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Null space of $$$\left[\begin{array}{cc}- \frac{i \left(7 - \sqrt{229}\right)}{2} + 8 i & -1 + 6 i\\1 + 6 i & - i - \frac{i \left(7 - \sqrt{229}\right)}{2}\end{array}\right]$$$
Your Input
Find the null space of $$$\left[\begin{array}{cc}- \frac{i \left(7 - \sqrt{229}\right)}{2} + 8 i & -1 + 6 i\\1 + 6 i & - i - \frac{i \left(7 - \sqrt{229}\right)}{2}\end{array}\right]$$$.
Solution
The reduced row echelon form of the matrix is $$$\left[\begin{array}{cc}1 & \frac{12 + 2 i}{9 + \sqrt{229}}\\0 & 0\end{array}\right]$$$ (for steps, see rref calculator).
To find the null space, solve the matrix equation $$$\left[\begin{array}{cc}1 & \frac{12 + 2 i}{9 + \sqrt{229}}\\0 & 0\end{array}\right]\left[\begin{array}{c}x_{1}\\x_{2}\end{array}\right] = \left[\begin{array}{c}0\\0\end{array}\right].$$$
If we take $$$x_{2} = t$$$, then $$$x_{1} = \frac{t \left(-12 - 2 i\right)}{9 + \sqrt{229}}$$$.
Thus, $$$\mathbf{\vec{x}} = \left[\begin{array}{c}\frac{t \left(-12 - 2 i\right)}{9 + \sqrt{229}}\\t\end{array}\right] = \left[\begin{array}{c}\frac{-12 - 2 i}{9 + \sqrt{229}}\\1\end{array}\right] t.$$$
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 $$$\left\{\left[\begin{array}{c}\frac{-12 - 2 i}{9 + \sqrt{229}}\\1\end{array}\right]\right\}\approx \left\{\left[\begin{array}{c}-0.497249671655802 - 0.082874945275967 i\\1\end{array}\right]\right\}.$$$A
The nullity of the matrix is $$$1$$$A.