# Linear algebra A^tb=0

1. Dec 12, 2012

### ji707

hi all, i was given a take home exam for my linear algebra course and i cant seem to find the answer to this problem.
1. The problem statement, all variables and given/known data
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
$\vec{b}$.

2. Relevant equations
$A^T A \vec{x} = A^T\vec{b}$

3. 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 dont know how to go about finding this.

2. Dec 12, 2012

### 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.

3. Dec 12, 2012

### 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?

4. Dec 13, 2012

### 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

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