2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Show us obtained matrices, please. $\endgroup$ – echzhen Feb 26 '16 at 0:15
  • $\begingroup$ Matrices will do it. Do you know how to decide whether three $2\times2$ matrices are independent? $\endgroup$ – David Feb 26 '16 at 0:17
1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ The row reduced matrix ends up having a row of zeroes; with three unknowns that implies a free variable. That means infinitely many solutions and therefore the set is linearly dependent? $\endgroup$ – AlgebraQuestions Feb 26 '16 at 1:20
  • $\begingroup$ @AlgebraQuestions You nailed it $\endgroup$ – egreg Feb 26 '16 at 9:12
  • $\begingroup$ Thank you so much for the help. :) $\endgroup$ – AlgebraQuestions Feb 27 '16 at 15:14
1
$\begingroup$

Hint:

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.