# Find the linear transformation associated with the matrix

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?

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 $\mathbb{R}^3$.

• so i don't have to bother row reducing like I was trying to do? Oct 10 '17 at 22:05
• @Aldrec No. Why would you want to do that? Oct 10 '17 at 22:06
• Thats the first way i thought about doing it, can you explain to me why thats wrong? i'm a bit confused Oct 10 '17 at 22:07
• @Aldrec No, I can't, since I don't understand what you were trying to do. Oct 10 '17 at 22:08
• I created an augmented matrix with a vector in $R^{3}$ and then row reduce that matrix Oct 10 '17 at 22:09

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

• how would I find that? Oct 10 '17 at 22:03
• It is just the matrix $M$ times a generic vector in $V$ Oct 10 '17 at 22:04