How Do Matrices and Determinants Relate to Each Other?

[rohan03]

please look at the attachement and my attempt at the solution - hope you can help.

Thanks

 

[alfredska]

I see no attachment.

 

[rohan03]

sorry I amended the post and forgot to reattached my document - here it is please look at it.

 

[alfredska]

After "This gives:", you need to check your math. I see (1-\lambda)^2(-1-\lambda)-4(1-\lambda)-4(1-\lambda)

To find eigenvectors, see if this resource helps: http://www.sosmath.com/matrix/eigen2/eigen2.html

Essentially, you will set up A \left(\begin{matrix} x \\ y \\ z \end{matrix}\right) = \lambda \left(\begin{matrix} x \\ y \\ z \end{matrix}\right) where you substitute in each \lambda found earlier. Then you'll have 3 equations and 3 unknowns for each \lambda.

 

[Office_Shredder]

What does it mean for there to be an eigenvalue of -1?
If you write down the equation that the eigenvector satisfies, you should get three equations and three unknowns - note in general the solution for the eigenvector will be non-unique since you can scale it to get a new eigenvector, but you should be able to get a one-dimensional set of vectors that works

 

[rohan03]

ok I had anotherlook at my simplification and I have spotted some error while finding eigen values using 3x3 matrix

what I get is :
( i am using y instead of lambda as I am typing here so apologies)

(1-y){(1-y)(-1-y)-(-2x-2)} -2{(0x-2) -(-2)(1-y)}
=(1-y){-1-y+y+y^2-4} -2 { 0-(-2)(1-y)}
=(1-y){-1+y^2-4-4(1-y)}
=(1-y)(y^2-9)
=(1-y) (y-3) (y+3)
hence giving eigen values of y=1, y=-3 and y=3

is this correct?

 

[alfredska]

hence giving eigen values of y=1, y=-3 and y=3

is this correct?

That looks correct.

 

[rohan03]

thank you so much for all your help.

 

[Ray Vickson]

sorry I amended the post and forgot to reattached my document - here it is please look at it.

I still do not see an attachment.

RGV