
#1
Nov1212, 06:41 AM

P: 119

1. The problem statement, all variables and given/known data
Let W be a 1dimensional subspace of V that is Ainvariant. Show that every non zero vector in W is a eigenvector of A. [A element of M_{n}(F)] 3. The attempt at a solution We know W is Ainvariant 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? 



#2
Nov1212, 06:57 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,904

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 "1dimensional" 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? 



#3
Nov1212, 07:55 AM

P: 119

Thanks for the help. I'm still a little confused though.



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