**A proof from matrix theory**

Let $V=(v_1,...,v_m)\in\Bbb R^{n\times m}$ be the matrix with the $v_i$ as columns.
The matrix $X:=V^\top V\in\Bbb R^{m\times m}$ has the following properties:

- $X$ is positive semi-definite, that is, all eigenvalues are real and $\ge 0$. Furthermore, the multiplicity of the eigenvalue zero is equal $m-d$, wehre $d$ is the rank of $X$ and equals $d:=\dim \mathrm{span}\{v_1,...,v_m\}\le n$.
- $X_{ij}=v_i\cdot v_j$, in particular, all off-diagonal entries of $X$ are negative.

The matrix $Y:=I-X$ has only positive entries, and by Perron-Frobenius, its maximal eigenvalue has therefore multiplicity one.
Hence, the minimal eigenvalue of $X$ has multiplicity one.
Thus, the multiplicity of zero as eigenvalue of $X$ is *at most* one (it could be that zero is not an eigenvalue at all, but if it is one, it is the smallest).

Conclusively $m-d\le1\implies m\le d+1\le n+1$.

`LaTeX block`

and see if that helps? $\endgroup$ – Yemon Choi Jul 11 '10 at 18:09`\{`

and converts it to`{`

, and then jsMath sees`{`

and thinks its just a brace. Another reason for the habit of protecting TeX in backticks is that Markdown and jsMath have very different ideas about what an underscore`_`

means, and can get into conflict. $\endgroup$ – Theo Johnson-Freyd Jul 11 '10 at 18:38