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I have vectors $v_1, v_2, v_3$ in $\Bbb{R}^3$ and vectors $w_1, w_2, w_3$ in $\Bbb{R}^4$. There are three mappings from $v_1$ to $w_1$, from $v_2$ to $w_2$, from $v_3$ to $w_3$.

The question is, is there a linear map $\phi$ that maps those vectors in that way?

My first thought is, if $v_1, v_2, v_3$ are linearly independent and $w_1, w_2, w_3$ are linearly independent, then we can have just the transformation matrix $4 \times 3$ with the vectors w as columns.

But what if the $v$ - vectors are not linearly independent? What if the $w$ - vectors are not linearly independent?

How do I find a transformation matrix?

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If $v_1,v_2,v_3$ are linearly independent then irrespective of choice of $w_1,w_2,w_3$, there is such a linear map $\phi$. But if $v_1,v_2,v_3$ are not linearly independent, then it depends on $w_1,w_2,w_3$.
Edit Suppose $v_1=(1,0,0)$ and $v_2=(2,0,0)$ then $v_1$ and $v_2$ are linearly dependent because $v_2=2v_1$. So now if there exists a such linear map $\phi$ then $\phi(v_2)=\phi(2v_1)=2\phi(v_1)$. This gives $w_2$ has to be equal with $2w_1$. So in this example if your chosen $w_2\neq 2w_1$ then there will be no such linear map $\phi$.

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  • $\begingroup$ Thank You for Your answer. But can You elaborate on "it depends on W"? $\endgroup$ – Georgia Jul 18 '20 at 19:32
  • $\begingroup$ Great, I understood that, Thank You. What is if the w - vectors are not linearly independent. Does that change something? $\endgroup$ – Georgia Jul 18 '20 at 20:53
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    $\begingroup$ No, it only depends of dependency of $v$ vectors. If they are linearly independent then no problem. If they are dependent then investigation (as I did in the answer for an example) is needed. So focus of concern is on $v$ vectors. $\endgroup$ – user598858 Jul 19 '20 at 3:12

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