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    Consider the subset

    X = \{(x_1, x_2, x_1-x_2, 3x_2) : x_1, x_2 \in \mathbb{R}\}

    of \mathbb{R}^4 and the function f:X\rightarrow X

    given by f(x_1, x_2, x_1-x_2, 3x_2) = (x_1, x_1, 0, 3x_1)

    1/ Find a basis of X
    2/ Find the matrix for f wrt this basis.

    I got a basis of (1,1,0,3),(1,0,1,0)

    But I'm getting really confused about finding the matrix because the vector space only has 2 vectors in the basis but the vectors themselves have 4 entries. Can someone explain how to do this?
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    (Original post by TheJ0ker)
    Consider the subset

    X = \{(x_1, x_2, x_1-x_2, 3x_2) : x_1, x_2 \in \mathbb{R}\}

    of \mathbb{R}^4 and the function f:X\rightarrow X

    given by f(x_1, x_2, x_1-x_2, 3x_2) = (x_1, x_1, 0, 3x_1)

    1/ Find a basis of X
    2/ Find the matrix for f wrt this basis.

    I got a basis of (1,1,0,3),(1,0,-1,0)
    (1,0,-1,0)\not\in X

    But I'm getting really confused about finding the matrix because the vector space only has 2 vectors in the basis but the vectors themselves have 4 entries. Can someone explain how to do this?
    If e1,e2 are your two basis vectors, then any vector can be written in the form ae1+be2

    Then you are looking to find a 2x2 matrix that takes (a,b)^T and transforms it into the appropriate new vector ce1+de2

    So you need to find what f does to e1, e2, and express the result in terms of e1 and e2, in each case.
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    (Original post by ghostwalker)
    (1,0,-1,0)\not\in X



    If e1,e2 are your two basis vectors, then any vector can be written in the form ae1+be2

    Then you are looking to find a 2x2 matrix that takes (a,b)^T and transforms it into the appropriate new vector ce1+de2

    So you need to find what f does to e1, e2, and express the result in terms of e1 and e2, in each case.
    Yeah sorry, I meant (1,0,1,0);

    so let e1 = (1,1,0,3), e2 = (1,0,1,0)

    then f(e1) = (1,1,0,3) = (1)e1 + 0e2
    and f(e2) = (1,1,0,3) = (1)e1 + 0e2

    then how would I find the matrix? Is it
    \begin{pmatrix} 1 & 0 \\1 & 0 \end{pmatrix} ?

    And why is it a 2x2 surely I need a 4x4 if I want to apply the matrix transformation to a vector in X?
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    It's not that matrix actually is it
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    (Original post by TheJ0ker)
    Yeah sorry, I meant (1,0,1,0);

    so let e1 = (1,1,0,3), e2 = (1,0,1,0)

    then f(e1) = (1,1,0,3) = (1)e1 + 0e2
    and f(e2) = (1,1,0,3) = (1)e1 + 0e2

    then how would I find the matrix? Is it
    \begin{pmatrix} 1 & 0 \\1 & 0 \end{pmatrix} ?
    Close. You need to transpose that.

    And why is it a 2x2 surely I need a 4x4 if I want to apply the matrix transformation to a vector in X?
    Your matrix is relative to a basis, in this case your chosen one.
    The domain is two dimensional, any vector is represented in the form ae1+be2, relative to that basis. I.e. there are only two co-ordinates.
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    (Original post by ghostwalker)
    Close. You need to transpose that.



    Your matrix is relative to a basis, in this case your chosen one.
    The domain is two dimensional, any vector is represented in the form ae1+be2, relative to that basis. I.e. there are only two co-ordinates.
    Ah thank you so so much, I've been getting so frustrated over this all day and I understand the whole thing now.
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    (Original post by TheJ0ker)
    Ah thank you so so much, I've been getting so frustrated over this all day and I understand the whole thing now.
    :cool:
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    Isn't (1,0) , (0,1) a basis here given that everything is basically a linear combination of these two?
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    (Original post by Zilch)
    Isn't (1,0) , (0,1) a basis here given that everything is basically a linear combination of these two?
    (1,0),(0,1)\not\in\mathbb{R}^4  \text{, let alone in X.}
 
 
 
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