Question

Linear algebra help?

Answer 1

I know the formula is Au=lambda*u

Answer 2

I am not really sure how to check it though...

Answer 3

Just a thought, do you need to find the eigenvalues first?

Answer 4

I think I do but I am not really sure :/ . I would assume so though or else I can't check the inequality.

Answer 5

equality*

Answer 6

Thing is though, it's only 2 marks so it seems like it should be easy.

Answer 7

For the left, just multiply the matrix A with u1 and see if you can get (lambda)(u1)

Answer 8

So I still have to find the eigenvalues eh?

Answer 9

No!

Answer 10

But lambda are the eigenvalues which I could find.

Answer 11

For u1,
\[Au_1=\left[\begin{matrix}1 & 1 \\ 4 & -2\end{matrix}\right]\left[\begin{matrix}1 \\ 1\end{matrix}\right] = \left[\begin{matrix}2 \\ 2\end{matrix}\right]=2\left[\begin{matrix}1 \\ 1\end{matrix}\right]=2u_1\]
So, u1 is a eigenvector of A with lambda=2.
Got it? Does it make sense?

Answer 12

Ohh yeah, that makes sense :P .

Answer 13

And you can check u2 :)

Answer 14

Same logic with u2 right?

Answer 15

Yes :)

Answer 16

Thanks! COULD I use the characteristic equation if I wanted though?

Answer 17

Would that be complicated?

Answer 18

Na not really.

Answer 19

But this is significantly easier :P .

Answer 20

So thanks :) .

Answer 21

I think it'll work. You can try :)

Answer 22

@Callisto : I believe u2 is NOT an eigenvector of A right?

Answer 23

Hmm.. Can you show your work?

Answer 24

I don't know how to type matrices in LATEX :P .

Answer 25

Draw it :)