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Invariant spaces and eigenvector problem

by gottfried
Tags: eigenvector, invariant, spaces
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gottfried
#1
Nov12-12, 06:41 AM
P: 119
1. The problem statement, all variables and given/known data
Let W be a 1-dimensional subspace of V that is A-invariant. Show that every non zero vector in W is a eigenvector of A. [A element of Mn(F)]

3. The attempt at a solution

We know W is A-invariant therefore for all w in W A.w is in W. W is one dimensional which implies to me that A must therefore be a one by one matrix with an entry from F. Is this a correct assumption?

If so then A.w=λ.w where λ is an element of F which implies that all w in W are eigenvectors of A.

I'm new to this sort of linear algebra and therefore can't tell if I've made a blatant mistake?
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HallsofIvy
#2
Nov12-12, 06:57 AM
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Quote Quote by gottfried View Post
1. The problem statement, all variables and given/known data
Let W be a 1-dimensional subspace of V that is A-invariant. Show that every non zero vector in W is a eigenvector of A. [A element of Mn(F)]

3. The attempt at a solution

We know W is A-invariant therefore for all w in W A.w is in W. W is one dimensional which implies to me that A must therefore be a one by one matrix with an entry from F. Is this a correct assumption?
No, in fact it makes no sense at all. Any n by n matrix can have "1-dimensional invarient subspace".

If so then A.w=λ.w where λ is an element of F which implies that all w in W are eigenvectors of A.
How did you arrive at that? If w is in the invariant subspace, W, then Aw is also in W and, since W is "1-dimensional", it must be a multiple of A. But it does not immediately follow that that multiple is the same for all vectors in W.

Suppose that u and v are different vectors in W. Then we can say that [itex]Au= \lambda_1 u[/itex] and that [itex]Av= \lambda v[/itex] but we cannot yet say that [itex]\lambda_1= \lambda_2[/itex]. To do that, we need to use the fact that W is "1-dimensional" again. Because of that, any vector in the subspace is a mutiple of any other (except 0). We can write u= xv so that Au= x(Av). What does that give you?

I'm new to this sort of linear algebra and therefore can't tell if I've made a blatant mistake?
gottfried
#3
Nov12-12, 07:55 AM
P: 119
Thanks for the help. I'm still a little confused though.

Quote Quote by HallsofIvy View Post

Suppose that u and v are different vectors in W. Then we can say that [itex]Au= \lambda_1 u[/itex] and that [itex]Av= \lambda v[/itex] but we cannot yet say that [itex]\lambda_1= \lambda_2[/itex].
Why do we need to show that the two λ's are the same. Isn't that [itex] Au= \lambda_1 u[/itex] sufficient because that is the definition of an eigenvector? Unless eigenvalues have to unique for each matrix?


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