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Math Help - [SOLVED] Onto linear transfomation

  1. #1
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    [SOLVED] Onto linear transfomation

    Let V be the real vector space of all real 2x3 matrices, and let W be the real vector space of all real 4x1 column vectors. If T is a linear transformation from V onto W, what is the dimension of the subspace {v \in V: T(v) = 0}?

    ker(v)=nullity; therefore, the range of v=6.

    \begin{bmatrix}<br />
a & b & c\\ <br />
d & e & f<br />
\end{bmatrix}

    I know the answer is 2 but how do I show it?
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    6 = \mbox{dim }V = \mbox{dim } T(V) + \mbox{dim ker }V = 4 + \mbox{dim ker }V
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    Quote Originally Posted by Bruno J. View Post
    6 = \mbox{dim }V = \mbox{dim } T(V) + \mbox{dim ker }V = 4 + \mbox{dim ker }V
    Then that leaves us with:

    6 = dim V = dim T(v) + 0 = 4 + 0

    6 = dim V = dim T(v) = 4

    Will this equality always work for any transformation?
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  4. #4
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    Quote Originally Posted by dwsmith View Post
    Then that leaves us with:

    6 = dim V = dim T(v) + 0 = 4 + 0

    6 = dim V = dim T(v) = 4

    Will this equality always work for any transformation?
    6=4?


    Ker(T) = \{v\in V : T(v) = 0\}. You want to find dim ~ Ker(T). By a theorem (it is rather easy to prove), as Bruno had mentioned, we have that dim ~ V = dim ~ Im(T) + dim ~ Ker(T). Since T is onto a space of dimension 4, we get that dim ~ Im(T) = 4. However, dim ~ V = 6 and so: dim ~ V = dim ~ Im(T) + dim ~ Ker(T) \Rightarrow 6 = 4 + dim ~ Ker(T)  \Rightarrow dim ~ Ker(T) = 6-4 = 2
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  5. #5
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    Quote Originally Posted by Defunkt View Post
    6=4?


    Ker(T) = \{v\in V : T(v) = 0\}. You want to find dim ~ Ker(T). By a theorem (it is rather easy to prove), as Bruno had mentioned, we have that dim ~ V = dim ~ Im(T) + dim ~ Ker(T). Since T is onto a space of dimension 4, we get that dim ~ Im(T) = 4. However, dim ~ V = 6 and so: dim ~ V = dim ~ Im(T) + dim ~ Ker(T) \Rightarrow 6 = 4 + dim ~ Ker(T)  \Rightarrow dim ~ Ker(T) = 6-4 = 2

    6= dim V

    6 = dim T(v) = 4.... I don't know how you came up with 6=4 from that because 6-4=2 = dim T(v) = 4-4=0
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  6. #6
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    6 = dim T(v) = 4
    :P
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