# How to determine if set of linear mappings are linearly independent or dependent?

I missed a couple of my Linear algebra classes, so I'm a little lost on this question...

Given $S_1$, $S_2$, $S_3 : \mathbf{R}^2\to \mathbf{R}^2$ are linear mappings defined by:

$S_1(x_1, x_2) = (x_1-x_2, -2x_1+x_2)$

$S_2(x_1,x_2) = (2x_1- x_2, -4x_1+ x_2)$

$S_3(x_1,x_2) = (-x_1+ x_2 , 2x_1+x_2)$

And I need to figure out whether the set $\{S_1, S_2, S_3\}$ is linearly independent or dependent.

I began to try to use constants $c_1$, $c_2$,$c_3$ and made each mapping a matrix, but that did not seem to work. Any help is appreciated.

• Show us obtained matrices, please. – echzhen Feb 26 '16 at 0:15
• Matrices will do it. Do you know how to decide whether three $2\times2$ matrices are independent? – David Feb 26 '16 at 0:17

Saying $c_1S_1+c_2S_2+c_3S_3=0$ means that, for all $(x_1,x_2)\in\mathbf{R}^2$, $$c_1S_1(x_1,x_2)+c_2S_2(x_1,x_2)+c_3S_3(x_1,x_2)=0$$ that is, $$c_1(x_1-x_2, -2x_1+x_2)+c_2(2x_1−x_2,−4x_1+x_2)+c_3(−x_1+x_2,2x_1+x_2)=0$$ In particular, if $x_1=1$ and $x_2=0$, you get $$c_1(1,-2)+c_2(2,-4)+c_3(-1,2)=0$$ You also get, for $x_1=0$ and $x_2=1$, $$c_2(-1,1)+c_2(-1,1)+c_3(1,1)=0$$ and similarly for $x_1=2$, $x_2=1$.
However, there is a simpler method. The matrices associated to $S_1$, $S_2$ and $S_3$ are $$\begin{bmatrix} 1 & -1\\ -2 & 1 \end{bmatrix}, \qquad \begin{bmatrix} 2 & -1\\ -4 & 1 \end{bmatrix}, \qquad \begin{bmatrix} 1 & 1\\ -2 & 1 \end{bmatrix}.$$ Take a linear combination which is $0$: $$a\begin{bmatrix} 1 & -1\\ -2 & 1 \end{bmatrix}+ b\begin{bmatrix} 2 & -1\\ -4 & 1 \end{bmatrix}+ c\begin{bmatrix} 1 & 1\\ -2 & 1 \end{bmatrix}= \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix}$$ which gives $$\begin{cases} a+2b+c=0\\ -a-b+c=0\\ -4a-2b-2c=0\\ a+b+c=0 \end{cases}$$ Does this system have non trivial solutions?
Consider the isomorphism \;\begin{aligned}\mathcal M_{2\times 2}(\mathbf R)&\longrightarrow \mathbf R^4\\\begin{pmatrix}a&b\\c&d\end{pmatrix} &\longmapsto (a,b,c,d)\end{aligned},\enspace and use row reduction.