Null space of $$$\left[\begin{array}{cccc}\sqrt{6} & 0 & 1 & 0\\0 & \sqrt{6} & 1 & 0\\1 & 1 & \sqrt{6} & 2\\0 & 0 & 2 & \sqrt{6}\end{array}\right]$$$
Your Input
Find the null space of $$$\left[\begin{array}{cccc}\sqrt{6} & 0 & 1 & 0\\0 & \sqrt{6} & 1 & 0\\1 & 1 & \sqrt{6} & 2\\0 & 0 & 2 & \sqrt{6}\end{array}\right]$$$.
Solution
The reduced row echelon form of the matrix is $$$\left[\begin{array}{cccc}1 & 0 & 0 & - \frac{1}{2}\\0 & 1 & 0 & - \frac{1}{2}\\0 & 0 & 1 & \frac{\sqrt{6}}{2}\\0 & 0 & 0 & 0\end{array}\right]$$$ (for steps, see rref calculator).
To find the null space, solve the matrix equation $$$\left[\begin{array}{cccc}1 & 0 & 0 & - \frac{1}{2}\\0 & 1 & 0 & - \frac{1}{2}\\0 & 0 & 1 & \frac{\sqrt{6}}{2}\\0 & 0 & 0 & 0\end{array}\right]\left[\begin{array}{c}x_{1}\\x_{2}\\x_{3}\\x_{4}\end{array}\right] = \left[\begin{array}{c}0\\0\\0\\0\end{array}\right].$$$
If we take $$$x_{4} = t$$$, then $$$x_{1} = \frac{t}{2}$$$, $$$x_{2} = \frac{t}{2}$$$, $$$x_{3} = - \frac{\sqrt{6} t}{2}$$$.
Thus, $$$\mathbf{\vec{x}} = \left[\begin{array}{c}\frac{t}{2}\\\frac{t}{2}\\- \frac{\sqrt{6} t}{2}\\t\end{array}\right] = \left[\begin{array}{c}\frac{1}{2}\\\frac{1}{2}\\- \frac{\sqrt{6}}{2}\\1\end{array}\right] t.$$$
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 $$$\left\{\left[\begin{array}{c}\frac{1}{2}\\\frac{1}{2}\\- \frac{\sqrt{6}}{2}\\1\end{array}\right]\right\}\approx \left\{\left[\begin{array}{c}0.5\\0.5\\-1.224744871391589\\1\end{array}\right]\right\}.$$$A
The nullity of the matrix is $$$1$$$A.