Strang, Introd. to Linear Algebra: find all matrices that have a specific vector space as their nullspace.

Consider the vectors $$v_1=(1,2,0)$$ and $$v_2=(2,3,0)$$. These vectors are linearly independent and span a vector subspace we can call $$V$$.

In the worked examples of Section 3.5 of Strang's Introduction to Linear Algebra he asks us

Which matrices have $$V$$ as their nullspace?

This $$V$$ is the nullspace of any $$m$$ by $$3$$ matrix $$B$$ of rank $$1$$, if every row is a multiple of $$(0,0,1)$$. In particular, take $$B=[0,0,1]$$. Then $$Bv_1=0$$ and $$Bv_2=0$$.

Such matrices $$B$$ work because the first two column vectors are the zero vector. Vectors in $$V$$ have third coordinate equal to zero, so they effectively take a linear combination of two zero vectors plus zero times a non-zero vector, giving the zero vector.

Furthermore, $$B$$ has rank 1 and so the nullspace has dimension $$2$$. Hence $$V$$ is the entire nullspace of $$B$$.

My question is: how do we know such matrices are the only ones with $$V$$ as their nullspace?

• Solve the linear equation \begin{align}x_1+2x_2&=0\\2x_1+3x_2&=0\end{align}. There is only one solution. Dec 24, 2022 at 16:33
• Doesn't that only show that the vectors $v_1$ and $v_2$ are independent, ie they are a basis for $V$? We want $V$ itself to be the nullspace of a matrix.
– xoux
Dec 24, 2022 at 17:00
• Not onlybthat, it also shows that if $(x_1,x_2,x_3)\cdot(1,2,0)=0=(x_1,x_2,x_3)\cdot(2,3,0)$, then $x_1=x_2=0$ and $x_3$ is any number. Dec 24, 2022 at 17:02

Recall that, given a matrix $$A$$ and its adjoint/Hermitian transpose $$A^*$$, we have $$\operatorname{colspace} A^* = (\operatorname{null} A)^\perp$$. In this case, the perpendicular complement of $$\operatorname{span}((1, 2, 0), (2, 3, 0))$$ is just $$\operatorname{span}((0, 0, 1))$$; we know it's one-dimensional, and $$(0, 0, 1)$$ belongs to the perpendicular complement, so it must be the whole thing.
Thus, every column in $$A^*$$ must be a scalar multiple of $$(0, 0, 1)$$. In order to reach the full span, at least one column must be non-zero (a point that the solution skipped over). So, taking the Hermitian transpose again, we see that all rows must be scalar multiples, not all $$0$$, of $$(0, 0, 1)$$.