Show that these linear maps are linearly independent Let $f, g, h \in L$, where L is the vector space of all linear maps that map from $\mathbb{R}^3 \rightarrow \mathbb{R}^2$.
$ f \left(\left( \begin{array}{ccc}
x_1 \\
x_2 \\
x_3 \end{array} \right) \right) = \left( \begin{array}{cc}
x_1 + x_2 + x_3 \\
x_1 + x_2 \\ \end{array} \right)$
$ g \left(\left( \begin{array}{ccc}
x_1 \\
x_2 \\
x_3 \end{array} \right) \right) = \left( \begin{array}{cc}
2x_1 + x_3 \\
x_1 + x_2 \\ \end{array} \right)$
$ h \left(\left( \begin{array}{ccc}
x_1 \\
x_2 \\
x_3 \end{array} \right) \right) = \left( \begin{array}{cc}
2x_2 \\
x_1 \\ \end{array} \right)$
Show that f, g and h are linearly independent.
Could I show this by using the matrices of those linear maps, which are uniquely determined, and then show that there aren't two of those matrices that are equal, proving that there isn't a linear combination of f, g and h, besides the trivial one, that is equal to 0?
 A: You can show that the matrices of $f$, $g$ and $h$ are linearly independent as elements of the vector space $M_{2\times3}(\mathbb{R})$ of $2\times 3$ matrices.
Namely, suppose
$$
a\begin{bmatrix}
1 & 1 & 1\\ 
1 & 1 & 0
\end{bmatrix}
+
b\begin{bmatrix}
2 & 0 & 1\\
1 & 1 & 0
\end{bmatrix}
+
c\begin{bmatrix}
0 & 2 & 0\\
1 & 0 & 0
\end{bmatrix}
=O
$$
where $O$ is the null matrix and prove that $a=b=c=0$. This is equivalent to your claim, because the map from the vector space $\hom(\mathbb{R}^3,\mathbb{R}^2)\to M_{2\times 3}(\mathbb{R})$ that to any linear map associates its matrix relative to the canonical bases is an isomorphism of vector spaces.
Can you go on from here?
Note that being different is not sufficient for the matrices to be linearly independent. For instance, a set containing the zero linear map is never linearly independent.

A slightly different approach is to consider a zero linear combination:
$$
af+bg+ch=0
$$
(the zero map) and apply this to the vectors in a basis of $\mathbb{R}^3$, for instance those of the canonical basis; by definition,
$$
(af+bg+ch)(v)=af(v)+bg(v)+ch(v)
$$
so, computing for $v=e_1$, $v=e_2$ and $v=e_3$, we get
\begin{gather}
a\begin{bmatrix}1\\1\end{bmatrix}+
b\begin{bmatrix}2\\1\end{bmatrix}+
c\begin{bmatrix}0\\1\end{bmatrix}=
\begin{bmatrix}0\\0\end{bmatrix}
\\[2ex]
a\begin{bmatrix}1\\1\end{bmatrix}+
b\begin{bmatrix}0\\1\end{bmatrix}+
c\begin{bmatrix}2\\0\end{bmatrix}=
\begin{bmatrix}0\\0\end{bmatrix}
\\[2ex]
a\begin{bmatrix}1\\0\end{bmatrix}+
b\begin{bmatrix}1\\0\end{bmatrix}+
c\begin{bmatrix}0\\0\end{bmatrix}=
\begin{bmatrix}0\\0\end{bmatrix}
\end{gather}
which is exactly the same as before.
A: It's not enough just to show that there aren't two of the matrices $F, G, H$ (corresponding to maps $f, g, h$) that are equal, you need to show that the matrices $F, G, H$ are linearly independent. The reason linear independence doesn't follow from the fact that no two are equal is because, for example, we may have $H = aF + bG$, so that $aF + bG - H = 0$, for nonzero $a,b$, even if no two of $F, G, H$ are equal.
Note also that the matrix representations of the maps are only uniquely determined up to a choice of bases, but that this doesn't matter so much because if the matrices are linearly independent with respect to one choice of bases it follows that they're linearly independent with respect to any choice of basis.
To show the linear independence of the matrices, set an arbitrary linear combination of the matrices equal to zero and show the resulting system of equations has no solution. It will be a system of six equations in three unknowns, so it won't be particularly tough to show that it's overdetermined, for example by Guassian elimination.
