Question

Find the EigenVectors of \[\textbf{T}\]

Answer 1

\[ \text{Let } \textbf{T}= \begin{pmatrix} 1 & 1-i \\ 1+i & 0 \\ \end{pmatrix} \]

Answer 2

i have shown the eigenvalues are
\[\lambda_{1,2}=2,-1\]

Answer 3

you just need to solve \[\left( T -\lambda I\right)x =0\] x is the eigenvector

Answer 4

where am i going wrong

Answer 5

\[\left( T-\lambda I \right)x=0\]\[\left[\begin{matrix}1-\lambda & 1-i \\ 1+i & -\lambda\end{matrix}\right]
\left(\begin{matrix}y \\ z\end{matrix}\right)=\left(\begin{matrix}0 \\ 0\end{matrix}\right)\]\[\lambda=2\]\[\left[\begin{matrix}-1 & 1-i \\ 1+i & -2\end{matrix}\right]
\left(\begin{matrix}y \\ z\end{matrix}\right)=\left(\begin{matrix}0 \\ 0\end{matrix}\right)\]we rewrite as a system of equations,
\[-y + (1 - i)z = 0\]\[(1+i)y - 2z = 0\]from the first equation we get,
\[y=\left( 1-i \right)z\]so the vector x,
\[x=z\left(\begin{matrix}1-i \\ 1\end{matrix}\right)\]z is a constant, therefore the eigenvector,
\[x=\left(\begin{matrix}1-i \\ 1\end{matrix}\right)\]

Answer 6

repeat the steps for \[\lambda=-1\]

Answer 7

In your posted pdf document you get down to 2 equations
\[ a_1 + \frac{1-i}{2} a_2=0 \]
\[ a_2=0 \]
The second equation should be
\[ 0a_1+0a_2=0 \]
Standard procedure is to say that a_2 is a free variable, and can take any value (though a_2= 0 leads to the (0,0) solution which we do not want) Typically people assign a_2= 1
with a_2= 1, a1 becomes (i-1)/2 and you eigenvector is

v1= [(i-1)/2 1]
if you normalize, v1= v1/sqrt( conj(v1) dot v1) you get
v1n= [ (i-1)/sqrt(6) sqrt(2/3) ]