# Eigenvectors and Eigenvalues of a Linear Map

• Oct 2nd 2010, 10:05 PM
Anthonny
Eigenvectors and Eigenvalues of a Linear Map
Suppose I had a linear map $\displaystyle T:\mathbb{R}^2\rightarrow\mathbb{R}^2$ such that $\displaystyle T(x,y)=(10x+y,10y+x)$ and I want to find the eigenvalues and eigenvectors of T.

I'm doing this without determinants. I have it that
$\displaystyle 10x+y=\lambda{x}$
$\displaystyle 10y+x=\lambda{y}$

After rearranging terms and doing simple elimination I have
$\displaystyle -(10-\lambda)(10-\lambda)=0$

I believe the eigenvalue is then $\displaystyle \lambda=10$. Next to find the eigenvectors I would find $\displaystyle null(T-10I)$

To find $\displaystyle null(T-10I)$ I would end up having that
$\displaystyle 10x+y-10x=0$
$\displaystyle 10y+x-10y=0$

and so I have $\displaystyle x=y$. I believe this means that all eigenvectors would be in the form $\displaystyle (x,x)$ (i.e. (1,1),(2,2)...)

Was everything I did correct?

Thank you.
• Oct 3rd 2010, 01:11 AM
HappyJoe
Check the step again, where you find $\displaystyle \lambda$.

You are supposed to get that $\displaystyle \lambda$ is 9 or 11 (10 is not an eigenvalue).
• Oct 3rd 2010, 03:47 AM
HallsofIvy
Quote:

Originally Posted by Anthonny
Suppose I had a linear map $\displaystyle T:\mathbb{R}^2\rightarrow\mathbb{R}^2$ such that $\displaystyle T(x,y)=(10x+y,10y+x)$ and I want to find the eigenvalues and eigenvectors of T.

I'm doing this without determinants. I have it that
$\displaystyle 10x+y=\lambda{x}$
$\displaystyle 10y+x=\lambda{y}$

After rearranging terms and doing simple elimination I have
$\displaystyle -(10-\lambda)(10-\lambda)=0$

No, you don't. Multiply the second equation by $\displaystyle 10- \lambda$ and the subtract it from the first equation. I suspect you did that but forgot the x in the second equation. You get $\displaystyle ((10-\lambda)^2-1)x= 0$. In order that that equation not have x= 0 as its only solution we must have $\displaystyle (10- \lambda)^2- 1= 0$

Quote:

I believe the eigenvalue is then $\displaystyle \lambda=10$. Next to find the eigenvectors I would find $\displaystyle null(T-10I)$

To find $\displaystyle null(T-10I)$ I would end up having that
$\displaystyle 10x+y-10x=0$
$\displaystyle 10y+x-10y=0$

and so I have $\displaystyle x=y$. I believe this means that all eigenvectors would be in the form $\displaystyle (x,x)$ (i.e. (1,1),(2,2)...)

Was everything I did correct?

Thank you.
• Oct 3rd 2010, 11:31 AM
Anthonny
Quote:

Originally Posted by HallsofIvy
No, you don't. Multiply the second equation by $\displaystyle 10- \lambda$ and the subtract it from the first equation. I suspect you did that but forgot the x in the second equation. You get $\displaystyle ((10-\lambda)^2-1)x= 0$. In order that that equation not have x= 0 as its only solution we must have $\displaystyle (10- \lambda)^2- 1= 0$

Thanks. I see what I did wrong, I did forget about the x.

For the eigenvectors for $\displaystyle \lambda=9$ I would have to find $\displaystyle null(T-9I)$ which would be $\displaystyle null((10x+y,10y+x)-(9x,9y))=null(x+y,y+x)$.

Setting (x+y,y+x)=(0,0) I would have x+y=0 and y+x=0 and so x=-y and y=-x. I believe the eigenvectors are then in the form (x,-x) i.e. (1,-1),(2,-2),...

Is how I found the eigenvectors correct?
• Oct 3rd 2010, 11:55 AM
HappyJoe
It is indeed correct that all eigenvectors corresponding to the eigenvector 9 are of the form (x,-x), but do notice that this also implies that stuff like (1/2,-1/2) and (pi,-pi) are eigenvectors - not just (1,-1), (2,-2), and so on. :)