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Show or give counterexample: Every subspace of $\mathbb{R}^4$ is the nullspace of some matrix.

Would it be valid to just state that every subspace of $\mathbb{R}^n$ can be described as the null space of some matrix. Is there a counterexample?

I attempted showing the statement true in a concise manner. Does this work?

Let the basis $\{a_1$,$a_2$,$a_3$,$a_4\}$ be subspace S of $\mathbb{R}^4$.

Then let some matrix $A= \begin{bmatrix} a_1 \\ a_2 \\ a_3 \\ a_4 \end{bmatrix}$

The row vectors of A are a basis for the null space of subspace S.

null(S)= A

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    If $W\subset \mathbb{R}^4$ is a subspace, take the $\mathbb{R}$-vector space $\mathbb{R}^4/W$ and canonical map between $\mathbb{R}^4\to \mathbb{R}^4/W$. The nullspace of the linear transformation (and the associated matrix once you choose a basis) is $W$. – 2017-02-04
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    your last sentence makes no sense. i think you meant to say $Null(S)=R(A^T)$. – 2018-04-08

2 Answers 2

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You need a theorem:

Every Linear Transformation $T: \mathbb R^n \to \mathbb R^m$ has matrix representation. This can be proven just by specifying a basis. A proof can be found here.

From this, just let $W \subset \mathbb R^n$ be a subspace. From this, consider the natural surjection $\phi: \mathbb R^n \to \mathbb R^n/W$, specified by $x \mapsto x+W$.

The kernel of $\phi$ is exactly $W$, so you take its matrix representation (whatever it may be.)

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Let $V$ be a vector space and $U$ a subspace of $V$. Then there is a subsace $W$ such that

$V= U \oplus W$. Define the linear mapping $P:V \to V$ as follows:

for $v \in V$ there are unique $u \in U$ and $w \in W$ such that $v=u+w$; put

$$Pv=w.$$

Then: $ker(P)=U$.

Can you get it from here ?