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 $b_1\ne0$.

  • $\begingroup$ 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? $\endgroup$ – JKH May 18 '13 at 15:55
  • $\begingroup$ $\gamma_\epsilon \in L^\infty(\Omega)$, and $g\in L^2(\partial\Omega)$ $\endgroup$ – JKH May 18 '13 at 15:57
  • $\begingroup$ @JKH: I'm not sure I understand what you're saying. If your question is as you wrote it, about the implications of a rectangular matrix being of maximal rank, then my answer stands. If you have a completely different question, about what inferences can be drawn from the properties of certain functions and integrals, then of course the answer might be completely different. In that case, I suggest that you post a new question, including sufficient information about the functions and integrals. $\endgroup$ – joriki May 18 '13 at 16:01
  • $\begingroup$ Well I thought my question was as I wrote it, because that's how my professor said it, but I guess my problems with understanding this were reasonable, and there must be some other reason I have the existance of a solution. Perhaps my prof forgot we were going from dimension m to m+1. $\endgroup$ – JKH May 18 '13 at 16:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.