# Is there a transformation matrix for given mapping of vectors

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?

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 $$\phi$$.
• 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. – user598858 Jul 19 '20 at 3:12