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Ask a dumb question:

We all know the following:

$\sum_{i=1}^m |a_i^Tx-b_i| = ||Ax-b||_1$

How about

$\sum_{i=1}^m \frac{1}{a_i^Tx-b_i}$?

I think it is definitely not $ ||(Ax-b)^{-1}||_1$

sleeve chen
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    You are missing $|\cdot|$ and what do you mean by the inverse of a vector? – copper.hat Feb 14 '15 at 06:19
  • Yes, this is what I am confused. Is there any other way to define the inverse of a vector? I am not good at this part. – sleeve chen Feb 14 '15 at 06:24
  • I have no idea what you are trying to do? – copper.hat Feb 14 '15 at 06:26
  • @copper.hat: I think everything is fine, $b_i$ is the i'th component of $b$, while $a_i$ is meant to be the i'th row of $A$. – Alex R. Feb 14 '15 at 06:29
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    This probably makes some minor sense. For example, there's this, while not the same, of similar spirit: https://terrytao.wordpress.com/2008/10/18/from-the-littlewood-offord-problem-to-the-circular-law-universality-of-the-spectral-distribution-of-random-matrices/ – Alex R. Feb 14 '15 at 06:39
  • @AlexR.Yes, it is what I mean. I think vectors cannot be inverted easily. As in this discussion: http://math.stackexchange.com/questions/208447/can-vectors-be-inverted – sleeve chen Feb 14 '15 at 06:50
  • I guess that $\sum_{i=1}^m (a_i^Tx-b_i) = ||Ax-b||1$ should be $\sum{i=1}^m |a_i^Tx-b_i| = ||Ax-b||1$. If one consider something like $(Ax-b)^{-1}$, then there should be some multiplication with a unit element. But what is it here? Why to complicate with $(Ax-b)^{-1}$, let us ask what is the meaning of $\sum{i=1}^{m}\frac{1}{|x_i|}$ in the case when all components of vector $x$ are non-zero. – Janko Bracic Feb 14 '15 at 07:04
  • It is $||[x_1^{-1},...,x_m^{-1}]^T||_1$. – sleeve chen Feb 14 '15 at 07:36

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