Subspace of $\Bbb R^m$ and $Col(A)$ I have the following question:
Let $W$ be a subspace of $\Bbb R^m$. Prove that there exists $B \in M_{m\times n}$ such that
$W = \{Bx | x \in \Bbb R^n\}.$
I thought of a solution but it seems too obvious to me.
What I wanted to say is that there is a matrix $B$ such that $Col(B) = W$, Because $Bx$ is exactly the definition of $Col(B)$.
What am I doing wrong?
 A: Here's an example of such a computation. Let's say I'm given
$$ W = \{(x, y, z) \in \mathbf{R}^3 : x = 2y - z \} = \operatorname{Null}\begin{pmatrix} 1 & -2 & 1 \end{pmatrix}.$$
I want to write this $W$ as $\operatorname{Col}(B)$ for an appropriate matrix $B$.
What I need is a matrix:


*

*whose columns belong to $W$

*and whose columns span $W$.


The first condition says that says that $\{ Bx : x \in \mathbf{R}^n \} \subseteq W$ because $Bx$ is a linear combination of columns of $B$ and if each column is in $W$, then the whole linear combination is in $W$ (since $W$ is a subspace).
The second condition says that $W \subseteq \{ Bx : x \in \mathbf{R}^n \}$. To say that a set of vectors span some subspace, means that the vectors in that subspace can be written as a linear combination of that set of vectors. So if I have a vector in $W$, I can write it as a linear combination of the columns of $B$. But a linear combination of the columns of $B$ is nothing but $Bx$ for some $x \in \mathbf{R}^n$. So, if the columns span $W$ then every vector in $W$ is of the form $Bx$ for some $x$.
Putting these two together, we get $W = \operatorname{Col}(B)$.
So we simply look for a set of vectors in $W$ that span $W$. I claim, and you can check, that
$$ \left\{\begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix},\; \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} \right\} \tag{$*$}$$
is such a set. Therefore, the columns of
$$ B \overset{\rm def}{=} \begin{pmatrix} 1 & 1 \\ 2 & 0 \\ 0 & -1 \end{pmatrix}$$
belong to $W$ and span $W$. The columns of $B$ is nothing but the set $(*)$. So because the set $(*)$ has the properties 1. and 2. that I need, we have $W = \operatorname{Col}(B)$.
I can also replace $(*)$ with other sets that span to get different matrices:
$$ B' = \begin{pmatrix} 1 & 1 & 1 \\ 2 & 0 & 2 \\ 0 & -1 & -1 \end{pmatrix},\quad W = \operatorname{Col}(B').$$
A: Since $ W $ is given. Let the basis of $W $ be $w_1,w_2,....w_r$
Define $B=[w_1 \space w_2... w_r ....w_n]$, where $w_{r+1},...w_n$ are vectors in W.
Now, if $y \in Col(B)$
$ y $ is a linear comb. of $w_1, w_2,...,w_r$, which are basis vectors of $W$.
Hence $y\in Col(B) \implies$$y \in W $ .....(1)
Similarly, $y \in W$ $\implies y \in Col(B)$.....(2)
From (1) and (2):
$Col(B)=W $
