The eigenspinors for the spin operator in the $x$-direction? $$S_x= \frac{\hbar}{2}\quad\begin{pmatrix}0&1\\1&0\end{pmatrix}\quad$$
$$S_x{X_+}^{(x)}=\frac{\hbar}{2}{X_+}^{(x)}$$
How can I find the eigenvalue of $S_x$?
My book says
$$
  \left| \begin{array}{cc}
-\lambda &  \frac{\hbar}{2} \\
\frac{\hbar}{2} & -\lambda\\\end{array} \right|=0 $$
So $\lambda=\frac{\hbar}{2} or \frac{-\hbar}{2}$ and therefore
$$S_x{X_+}^{(x)}=\frac{\hbar}{2}{X_+}^{(x)}$$ and $$S_x{X_-}^{(x)}=\frac{-\hbar}{2}{X_-}^{(x)}$$
My question is why do we need determinant=0? And what does the ${-\lambda}$ in the diagonal stand for?
 A: Take case for an $n\times n$ matrix $A$. To find its eigenvalues, first you write the eigenvalue equation for it. 
$$Au=\lambda u$$
where $u$ are its eigenvectors. This can be rewritten in the following way
$$Au-\lambda u=(A-\lambda I)u=0$$
with $I$ the identity matrix. Let $A-\lambda I=B$, and we know that the equation $Bu=0$ has a non zero solution $u$ if and only if $\mathrm{det}B=0$. From this we find that the eigenvalues of $A$ are the $\lambda 's$ which satisfy the following equation
$$\mathrm{det}(A-\lambda I)=0$$ 
A: $\lambda$ stands for the eigenvalue.
Eigenvalue equation is:
$S_xX=\lambda X$
$S_xX-\lambda X=0$
$(S_x-\lambda I)X=0$
Since X is eigenfunction, we seek solutions for $det(S_x-\lambda I)=0$
\begin{align}
(S_x-\lambda I)=
\begin{bmatrix}
 0 & \frac{\hbar}{2} \\
 \frac{\hbar}{2} &0 
\end{bmatrix}
-
\begin{bmatrix}
 \lambda & 0 \\
 0 & \lambda 
\end{bmatrix}
=
\begin{bmatrix}
 - \lambda & \frac{\hbar}{2} \\
 \frac{\hbar}{2} & - \lambda 
\end{bmatrix}
\end{align}
So to find eigenvalues you should just solve this:
\begin{align}
\begin{vmatrix}
 - \lambda & \frac{\hbar}{2} \\
 \frac{\hbar}{2} & - \lambda 
\end{vmatrix}=0
\end{align}
