Eigenvectors and Eigenvalues of a Linear Map

**Anthonny** · October 2nd 2010, 10:05 PM

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

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

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

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

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

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

Was everything I did correct?

Thank you.

**HappyJoe** · October 3rd 2010, 01:11 AM

Check the step again, where you find $\lambda$ .

You are supposed to get that $\lambda$ is 9 or 11 (10 is not an eigenvalue).

**HallsofIvy** · October 3rd 2010, 03:47 AM

Originally Posted by Anthonny

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

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

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

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

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

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

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

Was everything I did correct?

Thank you.

**Anthonny** · October 3rd 2010, 11:31 AM

Originally Posted by HallsofIvy

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

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

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

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?

**HappyJoe** · October 3rd 2010, 11:55 AM

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.

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