Problem 3. Find the unitary operator hat{U} which diagonalizes one of the Pauli matrices hat{sigma}_x = begin{pmatrix}0 & 1 1 & 0 end{pmatrix} Obtain the eigenvalues and eigenvectors of hat{sigma}_x

Step 1
Given
Pauli matrices
${\hat{\sigma }}_x=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$
According to question
$\frac{\hat{U}}{{\hat{\sigma }}_x}=\left[\begin{array}{cc}{\lambda }_1& 0\\ 0& {\lambda }_2\end{array}\right]$
Where $\hat{U}$ is the unitary operator which diagonalizes the the Pauli matrices.
Step 2
Now Eigenvalues of Pauli matrix.
$|{\hat{\sigma }}_x-\lambda I|=\left[\begin{array}{cc}-\lambda & 1\\ 1& -\lambda \end{array}\right]=0$
$$\begin{gathered} \Rightarrow \left[\begin{array}{cc}-\lambda & 1\\ 1& -\lambda \end{array}\right]=0 \\ \Rightarrow {\lambda }^2-1=0 \end{gathered}$$
$\Rightarrow \lambda =1,-1$
Now Eigenvector,
$\lambda =1$
$\left[\begin{array}{cc}-1& 1\\ 1& -1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=0$
$\Rightarrow x=y$
Hence Eigenvector for $\lambda =1$ is
$X=\left[\begin{array}{c}1\\ 1\end{array}\right]$
Eigenvector for
$\lambda =-1$
$\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=0$
$\Rightarrow x=-y$
Hence Eigenvector for $\lambda =-1$ is
$Y=\left[\begin{array}{c}1\\ -1\end{array}\right]$
Let for unitary operator,
$\hat{U}=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\text{ then}$
$\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]$
$\Rightarrow \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]{\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]}^{-1}$
$\Rightarrow \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$
$\Rightarrow \hat{U}=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$