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Math Help - Finding a basis.

  1. #1
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    Finding a basis.

    I have a question here that asks to find a basis, which states:

    Find a basis for V, where V = {T belongs to L(R,R)| T(1,2) = (0,0)}

    Could anyone offer any ideas on how to tackle this question? I appreciate any help. Thanks.
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  2. #2
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    Quote Originally Posted by GreenDay14 View Post
    I have a question here that asks to find a basis, which states:

    Find a basis for V, where V = {T belongs to L(R,R)| T(1,2) = (0,0)}

    Could anyone offer any ideas on how to tackle this question? I appreciate any help. Thanks.
    Are you sure you got the complete question?
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  3. #3
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    Actually, the problem is not that there is not enough information but that what you give is contradictory. You say T is from R^2 to R^3 but then you say T((1, 2))= (0, 0) which is a transformation from R^2 to R^3.

    Is the subspace the set of all linear transformation from R^2 to R^3 such that T((1,2))= (0,0,0)? Or did you mean that T is from R^2 to R^2?

    Assuming you meant the former, any linear transformation from R^2 to R^3 can be written as a matrix:
    \begin{bmatrix}a & b \\ c & d \\ e & f\end{bmatrix}
    and your condition is
    \begin{bmatrix}a & b \\ c & d \\ e & f\end{bmatrix}\begin{bmatrix}1 \\ 2\end{bmatrix}= \begin{bmatrix}a+ 2b \\ c+ 2d \\ e+ 2f\end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}
    so you must have a+ 2b= 0, c+ 2d= 0, and e+ 2f= 0. Those, of course, give a= -2b, c= -2d, and e= -2f. The matrix can be written
    \begin{bmatrix}-2b & b \\ -2d & d \\ -2f & f\end{bmatrix}= b\begin{bmatrix}-2 & 1 \\ 0 & 0 \\ 0 & 0\end{bmatrix}+ d\begin{bmatrix}0 & 0 \\ -2 & 1 \\ 0 & 0\end{bmatrix}+ f\begin{bmatrix}0 & 0 \\ 0 & 0 \\ -2 & 1\end{bmatrix}.

    If you mean R^2 to R^2 just follow the same ideas.
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  4. #4
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    aghh, thanks both of your for the help and I aplogize. Yes I did make a mistake, i meant for it to be T is from R to R. Sorry again.
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    Actually, the problem is not that there is not enough information but that what you give is contradictory. You say T is from R^2 to R^3 but then you say T((1, 2))= (0, 0) which is a transformation from R^2 to R^3.

    Is the subspace the set of all linear transformation from R^2 to R^3 such that T((1,2))= (0,0,0)? Or did you mean that T is from R^2 to R^2?

    Assuming you meant the former, any linear transformation from R^2 to R^3 can be written as a matrix:
    \begin{bmatrix}a & b \\ c & d \\ e & f\end{bmatrix}
    and your condition is
    \begin{bmatrix}a & b \\ c & d \\ e & f\end{bmatrix}\begin{bmatrix}1 \\ 2\end{bmatrix}= \begin{bmatrix}a+ 2b \\ c+ 2d \\ e+ 2f\end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}
    so you must have a+ 2b= 0, c+ 2d= 0, and e+ 2f= 0. Those, of course, give a= -2b, c= -2d, and e= -2f. The matrix can be written
    \begin{bmatrix}-2b & b \\ -2d & d \\ -2f & f\end{bmatrix}= b\begin{bmatrix}-2 & 1 \\ 0 & 0 \\ 0 & 0\end{bmatrix}+ d\begin{bmatrix}0 & 0 \\ -2 & 1 \\ 0 & 0\end{bmatrix}+ f\begin{bmatrix}0 & 0 \\ 0 & 0 \\ -2 & 1\end{bmatrix}.

    If you mean R^2 to R^2 just follow the same ideas.
    @HallsofIvy - Thanks for the explanation. In the case you have illustrated, dimension of T such that T((1,2))= (0,0,0) is obviously 3. You in fact have shown us a basis. Is there an easy way to compute the dimension without actually solving equations?
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  6. #6
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    I am very lost in this question lol.
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