Seemingly simple linear algebra

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Why is it that for arbitrary z, A[itex]^{T}[/itex]z = 0 and b[itex]^{T}[/itex]z ≠ 0 when there does not exist an x such that Ax = b, i.e. that b is not in the range space of A, where A is an n x m matrix?
 
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  • #2
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"Range space" ?? Do you mean "row space" ??
 
  • #3
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Actually, it means column space. This means that there is no linear combination of A's columns that gives b. This must mean that there is no x for which Ax = b, right?
 
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Ray Vickson
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Why is it that for arbitrary z, A[itex]^{T}[/itex]z = 0 and b[itex]^{T}[/itex]z ≠ 0 when b is not in the range space of A when Ax = b, where A is an n x m matrix?
Huh? How can 'b' not be in the range space of A when AX=b? That is contradictory!
 
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Edited the post so that it makes more sense. Does it now?
 
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Ray Vickson
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Edited the post so that it makes more sense. Does it now?
Yes.
 
  • #7
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Actually, it means column space. This means that there is no linear combination of A's columns that gives b. This must mean that there is no x for which Ax = b, right?
Okay, sorry. I am just not used to using that term. I usually use the following for the 4 fundamental spaces

C(A) - column space (of A)
[itex] C(A^T) [/itex] - row space (of A)
N(A) - null space (of A)
[itex] C(A^T) [/itex] - left handed nullspace of A (i.e nullspace of A transpose)
 
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  • #8
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So the column space of A transpose is the null space of A?
 
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Ray Vickson
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Why is it that for arbitrary z, A[itex]^{T}[/itex]z = 0 and b[itex]^{T}[/itex]z ≠ 0 when there does not exist an x such that Ax = b, i.e. that b is not in the range space of A, where A is an n x m matrix?
What have you tried so far? Show your work.
 
  • #10
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So the column space of A transpose is the null space of A?
Whoops... My error. No it is not. The nullspace of A transpose is NOT the column space of A transpose I meant to write

[tex] N(A^T) [/tex]

The null space of A transpose is of course the set of all vectors y such that

[tex] A^T y \, = \, 0 [/tex]

We sometimes call it the "left null space" because is we take the transpose of both sides of the above equation.

[tex] (A^Ty)^T \, = \, 0^T [/tex]

we get

[tex] y^T A \, = \, 0^T [/tex]

Sorry for my mistake. I'm going to try to edit it and correct my typo before it causes confusion. Unfortunately I don't see an edit function available :(
 
  • #11
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This statement came from the book Convex Optimization by Boyd and Vandenberghe. I forget which page now, but the idea has come up again in Farkas' lemma, but in a different form.

http://en.wikipedia.org/wiki/Farkas'_lemma

Let A be an n × m matrix and b an n-dimensional vector. Then, exactly one of the following two statements is true:

(1) There exists an x ∈ Rm such that Ax = b and x ≥ 0.
(2) There exists a y ∈ Rn such that A[itex]^{T}[/itex]y ≥ 0 and b[itex]^{T}[/itex]y < 0.

This is close but the original statement involves equalities similar to (2).
 

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