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Null space of $$$\left[\begin{array}{cc}2 & 1 - i\\1 + i & 1\end{array}\right]$$$
Your Input
Find the null space of $$$\left[\begin{array}{cc}2 & 1 - i\\1 + i & 1\end{array}\right]$$$.
Solution
The reduced row echelon form of the matrix is $$$\left[\begin{array}{cc}1 & \frac{1}{2} - \frac{i}{2}\\0 & 0\end{array}\right]$$$ (for steps, see rref calculator).
To find the null space, solve the matrix equation $$$\left[\begin{array}{cc}1 & \frac{1}{2} - \frac{i}{2}\\0 & 0\end{array}\right]\left[\begin{array}{c}x_{1}\\x_{2}\end{array}\right] = \left[\begin{array}{c}0\\0\end{array}\right].$$$
If we take $$$x_{2} = t$$$, then $$$x_{1} = t \left(- \frac{1}{2} + \frac{i}{2}\right)$$$.
Thus, $$$\mathbf{\vec{x}} = \left[\begin{array}{c}t \left(- \frac{1}{2} + \frac{i}{2}\right)\\t\end{array}\right] = \left[\begin{array}{c}- \frac{1}{2} + \frac{i}{2}\\1\end{array}\right] t.$$$
This is the null space.
The nullity of a matrix is the dimension of the basis for the null space.
Thus, the nullity of the matrix is $$$1$$$.
Answer
The basis for the null space is $$$\left\{\left[\begin{array}{c}- \frac{1}{2} + \frac{i}{2}\\1\end{array}\right]\right\} = \left\{\left[\begin{array}{c}-0.5 + 0.5 i\\1\end{array}\right]\right\}.$$$A
The nullity of the matrix is $$$1$$$A.