Question

A= 4 -2
-1 5
λ=3, 6
Find the eigenvector that corresponds to each eigenvalue

Answer 1

Well I know we have to use this equation:
(A−λI)v=0
Where A is the matrix and λ is an eigenvalue and I is the identity matrix but I get lost in the last step.

Answer 2

|a-L b | rref = |1 b/(a-L)|
| c d-L| |0 0 |


ev = <-b/(a-L),1>

Answer 3

since evs aint unique; we can also write it as:

<-b, a-L>

<b, L-a>

<d-L, -c>

<L-d, c>

etc .....

Answer 4

you seem to have frozen up :)

Answer 5

I dont understand your notation

Answer 6

\[(\left[\begin{matrix}4 & -2 \\ -1 & 5\end{matrix}\right]-\left[\begin{matrix}3 & 0 \\ 0 & 3\end{matrix}\right])v=0\]

Answer 7

\[\left[\begin{matrix}1 & -2 \\ -1 & 2\end{matrix}\right]v=0\]

Answer 8

i just row reduced your generic 2x2 matrix with arbitrary eugene values

Answer 9

\[v=\left(\begin{matrix}v1 \\ v2\end{matrix}\right)\]
v1 - 2*v2=0
-v1+ 2*v2=0
What do I do from here

Answer 10

row reduce the matrix

Answer 11

I could of done that before -.- haha anyways so I get v1 - 2*v2=0
and 0=0
does that mean v2 is a free variable?

Answer 12

yes

Answer 13

Ok now Im confused on how to write the vector....

Answer 14

2/4-3 = 2/1

L=3, Ev = <2,1>

Answer 15

solve for v1 in terms of v2

and solve v2 in terms of v2

Answer 16

oh nevermind I get it!!

Answer 17

Thanks!!!

Answer 18

yw