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anonymous

  • one year ago

if I have an mxn matrix A that does NOT have to be linearly independent and an affine subspace that's defined by (A^T)x = b, can it be proven that there always exists some combination of the columns of A which is part of that subspace?

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  1. anonymous
    • one year ago
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    can you guys help me with this?

  2. anonymous
    • one year ago
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    What do you have so far?

  3. anonymous
    • one year ago
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    nothing

  4. anonymous
    • one year ago
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    i assume it is true, but i want to prove it

  5. Michele_Laino
    • one year ago
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    what is A^T ?

  6. anonymous
    • one year ago
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    the transpose of A

  7. Michele_Laino
    • one year ago
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    ok!

  8. anonymous
    • one year ago
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    I can think of a few examples where this statement is true, thus i assume it must be true for all examples, but this wouldn't be the mathematics category if I would leave it with this assumption.

  9. Michele_Laino
    • one year ago
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    from your equation: \[\Large {A^T}x = b\] we can say that the column b is a linear combination of the column of A^T

  10. Michele_Laino
    • one year ago
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    columns*

  11. Michele_Laino
    • one year ago
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    namely, the column b is a linear combination of the rows of A

  12. anonymous
    • one year ago
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    yes

  13. Michele_Laino
    • one year ago
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    now we can say that the rank of A is equal to: \[\Large rank\left( A \right) \leqslant \min \left\{ {m,n} \right\}\]

  14. anonymous
    • one year ago
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    is there something wrong with my browser or do you intend to use all these symbols? It is a pretty hard to read.

  15. Michele_Laino
    • one year ago
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    for example if I have a 3 x 2 matrix, then all what I can say is that the rank of my matrix is less or equal 2

  16. anonymous
    • one year ago
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    yes

  17. Michele_Laino
    • one year ago
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    in other words, exists some linearly independent rows of A, which correspond to the same number of ilinearly independent columns of A^T, such that they generate the I mage of A as linear map.

  18. Michele_Laino
    • one year ago
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    linearly*

  19. Michele_Laino
    • one year ago
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    and those rows of A are a basis of a linear subspace

  20. anonymous
    • one year ago
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    \[\left[\begin{matrix}a11 & a12 & ... & a1n \\ a21 & ... & ... & ... \\ ... & ... & ... & ... \\ an1 & ... & ... & ann\end{matrix}\right] *\left(\begin{matrix}x1 \\ x2 \\ ... \\ xn\end{matrix}\right)\]

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