Question

easy question on eigen values and vectors! see image.

Answer 1

um

Answer 2

\(\textbf v=\left[\begin{array}\ v_1\\v_2\\ v_3\end{array}\right]\) cannot be \(\left[\begin{array}\ 0\\0\\ 0\end{array}\right]\) because a zero-vector is not an eigen-vector

Answer 3

thats the reason!

Answer 4

lol where is the logic in that?

Answer 5

well which direction does a zero vector point?

Answer 6

heh well it points no where

also why didn't he apply A-landa*I before doing gaussian reduction?

Answer 7

*nowhere

Answer 8

ok so it has to point somewhere therefore if it equals 0 then it equals 1 if it is the last variable to find when using gaussian reduction

Answer 9

a= original matrix
landa= eigenvalue
I = identity matrix

Answer 10

the answer is the same anyway so i can see that the lecturer may have skipped the step because he could see that v1 = 0 still

Answer 11

the first page finds the eigen-values
then second page if finding the eigen-vector for the first eigen-value
{for lambda equals one

Answer 12

in another example he does it twice...i'll just get it...

Answer 13

the definition of the eigen vector is one that satisfies

\[\textbf {Av}=\textbf v\]
where\(\textbf v≠\mathbf0\)

Answer 14

sorry lost net connection

Answer 15

i ment to say the definition of the eigenvector/eigenvalue is \[\textbf {Av}=\lambda\textbf v\]

Answer 16

what bit are you up to now/

Answer 17

i uploaded the other file to show that in that example just before gaussian reduction he used Av=λv

Answer 18

but in the first image i uploaded he does not minus by an identity matrix.

Answer 19

do you see what i mean?

Answer 20

in the second the eigenvalue beng used is lambda equal negative one , this is why we have taken away an identity matrix, ,
in the first example the eigen value being used is lambda equals one

Answer 21

yep that is right but shouldn't the \[\left[\begin{matrix}1 & -1 & 0 \\ 0 & -4 & 2 \\ 0 & 0 & 2\end{matrix}\right] -(-4)\times \left[\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right] \]

Answer 22

\[= \left[\begin{matrix}1 & 5 & -1 \\ 0 & 0 & 2 \\ 0 & 0 & 6\end{matrix}\right] \]

Answer 23

oh sorry i did the next eigen value

Answer 24

\[\left[\begin{matrix}1 & -1 & 0 \\ 0 & -4 & 2 \\ 0 & 0 & 2\end{matrix}\right] -(-4)\times \left[\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right]= \left[\begin{matrix}5 & -1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 6\end{matrix}\right]\]

Answer 25

sorry disconnected again and made a mistake

Answer 26

yeh so now you do gaussian reduction and the eigen vector is \[= \left[\begin{matrix}1 \\ 0 \\ 0\end{matrix}\right] \]

Answer 27

again

Answer 28

in the 2 examples i gave you he did them differently slightly...so one must be wrong or i missed something

Answer 29

*different

Answer 30

i dont know
should we try another example/

Answer 31

yeh ok...something similar...