spinor representation I am trying to study Special unitary group of order 2 and some textbooks mention objects transform under special unitary group are called Spinors then  How can we represent a spinor using matrix ?
 A: $SU(2)$ can be defined as the group of determinant-$1$ transformations which leaves,
$$ | z_1 |^2 + |z_2|^2 = 1,$$
invariant.
The transformations themselves act on the two complex numbers $z_1$ and $z_2$. It is common to represent $z_1$ and $z_2$ as the two components of a complex column vector,
$$ \left(\begin{array} \ z_1 \\ z_2 \end{array}\right),$$
but there is nothing to prevent us from representing these complex numbers in a $2\times 2$ matrix.
In particular the actual elements of $SU(2)$ can be thought of as spinors. From wikipedia we have the group elements of $SU(2)$ can be represented as
$$ \left( \begin{array} \ \alpha & - \bar{\beta} \\ \beta & \bar{\alpha} \end{array}\right) \qquad |\alpha|^2 + |\beta|^2 = 1,$$
notice that this matrix is determined by only two complex numbers which satisfy the same normalization as the $z_1$ and $z_2$ indicated previously. In fact this matrix is completely determined by the spinor in its first column.

To see that the representation is really faithful let us compare the action of group elements when they act on our spinors.
$$ \left( \begin{array} \ \alpha & - \bar{\beta} \\ \beta & \bar{\alpha} \end{array}\right) \left( \begin{array} \ z_1 \\ z_2\end{array} \right) = \left( \begin{array} \ \alpha z_1 - \bar{\beta} z_2 \\ \beta z_1 + \bar{\alpha} z_2\end{array}\right)$$
Now consider the other transformation:
$$ \left( \begin{array} \ \alpha & - \bar{\beta} \\ \beta & \bar{\alpha} \end{array}\right) \left( \begin{array} \ z_1 & - \bar{z_2} \\ z_2 & \bar{z_1} \end{array}\right)$$
$$ = \left( \begin{array} \ \alpha z_1 - \bar{\beta} z_2 &  -\alpha \bar{z_2} - \bar{\beta} \bar{z_1} \\ \beta z_1 + \bar{\alpha} z_2 & -\beta \bar{z_2} + \bar{\alpha} \bar{z_1} \end{array}\right) $$
$$ = \left( \begin{array} \ \alpha z_1 - \bar{\beta} z_2 &  \overline{ -\bar{\alpha} z_2 - \beta z_1} \\ \beta z_1 + \bar{\alpha} z_2 & \overline{-\bar{\beta} z_2 + \alpha z_1 } \end{array}\right) $$
$$ = \left( \begin{array} \ \alpha z_1 - \bar{\beta} z_2 &  -\overline{\left( \bar{\alpha} z_2 + \beta z_1\right)} \\ \beta z_1 + \bar{\alpha} z_2 & \overline{ \alpha z_1 -\bar{\beta} z_2  } \end{array}\right)      $$
$$ = \left( \begin{array} \ \alpha z_1 - \bar{\beta} z_2 &  -\overline{\left(  \beta z_1 + \bar{\alpha} z_2 \right)} \\ \beta z_1 + \bar{\alpha} z_2 & \overline{ \alpha z_1 -\bar{\beta} z_2  } \end{array}\right) $$
You can see that the $z_1$ and $z_2$ are mixed in the same way for both transformations.

It doesn't cause any difficulty if we allow the quadratic form $|z_1|^2+|z_2|^2$ to equal any real number.
The transformation matrices would remain the same.
$$ \left( \begin{array} \ \alpha & - \bar{\beta} \\ \beta & \bar{\alpha} \end{array}\right) \qquad |\alpha|^2 + |\beta|^2 = 1,$$
The spinors would still be expressible as $2\times 2$ matrices.
$$ \left( \begin{array} \ z_1 & - \bar{z_2} \\ z_2 & \bar{z_1} \end{array}\right)$$
