I have a 2 x 2 vector for which I thought I had found the eigenvalues and eigenvectors, but I think I'm misunderstanding a step, because I'm coming up with a scalar version of the answer in the book. Could someone please take a look at my work? (Attached) Thanks! :)

Question

anonymous · Answer

attachment

anonymous · Answer

Thanks, that's what I thought. The notation in the book was a little confusing. It said $$\lambda _1=-2:v_2=(2,3);\lambda _2=-1:v_1=(3,4)  $$

anonymous · Answer

Ok I get it

anonymous · Answer

So basically t can really equal anything you would like except for 0

anonymous · Answer

So are my answer and the book answer essentially equivalent?

anonymous · Answer

Yes

anonymous · Answer

the just set t=4 to get rid of the fraction

anonymous · Answer

since we dont want a general vector so we can pick any value for t especially one that would make the vector look nice

anonymous · Answer

Ah, it's back to aesthetics. :) Well, thanks for your help!

anonymous · Answer

In this case when t=4 then we get rid of the fraction