Question:
L: $\mathbb R^3 \rightarrow \mathbb R^3$ defined by Basis $B=\{u_1=[1,1,1],u_2=[1,1,0],u_3=[1,0,0]\}$ and $L(c_1u_1+c_2u_2+c_3u_3)=(c_1+c_2+c_3)\vec u_1+(2c_1+c_3)\vec u_2-(2c_2+c_3)\vec u_3$
Find the matrix of L with respect to B.
The work I am provided is:
$L(u_1)=L(1*u_1+0*u_2+0*u_3)=1u_1+2u+0u^3 \therefore L(u_1)_B=[1,2,0]$ $L(u_2)=L(0*u_1+1*u_2+0*u_3)=1u_1+0u-2u^3 \therefore L(u_2)_B=[1,0,-2]$ $L(u_3)=L(0*u_1+0*u_2+1*u_3)=1u_1+1u-0u_3 \therefore L(u_3)_B=[1,1,0]$
$\therefore M_B=\begin{bmatrix} 1 & 1 & 1\\ 2 & 0 & 1\\ 0 & -2 & 0 \end{bmatrix}$
My question is how does $L(u_3)$ not equal $1u_1+1u-1u_3 \therefore L(u_3)_B=[1,1,-1]$? I get u1 and u2 but setting $c_3$ to 1 and all the other c variables should be what I just said above. Pleas help. This is how my teacher showed me to do it but it is not making sense.
For clarification for each u you are setting the corresponding c to 1 and all the other c values to 0.