1. ## Linear transformation problem

A linear transformation is given: $T:R^3 -> R^3$ such as:
$T(1,1,1)=(1,1,1)$
$T(0,1,0)=(0,1,0)$
$T(1,0,2)=(1,0,1)$
Find $T(x,y,z)$ and $[T]_E$ $(E={(1,0,0),(0,1,0),(0,0,1)})$
I know that first you must prove that (1,1,1),(0,1,0) and(1,0,2) are linearly independent which is not a problem my problem is finding $T(x,y,z)$

2. Originally Posted by wicked
A linear transformation is given: $T:R^3 -> R^3$ such as:
$T(1,1,1)=(1,1,1)$
$T(0,1,0)=(0,1,0)$
$T(1,0,2)=(1,0,1)$
Find $T(x,y,z)$ and $[T]_E$ $(E={(1,0,0),(0,1,0),(0,0,1)})$
I know that first you must prove that (1,1,1),(0,1,0) and(1,0,2) are linearly independent which is not a problem my problem is finding $T(x,y,z)$
Since (1,1,1), (0,1,0), and (1,0,2) are linearly independent (since you say "not a problem" I presume you have already proved that) and there are three vectors, they form a basis for $R^3$. In particular, that means that any vector (x,y,z)= a(1,1,1)+ b(0,1,0)+ c(1,0,2) where a, b, and c are numerical functions of x, y, and z. After you have found those numbers, T(x,y,z)= aT(1,1,1)+ bT(0,1,0)+ cT(1,0,2)= a(1,1,1)+ b(0,1,0)+ c(1,0,1).

3. Originally Posted by HallsofIvy
(x,y,z)= a(1,1,1)+ b(0,1,0)+ c(1,0,2) where a, b, and c are numerical functions of x, y, and z. After you have found those numbers, T(x,y,z)= aT(1,1,1)+ bT(0,1,0)+ cT(1,0,2)= a(1,1,1)+ b(0,1,0)+ c(1,0,1).
Okay,so this is what I get
$a+c=x => a=2x-z$
$a+b=y => b=y+z-2x$
$a+2c=z => c=z-x$

aT(1,1,1)+ bT(0,1,0)+ cT(1,0,2)= a(1,1,1)+ b(0,1,0)+ c(1,0,1) =
(2x-z)(1,1,1)+(y+z-2x)(0,1,0)+(z-x)(1,0,1)=(2x-z,2x-z,2x-z)+ (0,y+z-2x,0)+ (z-x,0,z-x)=(x,y,x) right?

So the $[T]_E$ matrix is :

( $[T(1,0,0)]_E$ $[T(0,1,0)]_E$ $[T(0,0,1)]_E$) =

1 0 0
0 1 0 (this is the $[T]^E_E matrix(=[T]_E)$)
1 0 0
Is it correct?