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Thread: Linear transformation problem

  1. #1
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    Linear transformation problem

    A linear transformation is given: $\displaystyle T:R^3 -> R^3$ such as:
    $\displaystyle T(1,1,1)=(1,1,1)$
    $\displaystyle T(0,1,0)=(0,1,0)$
    $\displaystyle T(1,0,2)=(1,0,1)$
    Find $\displaystyle T(x,y,z)$ and $\displaystyle [T]_E$ $\displaystyle (E={(1,0,0),(0,1,0),(0,0,1)})$
    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 $\displaystyle T(x,y,z)$
    Thanks in advance.
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  2. #2
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    Quote Originally Posted by wicked View Post
    A linear transformation is given: $\displaystyle T:R^3 -> R^3$ such as:
    $\displaystyle T(1,1,1)=(1,1,1)$
    $\displaystyle T(0,1,0)=(0,1,0)$
    $\displaystyle T(1,0,2)=(1,0,1)$
    Find $\displaystyle T(x,y,z)$ and $\displaystyle [T]_E$ $\displaystyle (E={(1,0,0),(0,1,0),(0,0,1)})$
    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 $\displaystyle T(x,y,z)$
    Thanks in advance.
    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 $\displaystyle R^3$. 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).
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    (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
    $\displaystyle a+c=x => a=2x-z$
    $\displaystyle a+b=y => b=y+z-2x$
    $\displaystyle a+2c=z => c=z-x$

    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 $\displaystyle [T]_E$ matrix is :

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

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