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Math Help - Linear transformations, vectors

  1. #1
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    Linear transformations, vectors

    Hi,

    can somebody check if this is correct:

    Problem:

    Given two bases Xi and Yi such that Xi=aji*Yi. (1)Prove that the matrix with element aij is non singular. (2)Hence find the inverse transformation.

    Solution (1)

    X=A*Y

    If X and Y are bases, then Vectors (x1, x2,...,xn) and (y1,y2,...yn) are linear independent. So det(X) and det(Y)≠0.

    det(X)=det(AY)=det(A)*det(Y)≠0 . If det(X), det(Y)≠0 then det(A)≠0.

    Conclusion: Matrix A is non singular.



    Can somebody give me a hint how to find inverse transormation?

    Thanks
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  2. #2
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    Re: Linear transformations, vectors

    Quote Originally Posted by FilipVz View Post
    Given two bases Xi and Yi such that Xi=aji*Yi.
    Do you mean Xi=aji*Yj? And does this mean X_i=\sum_{j=1}^n a_{ji}Y_j?

    Quote Originally Posted by FilipVz View Post
    X=A*Y
    What do you mean by this? What exactly are X and Y here?
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  3. #3
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    Re: Linear transformations, vectors

    X and y are nxn matrices.
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  4. #4
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    Re: Linear transformations, vectors

    Quote Originally Posted by FilipVz View Post
    X and y are nxn matrices.
    Matrices containing what? You start with some vectors X_i and Y_j; there is no talk about numbers! And please don't say "Coordinates of X_i" and stop there!
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  5. #5
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    Re: Linear transformations, vectors

    Hi emakarov,

    Yes, Xi=aji*Yj does mean X_i=\sum_{j=1}^n a_{ji}Y_j?



    Xi and Yi are two sets of linear independent vectors. Elements of Xi are vectors (x1, x2, ..., xn) and element of Yi are (y1, y2, ... , yn).

    Matrices X and Y are defined as nxn matrices. Matrix X contains components of x1, x2, ..., xn vectors, and matrix Y contains components of y1, y2, ..., yn vectors.

    Are these assumptions correct?

    Thanks
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  6. #6
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    Re: Linear transformations, vectors

    No, this is not correct.

    Quote Originally Posted by FilipVz View Post
    Yes, Xi=aji*Yj does mean X_i=\sum_{j=1}^n a_{ji}Y_j?
    Why are you ending your sentence with a question mark. Also, you can wrap this formula in the [tex]...[/tex] tags so that it looks better.

    Quote Originally Posted by FilipVz View Post
    Xi and Yi are two sets of linear independent vectors. Elements of Xi are vectors (x1, x2, ..., xn) and element of Yi are (y1, y2, ... , yn).
    No. The index i of Xi is not a fixed decoration like prime '. It's a number that ranges from 1 to some n. That is, there is not one Xi, but n X's: X1, X2, ..., Xn, and similarly for Y. Each Xi is a vector, an element of a basis. At least this is my understanding of the problem statement.

    Quote Originally Posted by FilipVz View Post
    Matrices X and Y are defined as nxn matrices. Matrix X contains components of x1, x2, ..., xn vectors, and matrix Y contains components of y1, y2, ..., yn vectors.
    I am not sure what a component of a vector is. As for coordinates, one can't refer to coordinates of a vector until a basis is specified. There are two bases here, but it is not clear with respect to which to consider coordinates. You can call the first basis (a sequence of vectors) X and the second one Y, but then X and Y are not matrices containing numbers; they are just sequences of vectors.
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  7. #7
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    Re: Linear transformations, vectors

    Hi emakarov,

    thank you for your considerations. Sorry about question mark.



    If I have understood correctly, i can write:

    Vn is a vector space over a field F.

    Two bases of Vn are sets X and Y.

    X={(X_1,X_2,,X_n)}

    Y={(Y_1,Y,,Y_n)}

    Any vector in set X can be represented by linear combination of vectors in set Y.

    X_1=a_11 Y_1+ +a_1n Y_n
    .
    .
    .
    X_n=a_n1 Y_1+ +a_nn Y_n

    where aij is element of F


    therefore a change basis matrix "A", from X to Y is nxn matrix:

    a11 ... a1n
    . .
    . .
    . .
    an1 ... ann

    Similar any vector in set Y can be represented as linear combination of vectors in X.
    So we can write: Yi=bji*Xj

    where B is a change matrix from Y to X.

    let vector v∈Vn

    v in respect to the basis X is equal to:

    [v]_X=A[v]_Y

    v in respect to the basis Y is equal to:

    [v]_Y=B[v]_X

    hence,

    [v]_X=AB[v]_X

    AB=I

    which implies that A is a nonsingular matrix, and matrix B is the inverse of A
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