|
Null space of $$$\left[\begin{array}{ccc}8 & 8 & 16\\4 & 4 & 8\\-4 & -4 & -8\end{array}\right]$$$
Your Input
Find the null space of $$$\left[\begin{array}{ccc}8 & 8 & 16\\4 & 4 & 8\\-4 & -4 & -8\end{array}\right]$$$.
Solution
The reduced row echelon form of the matrix is $$$\left[\begin{array}{ccc}1 & 1 & 2\\0 & 0 & 0\\0 & 0 & 0\end{array}\right]$$$ (for steps, see rref calculator).
To find the null space, solve the matrix equation $$$\left[\begin{array}{ccc}1 & 1 & 2\\0 & 0 & 0\\0 & 0 & 0\end{array}\right]\left[\begin{array}{c}x_{1}\\x_{2}\\x_{3}\end{array}\right] = \left[\begin{array}{c}0\\0\\0\end{array}\right].$$$
If we take $$$x_{2} = t$$$, $$$x_{3} = s$$$, then $$$x_{1} = - 2 s - t$$$.
Thus, $$$\mathbf{\vec{x}} = \left[\begin{array}{c}- 2 s - t\\t\\s\end{array}\right] = \left[\begin{array}{c}-1\\1\\0\end{array}\right] t + \left[\begin{array}{c}-2\\0\\1\end{array}\right] s.$$$
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 $$$2$$$.
Answer
The basis for the null space is $$$\left\{\left[\begin{array}{c}-1\\1\\0\end{array}\right], \left[\begin{array}{c}-2\\0\\1\end{array}\right]\right\}$$$A.
The nullity of the matrix is $$$2$$$A.