiplicity?Say we have an eigenvalue [itex]\lambda[/itex] and corresponding eigenvectors of the form [itex](x,x,2x)^T[/itex].
What is the geometric multiplicity?
If we have a matrix [itex]A[/itex] and eigenvalue [itex]\lambda[/itex] then by definition the geometric multiplicity of [itex]\lambda[/itex] is the dimension of [itex]\text{Ker}(A-\lambda I)[/itex] which is just the dimension of the eigenspace.Well, what's the definition of geometric multiplicity?
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?If we have a matrix [itex]A[/itex] and eigenvalue [itex]\lambda[/itex] then by definition the geometric multiplicity of [itex]\lambda[/itex] is the dimension of [itex]\text{Ker}(A-\lambda I)[/itex] which is just the dimension of the eigenspace.
So if we have found an eigenvector, say [itex]\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}[/itex] 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?
How do I write a basis for the eigenspace? Won't the eigenspace for an eigenvalue always be multiples of an eigenvector?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?
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?How do I write a basis for the eigenspace? Won't the eigenspace for an eigenvalue always be multiples of an eigenvector?
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?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 [tex]\begin{bmatrix} 1 \\ 4 \\ 2 \end{bmatrix}[/tex] for an eigenvalue (from the equation [itex]x=\frac{1}{4}y=\frac{1}{2}z[/itex]).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?
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.A question I've just done has an eigenvector [tex]\begin{bmatrix} 1 \\ 4 \\ 2 \end{bmatrix}[/tex] for an eigenvalue (from the equation [itex]x=\frac{1}{4}y=\frac{1}{2}z[/itex]).
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).
In another question all the eigenvector equations are multiples of [itex]2x-y-2z=0[/itex]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.
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.In another question all the eigenvector equations are multiples of [itex]2x-y-2z=0[/itex]
In this case the geometric multiplicity is 2 - how do you know there are 2 linearly independent eigenvectors?
[tex]\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}[/tex] and [tex]\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}[/tex] are 2 linearly independent eigenvectors but isn't[tex]\begin{bmatrix}0 \\ 2 \\ -1 \end{bmatrix}[/tex] a 3rd?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.
[0,2,-1]=(1)*[1,2,0]+(-1)*[1,0,1]. No, the third vector isn't linearly independent of the first two.[tex]\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}[/tex] and [tex]\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}[/tex] are 2 linearly independent eigenvectors but isn't[tex]\begin{bmatrix}0 \\ 2 \\ -1 \end{bmatrix}[/tex] a 3rd?
So it is![0,2,-1]=(1)*[1,2,0]+(-1)*[1,0,1]. No, the third vector isn't linearly independent of the first two.
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????So it is!
If you have a zero in an eigenvector, say[tex]\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}[/tex] then how does this affect things? Does it reduce the dimension by 1?