# Thread: span,eigenvector,dimension 1,prove help-exam upcoming!

1. ## span,eigenvector,dimension 1,prove help-exam upcoming!

Hi ,
I need help with this question.thanks.

Let A be an nxn matrix,and let v be a vector in R^n. Prove that v is an eigenvector of A if and only if Span({v,Av}) has dimension 1.

2. ## Re: span,eigenvector,dimension 1,prove help-exam upcoming!

Hey n22.

Hint: If v is an eigenvector then it means that it satisfies Av = lv where l is an eigen-value. This means that A only scales the vector and doesn't change its direction.

Given this information you have scalar multiples of the same vector. What does this imply for the dimension of the span of the two vectors?

3. ## Re: span,eigenvector,dimension 1,prove help-exam upcoming!

Originally Posted by chiro
Hey n22.

Hint: If v is an eigenvector then it means that it satisfies Av = lv where l is an eigen-value. This means that A only scales the vector and doesn't change its direction.

Given this information you have scalar multiples of the same vector. What does this imply for the dimension of the span of the two vectors?
Could you provide me more guidance please?

4. ## Re: span,eigenvector,dimension 1,prove help-exam upcoming!

With eigen-vectors you have the situation where Av = lambda*v which means Av generates a vector that is a scalar multiple of the original vector v.

Now you have two vectors: v and cv. If you row-reduce those you should get a row of 0's leaving one line and hence one dimension for the spanning set involving both vectors.