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Math Help - Linear Transformation, Nullity and rank...

  1. #1
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    Linear Transformation, Nullity and rank...

    Find bases for both N(T) & R(T). Then compute the nullity and rank of T, and verify the dimension theorem. Use the appropriate theorems to determine whether T is 1-1 or onto..

    T: R2 -> R3 defined by T(a1, a2) = (a1+ a2, 0, 2a1 - a2)

    I'm not too sure where to go from finding the N(T)...I think I have a clue about the R(T)...but not totally sure...

    All I got so far is that N(T) : a1 + a2 = 0
    2a1 - a2 = 0 which gives a matrix...

    | 1 1 |
    | 2 -1 | this matrix can be reduced to

    | 1 0 |
    | 0 1 | and we can see that a1 = 0 and a2 = 0? Which means that the Basis of N(T) is (0,0) meaning dim N(t) = 0, and thus nullity of T = 0....this is what I got so far
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by leungsta View Post
    Find bases for both N(T) & R(T). Then compute the nullity and rank of T, and verify the dimension theorem. Use the appropriate theorems to determine whether T is 1-1 or onto..

    T: R2 -> R3 defined by T(a1, a2) = (a1+ a2, 0, 2a1 - a2)

    I'm not too sure where to go from finding the N(T)...I think I have a clue about the R(T)...but not totally sure...

    All I got so far is that N(T) : a1 + a2 = 0
    2a1 - a2 = 0 which gives a matrix...

    | 1 1 |
    | 2 -1 | this matrix can be reduced to

    | 1 0 |
    | 0 1 | and we can see that a1 = 0 and a2 = 0? Which means that the Basis of N(T) is (0,0) meaning dim N(t) = 0, and thus nullity of T = 0....this is what I got so far
    note that this linear transformation is characterized by multiplication by A = \left( \begin{array}{cc} 1 & 1 \\ 0 & 0 \\ 2 & -1 \end{array} \right)

    now can you find N(T) and R(T)?

    note that T is one-to-one iff A is invertible
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    note that this linear transformation is characterized by multiplication by A = \left( \begin{array}{cc} 1 & 1 \\ 0 & 0 \\ 2 & -1 \end{array} \right)

    now can you find N(T) and R(T)?

    note that T is one-to-one iff A is invertible
    If I solve the matrix....I'll get

    | 1 0 | a1 = 0
    | 0 1 | a2 = 0
    | 0 0 |

    Basis of N(T) is (0, 0)
    dim N(T) = 0

    Basis of R(T) = { (1,0,0) (0,1,0) }
    dim R(T) = 2

    T is 1-1 since N(T) = {0v} if and only if dim N(T) = 0


    Am I correct?
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  4. #4
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    bumppp...anyone? I jus wanna know if I got the answer or not


    oh forgot to mention that T would not be onto since it cannot map all of R^3
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