Construct a $4\times4$ matrix $A$ such that $\{(1,2,3,4),(1,1,2,2)\}$ is a basis of $N(A)$.
So I know that $A$ will have two pivot columns and two free columns, but beyond this I'm not sure how to approach/solve.
[Math] How to construct a matrix given the null basis of A
linear algebramatricesvector-spaces
Best Answer
The row space of a matrix is the orthogonal complement of its null space. So, you can construct the required matrix by finding a basis for this orthogonal complement. In this case, this will give you two of the rows, and the other two rows can be any linear combinations of those two rows, including rows of all zeros.
Calling the two given vectors $\mathbf n_1$ and $\mathbf n_2$, the orthogonal complement of their span is the set of all vectors $\mathbf x$ that satisfy $\mathbf n_1\cdot\mathbf x=\mathbf n_2\cdot\mathbf x=0$. This is a pair of homogeneous linear equations in the components of $\mathbf x$, so $\mathscr N(A)^\perp$ is the null space of the matrix $\small{\begin{bmatrix}\mathbf n_1 & \mathbf n_2\end{bmatrix}}^T$. I’m sure you know how to compute that.