# Find the Matrix A of the Linear Transformation

Can anyone walk me through the steps to complete this problem? I am unsure of where to start to solve the problem. I get that the resulting matrix $A$ should be a $2 \times2$ matrix, should I be finding another vector in the basis to find $BAB^{-1}$?

Let $V$ be the plane with equation $x_1 - 4 x_2 + 2 x_3 =0$ in ${\mathbb R}^3$. Find the matrix $A$ of the linear transformation $T(x)= \displaystyle\left[\begin{array}{ccc} -6 &-12 &0 \cr -2 &-2 &1 \cr -1 &2 &2 \cr \end{array}\right] x$ with respect to the basis $\displaystyle\left[\begin{array}{c} 4 \cr 1 \cr 0 \cr \end{array}\right]$ , $\displaystyle\left[\begin{array}{c} -2 \cr 0 \cr 1 \cr \end{array}\right]$

Let $u=(4,1,0)^T$ and $v=(-2,0,1)^T$.
1. Before answering the question, do a sanity check first. Do $\mathbf{u}$ and $\mathbf{v}$ really lie on $V$? That is, do they satisfy the equation $x_1-4x_2+2x_3=0$? They seem so. Good. Let's proceed.
2. Now compute $\mathbf{p} = T(\mathbf{u})$.
3. Can you express $\mathbf{p}$ as a linear combination of $\mathbf{u}$ and $\mathbf{v}$? That is, can you find two numbers $a$ and $b$ such that $\mathbf{p}=a\mathbf{u}+b\mathbf{v}$? Note that no difficult calculation is needed; $a$ and $b$ can be found by simple inspection (hint: observe that $\mathbf{u}$ and $\mathbf{v}$ each has a zero entry). Verify that the $a$ and $b$ you found are correct.
4. Similarly, find $c$ and $d$ such that $T(\mathbf{v})=c\mathbf{u}+d\mathbf{v}$.
5. Now you may write out the matrix $A$ in terms of $a,b,c$ and $d$.
• @JaredTerrell I mean, compute $T(u)$. That is, multiply the matrix representation of $T$ by the vector $u$. Call the result $p$. Yes, it is a column vector. No, it is not a column of $T$. (I give this yes and no answer because I'm not sure what does your "column" refer to.) Dec 9 '12 at 19:19