UnkleRhaukus
  • UnkleRhaukus
Eigenvectors anybody/
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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UnkleRhaukus
  • UnkleRhaukus
\[ \textbf{T}= \begin{pmatrix} 1 & 1-i \\ 1+i & 0 \\ \end{pmatrix} \]\[\lambda_{1,2}=-1,2 \in \mathbb{R}\]
amistre64
  • amistre64
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
amistre64
  • amistre64
does that mean 0,0 is a eugene vector?

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UnkleRhaukus
  • UnkleRhaukus
i dont think that a zero vector is allowed to be an eigenvector
amistre64
  • amistre64
i dont think so either
amistre64
  • amistre64
did we get the Ls right?
amistre64
  • amistre64
i think i get L = -1 and 0
UnkleRhaukus
  • UnkleRhaukus
Ls?
UnkleRhaukus
  • UnkleRhaukus
of you mean lambdas yeah im sure they are right
UnkleRhaukus
  • UnkleRhaukus
\[\text{The Eigenvalues of }\textbf{T } \lambda_j \]\[\textbf{T}|\alpha \rangle=\lambda_j|\alpha\rangle \]\[\left( \textbf{T}-\lambda \textbf{I}\right) | \alpha \rangle =0\]\[\left| \begin{pmatrix} 1-\lambda & 1-i \\ 1+i & -\lambda \\ \end{pmatrix}\right|=0\]\[(1-\lambda)(-\lambda)-(1+i)(1-i)=0\]\[\lambda^2-\lambda-(1-i^2)=0\]\[\lambda^2-\lambda-2=0\]\[(\lambda+1)(\lambda-2)=0\]\[\lambda_{1,2}=-1,2 \in \mathbb{R}\]
UnkleRhaukus
  • UnkleRhaukus
tell me if i have made a mistake
amistre64
  • amistre64
-(1+i) = -1-i -1-i 1+i ------ -1 -i+i-i^2 = -1--1 = 0
amistre64
  • amistre64
1+i 1-i ---- 1 +i -i -i^2 = 1--1 = 2 so i have to wonder what it is ive forgotten
amistre64
  • amistre64
-i -i = -2i; i see that
UnkleRhaukus
  • UnkleRhaukus
\[\textbf{T}|\alpha \rangle=\lambda_1|\alpha\rangle =-1|\alpha\rangle\] \[ \begin{pmatrix} 1-(-1) & 1-i \\ 1+i & 0-(-1) \\ \end{pmatrix} \begin{pmatrix} \alpha_1\\ \alpha_2 \end{pmatrix} = -1\begin{pmatrix} \alpha_1\\ \alpha_2 \end{pmatrix}\] \[ \begin{pmatrix} 2 & 1-i \\ 1+i & 1 \\ \end{pmatrix} \begin{pmatrix} \alpha_1\\ \alpha_2 \end{pmatrix} = \begin{pmatrix} -\alpha_1\\ -\alpha_2 \end{pmatrix}\] \[2\alpha_1+(1-i)\alpha_2=-\alpha_1\quad{{(i)}}\] \[(1+i)\alpha_1+\alpha_2=-\alpha_2\quad{{(ii)}}\] \[{(i) \rightarrow }\quad (1-i)\alpha_2=\alpha_1\] \[{{(ii)}\rightarrow}\quad(1+i)\alpha_1=0\]
amistre64
  • amistre64
yours looks better :)
UnkleRhaukus
  • UnkleRhaukus
i have been working on it for quite some time
amistre64
  • amistre64
my innate ability to multiply incorrectly rears its head
UnkleRhaukus
  • UnkleRhaukus
i dont know if this help but the solution in the back of the book part (c)
1 Attachment
UnkleRhaukus
  • UnkleRhaukus
it looks like they have left something out to me
amistre64
  • amistre64
Ax = Lx Ax - Lx = 0 (A-L)x = 0 rref A-L to determine eugene vectors
UnkleRhaukus
  • UnkleRhaukus
so i have to get it in reduced row echelon form i guess i didn't actually do that
amistre64
  • amistre64
thats what i tried to do before i forgot how to multiply
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=rref%7B%7B2%2C1-i%7D%2C%7B1%2Bi%2C1%7D%7D Ev1 = [ (i-1)/2 ] [ 1 ]
amistre64
  • amistre64
the other one should reduce to i-1 , 1
UnkleRhaukus
  • UnkleRhaukus
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.

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