# Is a linear map $T: V \to W$ possible given $\{v_1,v_2,v_3\}\in V$ and $\{w_1,w_2,w_3\}\in W$ are both linearly dependent?

Given linear map from $V$ to $W$:

$T:\mathbb{R^4}\to\mathbb{R^4}$ and transformations:

$T(\stackrel{v_1}{-1,5,2,3})=(\stackrel{w_1}{3,7,0,-3})$

$T(\stackrel{v_2}{-1,1,0,1})=(\stackrel{w_2}{1,-1,2,1})$

$T(\stackrel{v_3}{1,1,1,0})=(\stackrel{w_3}{2,3,1,-1})$

give an example of the linear map if it exists.

After reducing the corresponding matrix for $\{v_1,v_2,v_3\}$ we can that see that $v_1$ is a linear combination of $\{v_2,v_3\}$: $$\begin{pmatrix} v_3:&1&1&1&0\\ v_2:&-1&1&0&1\\ v_1:&-1&5&2&3 \end{pmatrix}\\ v_2\to v_2+v_3\\ v_1 \to v_1+v_3\\ \begin{pmatrix} v_3:&1&1&1&0\\ v_2:&0&2&1&1\\ v_1:&0&6&3&3 \end{pmatrix}\\ v_1\to v_1 - 3v_2\\ \begin{pmatrix} v_3:&1&1&1&0\\ v_2:&0&2&1&1 \end{pmatrix}$$

Likewise after reducing the corresponding matrix for $\{w_1,w_2,w_3\}$ we can that see that $w_1$ is a linear combination of $\{w_2,w_3\}$: $$\begin{pmatrix} w_2:&1&-1&2&1\\ w_1:&3&7&0&-3\\ w_3:&2&3&1&-1 \end{pmatrix}\\ w_1\to w_1 -3w_2\\ w_3\to w_3-2w_2\\ \begin{pmatrix} w_2:&1&-1&2&1\\ w_1:&0&10&-6&-6\\ w_3:&0&5&-3&-3 \end{pmatrix}\\ w_1\to w_1-2w_3\\ \begin{pmatrix} w_2:&1&-1&2&1\\ w_3:&0&5&-3&-3 \end{pmatrix}\\$$

At this point (*) we know that $\{v_1,v_3\}$ and $\{w_2,w_3\}$ are linearly independent so we can add 2 more vectors $v_3=(0,0,1,0)$ and $v_4=(0,0,0,1)$ such that $\{v_1,v_2,v_3,v_4\}$ is still independent. At this point I guess I could just make up any image for $\{v_3,v_4\}$ for example: $$T(v_3)=(1,2,3,4)\\ T(v_4)=(4,5,6,7)$$

My question is at the point (*) I could proceed because the $dimSpan\{v_2,v_3\}=dimSpan\{w_2,w_3\}$ or there's another logic working in this case?

• this is hard to read. You have $\vec {v_1}=3\vec {v_2}+2\vec {v_3}$ But $\vec {w_1} \neq 3\vec {w_2}+2\vec {w_3}$, unless I have misread. – lulu Jun 19 '17 at 14:55
• Yes. If there was a linear map $T$ then we'd have $T(\vec {v_1})=T(3\vec {v_2}+2\vec {v_3})=3T(\vec {v_2})+2T(\vec {v_3})\implies \vec {w_1}=3\vec {w_2}+2\vec {w_3}$ – lulu Jun 19 '17 at 15:17
• As I say, what you wrote is very hard to read and I might have misunderstood it. You should not define your vectors inside the arguments of functions and you shouldn't write vectors without parentheses. I read: $\vec {v_1}=(-1,5,2,3),\,\vec {v_2}=(-1,1,0,1),\,\vec {v_3}=(1,1,1,0)$. Assuming I have this right, then $3\vec {v_2}=(-3,3,0,3)$ and $2\vec {v_3}=(2,2,2,0)$ and my claim is easily verified. – lulu Jun 19 '17 at 15:28
• Ok, but it still looks like what I wrote is correct, no? Are you asking how I got it? If so: you said they were dependent, so I looked for a dependence, $\vec {v_1}=a\vec {v_2}+b\vec {v_3}$. As the third coordinate of $\vec {v_2}$ is $0$ we must have $b=2$. Similarly, looking at the fourth coordinate, we must have $a=3$. But I got this only looking at two coordinates so I need to check the first two. If that failed, it would mean that the vectors were not dependent in this way. But, as it happens, it checked out. – lulu Jun 19 '17 at 15:46
• That's one method. You could also use the definition of the map on the $\vec {v_i}$ to try to define it on a basis for the space they generate. Once you have that you can ask whether the map you've produced satisfies all the properties you want, or just some of them. – lulu Jun 19 '17 at 15:52