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Homework Help: Solve for a surjective martix

  1. Sep 10, 2010 #1
    I saw this in a book as a proposition but I think it's an error:

    Assume that the (n-by-k) matrix, [tex]A[/tex], is surjective as a mapping,

    [tex]A:R^{k}\rightarrow R^{n}[/tex].

    For any [tex]y \in R^{n} [/tex], consider the optimization problem

    [tex]min_{x \in R^{k}}\left{||x||^2\right}[/tex]

    such that [tex] Ax = y[/tex].

    Then, the following hold:
    (i) The transpose of [tex]A[/tex], call it [tex]A^{T}[/tex] is injective.
    (ii) The matrix [tex]A^{T}A[/tex] is invertible.
    (iii) etc etc etc....

    I have a problem with point (ii), take as an example the (2-by-3) surjective matrix
    [tex]A = \begin{pmatrix}
    1 & 0 & 0\\
    0 & 1 & 0
    \end{pmatrix}[/tex]

    [tex]A^{T}A[/tex] in this case is not invertible.

    What am I doing wrong ?
     
    Last edited: Sep 10, 2010
  2. jcsd
  3. Sep 10, 2010 #2

    Dick

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    Science Advisor
    Homework Helper

    I don't think you are doing anything wrong. But I think there must be a typo in the theorem statement and they meant to write AA^(T) instead of A^(T)A.
     
  4. Sep 10, 2010 #3
    That would be fine, but the result is used to carry some analysis. The crux of it is:

    [tex]\sigma\lambda = \alpha - \bf{r}[/tex]

    where [tex]\sigma \in R^{n-by-k}[/tex] is surjective, and [tex]\lambda \in R^{k}[/tex], [tex]\alpha , r \in R^{n}[/tex].

    How would you solve for [tex]\lambda[/tex] ? Isn't is critical that the 'typo' has to be correct to be able to solve for this ?

    The author's solution as you might expect is [tex]\lambda = \left(\sigma^{T}\sigma\right)^{-1}\sigma^{T}[\alpha - \bf{r}][/tex].

    BTW, thanks for making the effort to look at the problem. Much appreciated.
     
    Last edited: Sep 10, 2010
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