Question

Consider the basis for , where , , and , and let t:R^3 -> R^2 be the linear transformation for which T(v1)=(-1,0), T(v2)=(-1,1), T(v3)=(0,1). Find a formula for T(x1, x2, x3), and use that formula to find T(7,13,7)

Answer 1

$T$ - linear transformation:

$T(u+v)=T(u)+T(v)$

$T(ku)=kT(u)$

1.

$v=(x_1,x_2) \in \mathbb{R} ^2$ - arbitrary vector.

It can be expressed as a linear combination of $v_1$ and $v_2$:

$v=av_1+bv_2$

$(x_1,x_2)=a(-2,1)+b(1,3)=(-2a+b,a+3b)$

$\begin{cases} x_1=-2a+b \\ x_2=a+3b \end{cases} \; \; \begin{cases} x_1+2x_2=7b\\ x_2=a+3b \end{cases} \begin{cases} b=\frac{x_1}{7}+\frac{2x_2}{7} \\a=\frac{-3x_1}{7}+\frac{x_2}{7} \end{cases}$

$T(v)=T(av_1+bv_2)=aT(v_1)+bT(v_2)=a(-1,2,0)+b(0,-3,5)=$

$=(\frac{-3x_1}{7}+\frac{x_2}{7})(-1,2,0)+(\frac{x_1}{7}+\frac{2x_2}{7} )(0,-3,5)=$

$=(\frac{3x_1}{7}+\frac{-x_2}{7} , \frac{-9x_1}{7}+\frac{-4x_2}{7} , \frac{5x_1}{7}+\frac{10x_2}{7})= \frac{1}{7}(3x_1-x_2,-9x_1-4x_2,5x_1+10x_2)$

$T(2,-3)=\frac{1}{7}(9,-6,-20)=(\frac{9}7,-\frac{6}7,-\frac{20}7)$

2.

$v=(x_1,x_2,x_3)\in \mathbb{R}^3$ - arbitrary vector

It can bee expressed as a linear combination of $v_1,v_2,v_3:$

$v=av_1+bv_2+cv_3$

​$(x_1,x_2,x_3)=a(1,1,1)+b(1,1,0)+c(1,0,0)=(a+b+c,a+b,a)$

$\begin{cases} x_1=a+b+c \\ x_2=a+b \\ x_3=a \end{cases} \begin{cases} a=x_3 \\ b=x_2-x_3 \\ c= x_1-x_2 \end{cases}$

$T(v)=T(av_1+bv_2+cv_3)=aT(v_1) +bT(v_2)+cT(v_3)=$

$=a(3,-1,6)+b(4,0,1)+c(-1,7,1)=$

$= x_3(3,-1,6)+(x_2-x_3)(4,0,1)+(x_1-x_2)(-1,7,1)=$

$=(-x_1+5x_2-x_3,7x_1-7x_2-x_3, x_1+5x_3)$

$T(3,6,-1)=(28,-20,-2)$

3.

$T(4v_1-5v_2+6v_3)=4T(v_1)-5T(v_2)+6T(v_3)=$

$=4(1,-1,2)-5(0,3,2)+6(-3,1,2)=(-14,-13,10)$