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Let $m,n,$ and $k$ be positive integers and assume that $A\in {\mathbb{R}^{m\times n}}$,$B\in {\mathbb{R}^{k\times n}}$,( that is, A and B are matrices with real entries of sizes $m\times n$ and $k\times n$, respectively). Define $C=\left[\frac{A}{B}\right]$. Prove that $rank(C)\ge rank(A)$.

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Note that the rank of $A$ also equals the dimension of its rowspace. And $C$ has more rows ... –  martini Jun 8 '13 at 22:39
    
@martini Can you make that an answer? –  Andrew Salmon Jun 8 '13 at 22:51

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The $\def\rk{\mathop{\rm rk}}$Rank of $A$ equals the dimension of the rowspace of $A$. Now the rowspae of $A$ is obviously a subspace of the rowspace of $[{A \atop B}] = C$, so $\rk A \le \rk C$.

If you want to argue with the column space, you can do as follows: Let $a_{i_1}, \ldots, a_{i_k}$ be independent columns of $A$, and $b_{i_1}, \ldots, b_{i_k}$ the corresponding columns of $B$. Suppose for some $\lambda_i \in K$ (the ground field), we have $$ \sum_{j=1}^n \lambda_j \binom{a_{i_j}}{b_{i_j}} = 0 $$ this gives $\sum_j \lambda_j a_{i_j} = 0$ and linear indepence of the $a_{i_j}$ gives $\lambda_j = 0$ for all $j$. So $[{a_{i_j}\atop b_{i_j}}]$ are independent columns of $C$. As the rank is the maximal number of independent columns, this number is for $C$ as least as large as for $A$.

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