Question

How to find the Eignvalues and Eignvector?

Answer 1

\[\text{The Eigenvalues of }\textbf{T }: \lambda_j \]
\[\textbf{T}|\alpha \rangle=\lambda_j|\alpha\rangle \]
\[\left( \textbf{T}-\lambda \textbf{I}\right) | \alpha \rangle =0\]
\[\left| \begin{pmatrix} t_{11}-\lambda & t_{12} \\ t_{21} & t_{22}-\lambda \\ \end{pmatrix}\right|=0\]

Answer 2

I don't know what is that mean?!

Answer 3

ok well \(\textbf T\) is the matrix \(\begin{pmatrix} t_{11} & t_{12}\\ t_{21} & t_{22} \\ \end{pmatrix}\)
\( \textbf I=\begin{pmatrix} 1 & 0\\ 0 & 1 \\ \end{pmatrix}\)
\(\lambda \) is the eigen vector

Answer 4

ok I got that!

Answer 5

So the Eigenvector is lamda and subtract it from the matrix , I will get the Eigenvalues?

Answer 6

sorry \(\lambda\) is the eigen value,
you find the lambdas by taking the determinant of the matrix minus the identify matrix times lambda and equation it to zero

Answer 7

Gotcha

Answer 8

thank you so much

Answer 9

to get the eigen Vector \(|\alpha \rangle\). use either the first or second equation

Answer 10

Ok!

Answer 11

with one of the eigen values at a time

Answer 12

What is this mean?

Answer 13

well find the lambdas first from the determinant, you should get something like this \((\lambda-a)(\lambda+b)=0\)
and so the eigen values \(\lambda_{1,2}\) are \(a\) and \(-b\)

use \(\lambda=a\) to solve for \(|\alpha_1\rangle\)

then use \(\lambda=b\) to solve for \(|\alpha_2\rangle\)

Answer 14

got it
Thank you!