# Consequences of a rectangular matrix being of maximal rank

I have a real matrix $A$, $(m+1) \times m$ and a vector $b \in \mathbb R^{m+1}$ such that $b_{m+1}=0$. For any vector $u\in \mathbb R^m$, $Au=0 \Rightarrow u=0$. This means that $A$ is a rectangular matrix of maximal rank, i.e. of rank $m$.

Since $m< \infty$, I'm told this means there is a solution $u\in \mathbb R^m$ to $Au=b$. I don't understand why: if $\operatorname{rk}(A)=m$, then the dimension of the image of the linear application defined by $A$ in $\mathbb R^{m+1}$ is $m$. So for $b\in\mathbb R^m$ we could guarantee a solution, but how can we guarantee the solution satisfies our $b\in\mathbb R^{m+1}$?

Is there something simple I'm missing? (I've asked a couple of classmates who didn't know either, and my professor who told me this in the first place told me to look it up in a book that's in a library that's closed until next week.)

-

You're right, your professor is wrong. In such a case, it's usually a good idea to look for a simple example. For $m=1$, the matrix $A=\pmatrix{1\\1}$ has full rank $1$. For any vector $u\in\mathbb R^1$, $Au=0$ implies $u=0$. Yet there is clearly no $\mathrm u\in\mathbb R^1$ such that $Au=\pmatrix{b_1\\0}$ for any $b_1\ne0$.
Maybe there's something particular to my case that changes the situation... this professor isn't one to be wrong. The problem that this matrix comes from is a bit too complicated to put on a forum, but the idea is that I have a vector space of continuous function of dimension m, and a basis of m functions, $f_i$. The components of A and b all come from integrals of these functions over a domain or the boundary of the domain. ($\int_\Omega \gamma_\epsilon \nabla f_i \nabla f_j dx$, or $\int_{\partial\Omega} f_i d\sigma$, or $\int_{\partial\Omega} g f_i d\sigma$,). Could that change the result? –  JKH May 18 '13 at 15:55
$\gamma_\epsilon \in L^\infty(\Omega)$, and $g\in L^2(\partial\Omega)$ –  JKH May 18 '13 at 15:57