Question

Use the definition of linear transformation to show that the function T:R^3=>R^2 given by the formula

T(X1,X2,X3)=(2X1-X2+X3,X2-4X3) is a linear transformation

NOTE: the terms X1 and so on 1 is a subscript value

Answer 1

what are the two requirements for a linear transformation?

Answer 2

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

cT(u)=T(cu)

Answer 3

whr c is any constant and u and v are elements of vector space

Answer 4

ok, so try it
let's do the scalar bit first:\[\vec u=\langle x_1,x_2,x_3\rangle\]\[\vec v=\langle y_1,y_2,y_3\rangle\]\[T(\vec u)=\langle2x_1-x_2+x_3,x_2-4x_3,\rangle\]\[c\vec u=\langle cx_1,cx_2,cx_3\rangle\]\[T(c\vec u)=\langle2cx_1-cx_2+cx_3,cx_2-4cx_3,\rangle\]is that equal to\[cT(\vec u)\] ?

Answer 5

that's true

Answer 6

right, since\[T(c\vec u)=\langle2cx_1-cx_2+cx_3,cx_2-4cx_3,\rangle=c\langle2x_1-x_2+x_3,x_2-4x_3\rangle\]jolly, so on to the next one...
why don't you try it yourself for a sec and see how it goes?

Answer 7

the confution was that we are transforming from R^3 to R2 and I was tempted to define U as an element of R3 and V as an elelment of R2 which is wrong but let me try it now

Answer 8

what if I say
To show that T(u+v)=T(u)+T(v)

since we hv u=(x1,x2,x3) v=(y1,y2,y3)
then u+v=(x1+y1,x2+y2,+x3+y3)
T(u+v)=(2(x1+y1)-(x2+y2)+(x3+y3),(x2+y2)-4(x3+y3))
=2x1-3x3,2y1-3y3
=T(u)+T(v)

Answer 9

yeah that's fine, though you may want to show T(u)+T(v) explicitly just to make it more clear

Answer 10

what is the second-to-last line though?

Answer 11

T(u+v)=(2(x1+y1)-(x2+y2)+(x3+y3),(x2+y2)-4(x3+y3))

T(u)=(2x1-x2+x3,x2-4x3)
T(v)=(2y1-y2+y3,y2-4y3)
add the components and you finish the proof

Answer 12

alryt tnx