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 $\phi$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.