Proof verification I have a problem and a proposed solution. I want to know if I have done it correctly.
Problem Statement: Let $V=F^n$ be the space of column vectors. Prove that every subspace $W$ of $V$ is the space of solutions of some system of homogeneous linear equations $AX=0$.
My solution: The null space of an $m$ x $n$ matrix $A$, written as $Nul A$, is the set of all solutions to the homogeneous equation $Ax = 0$. The null space of an $m$ x $n$ matrix $A$ is a subspace of $R^n$. Equivalently, the set of all solutions to a system $Ax = 0$ of $m$ homogeneous linear equations in $n$ unknowns is a subspace of $R^n$. 
Then I proceeded to prove that the properties of a subspace hold for the null space, and hence the null space is a subspace of $R^n$. Please tell me if this is right... this problem has been on my head all day!!
Thanks!
 A: You have proved that the solution set of the system $Ax = 0$ is a subspace of $F^n$. What the question asks you to do is show that any subspace is the solution set of some system of equations. That is, you have to show that, given a subspace $W$ of $F^n$, there is a matrix $A$ such that $\{x \in F^n \mid Ax = 0\} = \operatorname{Nul}A = W$.
A: Disclaimer: Not entirely sure about this.
Proof: Let $V = \mathbb{F^{n}}$. Let $W$ be a subspace of $V$. By definition, every subspace of $V = \mathbb{F}^n$ must contain the $0$ vector and must be closed under addition and scalar multiplication. Therefore, all linear combinations of the system $Ax = a_1x_1+\cdots+a_nx_n = 0$ are contained in $W$,  so it is clear that $AX = 0 \in W$, and as a result, $W$ can be represented as the solution space to the homogenous equation. $\Box$
A: Let $W$ be your subspace. If $W$ is all of the space, it is the kernel of the null matrix.  If it is $\{0\}$, it is the kernel of any nonsingular matrix. So assume your subspace is proper, and let $\{w_1,\ldots,w_s\}$ be a basis. Extend this to a basis $\{w_1,\ldots,w_s,w_{s+1},\ldots,w_n\}$ of your space, let $\{w_{s+1},\ldots,w_n\}=W'$. Define the transformation $f:V\to V$ as the projection onto $W'$ through $W$. Then $W=\ker f$, and you can consider the matrix of $f$ in the canonical basis to get an appropriate system of equations. 
Alternatively, use the canonical inner product: if $W$ is your subspace, consider $W^\perp$. Then use that $W=(W^\perp)^{\perp}$ to get a set of equations for $W$, namely of the form $w_i\cdot x=0$ for $i=1,2,\ldots, r$ where $\{w_1,\ldots,w_r\}$ is a base for $W^{\perp}$.
A: Disclaimer: Not entirely sure about this. 
Here is a totally different solution that I have thought up out of desperation, but I'm not sure about it.
Let $W$ be a proper subspace of $V$. Then, as $\dim(W) + \dim(W^\perp) = \dim(V)$, we have $W^\perp$ not equal to $0$. Construct the matrix $A^t$ by taking its columns to be a basis for $W^\perp$. Let $M=\{x \in V : Ax = 0\}$ be the solution set under consideration. Then $W=M$. Next assume that $W=V$. Taking $A$ to equal the $0$ matrix, we are done.
