Linear Algebra: A^tb=0 - Solving for Best Approximation of b in Col A

[ji707]

hi all, i was given a take home exam for my linear algebra course and i can't seem to find the answer to this problem.

Homework Statement

if A^T\vec{b} = \vec{0}
what can you say about $\hat{b}$ the vector in Col A which is the best approximation of
$\vec{b}$.

Homework Equations

$A^T A \vec{x} = A^T\vec{b}$

The Attempt at a Solution

I don't even know where to start,
I have a feeling that there is a way to show that $\hat{b}$ is equal to $\vec{b}$. but I don't know how to go about finding this.
I've been thinking about it for a day already. any hints or nudges in the right direction would be very helpful please.

 

[lanedance]

start by considering the columns of A. Each component of A^T.b is the same as the dot product of a column of A with b.

 

[ji707]

I tried this not sure if its correct. i used this theorem

$(col A)^\perp = Nul A^T$
since i was given $A^T\vec{b} = \vec{0}$
$Nul A^T = \vec{b}$
and this means that $\vec{b}$ is already orthogonal to $col A$
and because $\vec{b} - \hat{b} = \perp$ to $Col A$
$\hat{b} = \vec{0}$

right? or am I completely off?

 

[lanedance]

I think you're heading in the right direction - b is orthogonal to every column vector in A, so they are going to do a pretty poor job when used to approximate b

$Nul A^T = \vec{b}$

this is not quite true

$\vec{b} \in Nul(A^T)$
is more accurate

i think you've pretty much got it, but you need to outline how you measure how "good" an approximation is to tie it together