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In my book on linear algebra (Lay), Lay writes

A system of linear equations is said to be homogeneous if it can be written in the form $A\vec{x} = \vec{0}$, where $A$ is an m x n matrix and $\vec{0}$ is the zero vector in $\mathbb{R}^m$. Such a system always has at least one solution, namely, $\vec{x}=\vec{0}$ (the zero vector in $\mathbb{R}^n$).

Seeing as both $\vec{x}$ and $\vec{b}$ are column vectors, how come they're not both in $\mathbb{R}^m$, the number of rows in the coefficient matrix $A$? I tried to look back in the book for an explanation, but could not find the reason; have I completely misunderstood the concept of $\mathbb{R}^n$ and $\mathbb{R}^m$?

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up vote 3 down vote accepted

This is because of the way you multiply matrices. A column vector in $R^n$ is a $n$ by $1$ matrix, and a $m$ by $n$ matrix multiplied by a $n$ by $1$ matrix gives a $m$ by $1$ matrix (a column vector in $R^m$).

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Thank you! That makes sense! – Jodles Aug 23 '11 at 14:45
E.g. $$\begin{pmatrix} 1 & 0 & 1 & 3 \\ 5 & -1 & 4 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 2 \\ 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 16 \\ 19 \end{pmatrix}$$ – anon Aug 29 '11 at 8:16
Thank you for the example! I know how matrix multiplication works, but I was too nearsighted to see the relation to $\mathbb{R}^n$ and $\mathbb{R}^m$. – Jodles Aug 29 '11 at 8:33

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