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Null space of $$$\left[\begin{array}{cc}\frac{1}{2} - \frac{\sqrt{5}}{2} & 1\\1 & - \frac{\sqrt{5}}{2} - \frac{1}{2}\end{array}\right]$$$
Your Input
Find the null space of $$$\left[\begin{array}{cc}\frac{1}{2} - \frac{\sqrt{5}}{2} & 1\\1 & - \frac{\sqrt{5}}{2} - \frac{1}{2}\end{array}\right]$$$.
Solution
The reduced row echelon form of the matrix is $$$\left[\begin{array}{cc}1 & \frac{2}{1 - \sqrt{5}}\\0 & 0\end{array}\right]$$$ (for steps, see rref calculator).
To find the null space, solve the matrix equation $$$\left[\begin{array}{cc}1 & \frac{2}{1 - \sqrt{5}}\\0 & 0\end{array}\right]\left[\begin{array}{c}x_{1}\\x_{2}\end{array}\right] = \left[\begin{array}{c}0\\0\end{array}\right].$$$
If we take $$$x_{2} = t$$$, then $$$x_{1} = \frac{t \left(1 + \sqrt{5}\right)}{2}$$$.
Thus, $$$\mathbf{\vec{x}} = \left[\begin{array}{c}\frac{t \left(1 + \sqrt{5}\right)}{2}\\t\end{array}\right] = \left[\begin{array}{c}\frac{1 + \sqrt{5}}{2}\\1\end{array}\right] t.$$$
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 $$$\left\{\left[\begin{array}{c}\frac{1 + \sqrt{5}}{2}\\1\end{array}\right]\right\}\approx \left\{\left[\begin{array}{c}1.618033988749895\\1\end{array}\right]\right\}.$$$A
The nullity of the matrix is $$$1$$$A.