Injectivity of $g_{1}:M_{2 \times 5}(\Bbb{R}) \rightarrow M_{5 \times 5}(\Bbb{R})$? This is a follow up question to - Dimension of $W_{2}$?
Let us define $B = \begin{bmatrix} 2 & -1\\ -3 & 1\\ 1 & 0\\ 0 & -2\\ 0 & 1\\ \end{bmatrix}$
$C = \begin{bmatrix} -1 & -2 & -1 & 1 & 0\\ -2 & 1&1&0&1\\ \end{bmatrix}$.

Let us define $g_{1}: M_{2 \times 5}(\Bbb{R}) \rightarrow M_{5 \times 5}(\Bbb{R})$
  defined by $g_{1}(Z) = BZ$ defined by $2\times5$ matrix $Z$.

I was thinking -

1) Injectivity of $g_{1}$ - 

Let $g_{1}(P) = g_{1}(Q)$ then if we show $P =Q$ we are done, for $P,Q \in M_{2 \times 5}(\Bbb{R})$ 
$BP =BQ$ but I cannot apply inverse of $B$ as it is not a square matrix? 

2) I have to prove that image of $g_{1}$ is a subspace of $W_{1}$? 
  I thought of this as - 

if that happens then I have to prove that $g_{1}(Z) = BZ \in W_{1}$ that is $BZ(X) = 0$, but how to show this?
 A: $g_1$ is injective.
Let us try to find a $5\times 2$ matrix $D$ such that $DB=I_{2\times2}$. For the given $B$ there are several possible choices of $D$, for example this one is quite easy to see: $$D=\begin{pmatrix}0&0&1&0&0\\0&0&0&0&1\end{pmatrix}.$$
If we define $h_1(X)=DX$, then we get that $h_1\circ g_1=id$ since
$$h_1(g_1(Z))=DBZ=Z.$$
Since $h_1\circ g_1$ is injective, $g_1$ is also injective. (See, for example, Composite functions and one to one or Show that if $g \circ f$ is injective, then so is $f$..)
Is $\operatorname{Im} g_1$ a subspace of $W_1$?
I assume that as in the linked question you mean $W_1=\{X\in M_{5\times5}; AX=0\}$.
So this is basically the question whether $ABZ=0$ for every $Z\in M_{2\times5}$.
Here $AB$ is a $5\times 2$ matrix and $Z$ is a $2\times 5$ matrix.  Notice that by making choosing $Z$ appropriately you can achieve that $(AB)Z$ contains as one of the columns the first column of $AB$. For example, you can use $Z=\begin{pmatrix}1&0&0&0&0\\0&0&0&0&0\end{pmatrix}$. And since this product is zero, you get that the first column of $AB$ has to be zero. A similar argument works for other columns.
So this condition will be true if and only if $AB=0$.
A: To prove the first part, observe that $BX=O$, whenever $[Bx_1\ \ Bx_2\ \ Bx_3\ \ Bx_4\ \ Bx_5]$ is zero, where $x_1,\ldots x_5$ are $2\times 1 $ vectors. Now, observe that each of $x_1\ldots,x_5$ should be the zero vectors, as rank of $B$ is $2$ and we could use the rank nullity theorem, and each such vector should belong to the null space. Hence, the kernel of $g_1$ being trivial , we get $g_1$ to be injective.
To the second part, you have misunderstood it. We should verify whether $ABZ=0$. However, I got all entries zero in $ABZ$ except one. May be some print mistake in the entries of $B$ or $A$. That is, I got $AB$ close to zero, which would give the desired conclusion. 
