1
$\begingroup$

I am looking to find vector spaces V & W and bases, ordered bases $\beta$ and $\gamma$ and a transformation $T$, Where M is the matrix that the transformation is describing.

$ M= \left[ {\begin{array}{cc} 1 & -1 & 0 & 2 \\ 0 & 3 & 1 & 1 \\ 1 & 0 & 2 & -1 \\ \end{array} } \right] $

I understand that I have quite a couple options for the my choices for V and W I chose $R^{4}$ & $R^{3}$ as my vectors spaces and let $\beta$ and $\gamma$ bet the standard bases. Now I'm confused on how to find the linear transformation associated with the matrix.

My first thought was to find the representation of a generic vector by the matrix but I'm not sure if that would be write. What I mean by this is to solve the following

$ M= \left[ {\begin{array}{cc} 1 & -1 & 0 & 2 & a_1 \\ 0 & 3 & 1 & 1 & a_2 \\ 1 & 0 & 2 & -1 & a_3\\ \end{array} } \right] $

Would this be the correct way to go about this problem or do I have it all wrong?

$\endgroup$

2 Answers 2

Reset to default
1
$\begingroup$

Since$$\begin{bmatrix}1 & -1 & 0 & 2 \\ 0 & 3 & 1 & 1 \\ 1 & 0 & 2 & -1\end{bmatrix}.\begin{bmatrix}x\\y\\z\\t\end{bmatrix}=\begin{bmatrix}x-y+2t\\3y+z+t\\x+2z-t\end{bmatrix},$$you can define$$\begin{array}{rccc}T\colon&\mathbb{R}^4&\longrightarrow&\mathbb{R}^3\\&(x,y,z,t)&\mapsto&(x-y+2t,3y+z+t,x+2z-t).\end{array}$$The basis will be, of course, the canonical basis of $\mathbb{R}^4$ and $\mathbb{R}^3$.

$\endgroup$
8
  • $\begingroup$ so i don't have to bother row reducing like I was trying to do? $\endgroup$
    – Temirzhan
    Oct 10, 2017 at 22:05
  • $\begingroup$ @Aldrec No. Why would you want to do that? $\endgroup$
    – José Carlos Santos
    Oct 10, 2017 at 22:06
  • $\begingroup$ Thats the first way i thought about doing it, can you explain to me why thats wrong? i'm a bit confused $\endgroup$
    – Temirzhan
    Oct 10, 2017 at 22:07
  • $\begingroup$ @Aldrec No, I can't, since I don't understand what you were trying to do. $\endgroup$
    – José Carlos Santos
    Oct 10, 2017 at 22:08
  • $\begingroup$ I created an augmented matrix with a vector in $R^{3}$ and then row reduce that matrix $\endgroup$
    – Temirzhan
    Oct 10, 2017 at 22:09
0
$\begingroup$

$M\begin {bmatrix} v_1\\v_2\\v_3\\v_4 \end {bmatrix} = \begin {bmatrix} v_1 - v_2 + 2v_4\\v_2 + v_3 + v_4\\v_1 + 2v_3 -v_4\end {bmatrix}$

$\endgroup$
2
  • $\begingroup$ how would I find that? $\endgroup$
    – Temirzhan
    Oct 10, 2017 at 22:03
  • $\begingroup$ It is just the matrix $M$ times a generic vector in $V$ $\endgroup$
    – Doug M
    Oct 10, 2017 at 22:04

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.