Question

Eigenvectors anybody/

Answer 1

\[ \textbf{T}= \begin{pmatrix} 1 & 1-i \\ 1+i & 0 \\ \end{pmatrix} \]\[\lambda_{1,2}=-1,2 \in \mathbb{R}\]

Answer 2

2 1-i
1+i 1

1 (1-i)/2
1+i 1

-1-i 0
1+i 1

1 (1-i)/2
0 1

1 0
0 1

hmm

Answer 3

does that mean 0,0 is a eugene vector?

Answer 4

i dont think that a zero vector is allowed to be an eigenvector

Answer 5

i dont think so either

Answer 6

did we get the Ls right?

Answer 7

i think i get L = -1 and 0

Answer 8

Ls?

Answer 9

of you mean lambdas yeah im sure they are right

Answer 10

\[\text{The Eigenvalues of }\textbf{T } \lambda_j \]\[\textbf{T}|\alpha \rangle=\lambda_j|\alpha\rangle \]\[\left( \textbf{T}-\lambda \textbf{I}\right) | \alpha \rangle =0\]\[\left| \begin{pmatrix} 1-\lambda & 1-i \\ 1+i & -\lambda \\ \end{pmatrix}\right|=0\]\[(1-\lambda)(-\lambda)-(1+i)(1-i)=0\]\[\lambda^2-\lambda-(1-i^2)=0\]\[\lambda^2-\lambda-2=0\]\[(\lambda+1)(\lambda-2)=0\]\[\lambda_{1,2}=-1,2 \in \mathbb{R}\]

Answer 11

tell me if i have made a mistake

Answer 12

-(1+i) = -1-i

-1-i
1+i
------
-1 -i+i-i^2 = -1--1 = 0

Answer 13

1+i
1-i
----
1 +i -i -i^2 = 1--1 = 2

so i have to wonder what it is ive forgotten

Answer 14

-i -i = -2i; i see that

Answer 15

\[\textbf{T}|\alpha \rangle=\lambda_1|\alpha\rangle =-1|\alpha\rangle\]
\[ \begin{pmatrix} 1-(-1) & 1-i \\ 1+i & 0-(-1) \\ \end{pmatrix} \begin{pmatrix} \alpha_1\\ \alpha_2 \end{pmatrix} = -1\begin{pmatrix} \alpha_1\\ \alpha_2 \end{pmatrix}\]
\[ \begin{pmatrix} 2 & 1-i \\ 1+i & 1 \\ \end{pmatrix} \begin{pmatrix} \alpha_1\\ \alpha_2 \end{pmatrix} = \begin{pmatrix} -\alpha_1\\ -\alpha_2 \end{pmatrix}\]
\[2\alpha_1+(1-i)\alpha_2=-\alpha_1\quad{{(i)}}\]
\[(1+i)\alpha_1+\alpha_2=-\alpha_2\quad{{(ii)}}\]

\[{(i) \rightarrow }\quad (1-i)\alpha_2=\alpha_1\]
\[{{(ii)}\rightarrow}\quad(1+i)\alpha_1=0\]

Answer 16

yours looks better :)

Answer 17

i have been working on it for quite some time

Answer 18

my innate ability to multiply incorrectly rears its head

Answer 19

i dont know if this help but the solution in the back of the book
part (c)

Answer 20

it looks like they have left something out to me

Answer 21

Ax = Lx

Ax - Lx = 0

(A-L)x = 0

rref A-L to determine eugene vectors

Answer 22

so i have to get it in reduced row echelon form i guess i didn't actually do that

Answer 23

thats what i tried to do before i forgot how to multiply

Answer 24

http://www.wolframalpha.com/input/?i=rref%7B%7B2%2C1-i%7D%2C%7B1%2Bi%2C1%7D%7D

Ev1 = [ (i-1)/2 ]
[ 1 ]

Answer 25

the other one should reduce to i-1 , 1

Answer 26

so the back of the book has the right answer,
and i hadn't thought of using wolfram to reduce my matrices so that is helpful to know as well.