Question-72
Compute the dimension of the hyperplane formed by the intersection of the following hyperplanes of \(\displaystyle \mathbb{R}^{4}\):
\[ \begin{equation*} \begin{aligned} x_{1} -2x_{2} +x_{3} -x_{4} & =0\\ x_{1} +x_{2} -x_{3} +x_{4} & =0 \end{aligned} \end{equation*} \]
Solve \(Ax = 0\)
\(2\)
We can treat this as a system of homogeneous equations whose coefficient matrix is given by:
\[ \begin{equation*} A=\begin{bmatrix} 1 & -2 & 1 & -1\\ 1 & 1 & -1 & 1 \end{bmatrix} \end{equation*} \]
The dimension of the intersection of the two hyperplanes is simply the dimension of the solution space of \(\displaystyle Ax=0\). This is just the nullity of \(\displaystyle A\). To get there, we can simply find the rank and subtract it from the number of columns. Upon inspection, it is clear that the rank of \(\displaystyle A\) is \(\displaystyle 2\). Therefore, the nullity is \(\displaystyle 2\).