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Math Help - Solution for a particular column vector implies solution for all column vectors.

  1. #1
    Senior Member Pinkk's Avatar
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    Solution for a particular column vector implies solution for all column vectors.

    Let A be a square matrix. Show that if the system AX=B has a unique solution for some particular column vector B, then it has a unique solution for all B.

    So I'm not really sure how to go about this since there's no assumption that A is invertible, and I'm assuming I have to utilize row operations or the like, but I just can't make any connections. Any help would be appreciated, thanks.
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by Pinkk View Post
    Let A be a square matrix. Show that if the system AX=B has a unique solution for some particular column vector B, then it has a unique solution for all B.

    So I'm not really sure how to go about this since there's no assumption that A is invertible, and I'm assuming I have to utilize row operations or the like, but I just can't make any connections. Any help would be appreciated, thanks.
    Here is my thoughts on the question.

    Every solution to a linear system can be decomposed into two separate pieces.

    X=X_c+X_p where X_c is the complimentary solution (the solution to the homogeneous equation) and X_p is a particular solution.

    Since we know the solution is unique for some particular b

    We have
    Ax=A(x_c+x_p)=b Since the solution is unique The complementary solution cannot have any free parameters.

    e.g the kernel of this Matrix has only one vector in it and since \vec{0} \in x_c the only solution to the system

    Av=0 is v=0

    This will give the desired conclusion.
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  3. #3
    Senior Member Pinkk's Avatar
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    I'm not sure I exactly follow (we haven't covered kernel, etc).

    Are you saying that since the solution is unique, it must be the case that x_{c} is the zero column vector?

    And I'm not sure I make the connection to how it follows that there must be a unique solution for any B.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Pinkk View Post
    So I'm not really sure how to go about this since there's no assumption that A is invertible, and I'm assuming I have to utilize row operations or the like, but I just can't make any connections. Any help would be appreciated, thanks.

    If AX=B_0 has unique solution, then

    E_n\cdot\ldots\cdot E_1AX=E_n\cdot\ldots\cdot E_1B_0\Leftrightarrow UX=E_n\cdot\ldots\cdot E_1B_0

    (where U is a row echelon form) has unique solution. This implies that all leading coefficients of U are different from 0. Now, you can conclude.


    Fernando Revilla
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  5. #5
    Senior Member Pinkk's Avatar
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    Whew, I get it now. Thanks so much!
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