# Seemingly simple linear algebra

Why is it that for arbitrary z, A$^{T}$z = 0 and b$^{T}$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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"Range space" ?? Do you mean "row space" ??

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?

Ray Vickson
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Why is it that for arbitrary z, A$^{T}$z = 0 and b$^{T}$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!

Edited the post so that it makes more sense. Does it now?

Ray Vickson
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Edited the post so that it makes more sense. Does it now?
Yes.

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)
$C(A^T)$ - row space (of A)
N(A) - null space (of A)
$C(A^T)$ - left handed nullspace of A (i.e nullspace of A transpose)

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So the column space of A transpose is the null space of A?

Ray Vickson
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Why is it that for arbitrary z, A$^{T}$z = 0 and b$^{T}$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.

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

$$N(A^T)$$

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

$$A^T y \, = \, 0$$

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

$$(A^Ty)^T \, = \, 0^T$$

we get

$$y^T A \, = \, 0^T$$

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 :(

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$^{T}$y ≥ 0 and b$^{T}$y < 0.

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