3
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thanks for your answer. I'm not sure how the system of equations that would need to be solved, would look in this case. Would I e.g. have to solve, for the $A_{1,1}$ element, the equation $a+2b=0$ and then continue in this manner for all elements of the matrix? $\endgroup$ – eager2learn Jan 11 '14 at 22:26
  • $\begingroup$ @eager2learn Yes, by the definition of matrix equality. I added a different view of the solution, but, as you see, it's exactly the same. $\endgroup$ – egreg Jan 11 '14 at 22:34
  • $\begingroup$ I have one more question. If I set up this system with 6 equations. The last one is 0=0, so I would get infinitely many solutions. But I would need only the trivial solution to show that f,g and h are independent, right? $\endgroup$ – eager2learn Jan 12 '14 at 18:55
  • $\begingroup$ @eager2learn All equations must be satisfied, not just one. $\endgroup$ – egreg Jan 12 '14 at 19:04
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thanks for your answer. Why is it that if matrices are linearly independent with respect to one choice of basis they are linearly independent for any choice of basis? $\endgroup$ – eager2learn Jan 11 '14 at 22:28
  • $\begingroup$ That's a good question. There are two steps to the proof. First of all, invertible linear transformations preserve linear independence. That is to say, if $S$ is a linear independent set in $V$, and $L$ is an invertible linear transformation $L:V\rightarrow V$, then $L(S)$ is also linearly independent. $\endgroup$ – Nick Jan 12 '14 at 4:35
  • $\begingroup$ Next, if $W_1, W_2$ are finite-dim'l vector spaces (over $F$) of dim'ns $n_1, n_2$ respectively, and $V =$ the set of $n_1\times n_2$-matrices, then changes of basis on one or both of $W_1, W_2$ induce invertible linear operators on $V$. $\endgroup$ – Nick Jan 12 '14 at 4:39

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.