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?

and answers that

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?

  • $\begingroup$ Solve the linear equation \begin{align}x_1+2x_2&=0\\2x_1+3x_2&=0\end{align}. There is only one solution. $\endgroup$
    – Mittens
    Dec 24, 2022 at 16:33
  • $\begingroup$ 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. $\endgroup$
    – xoux
    Dec 24, 2022 at 17:00
  • $\begingroup$ 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. $\endgroup$
    – Mittens
    Dec 24, 2022 at 17:02

1 Answer 1


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 $(0, 0, 1)$.


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