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?
Best Answer
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.