Rank of a block matrix

Question

Given rank(A)=k, how do we prove that rank(B)=2k if $ B= \left[ {\begin{array}{cc} A & 0 \\ 0 & A \\ \end{array} } \right] $?

Answer 1

Let $A=(a_{ij})_{n\times n}$. $Rank(A)=k$ implies that there are exactly $k$ many independent columns in $A$, say $c_{1},\ldots,c_{k}$, where $c_{i}=[a_{1i},a_{2i},\ldots,a_{ni}]_{n\times 1}$. Define $\tilde c_{i}=[a_{1i},a_{2i},\ldots,a_{ni},0,0,\ldots,0]_{2n\times 1}$ and $\hat c_{i}=[0,0,\ldots,0,a_{1i},a_{2i},\ldots,a_{ni}]_{2n\times 1}$. Notice that $\tilde c_{i}$ is orthogonal to $\hat c_{j}$ for all $1\leq i, j\leq k$. In particular, $\tilde c_{1},\ldots,\tilde c_{k},\hat c_{1},\ldots,\hat c_{k}$ is a set of independent column vectors of $B$.

Since $c_{1},\ldots,c_{k}$ are the only independent columns of $A$ (i.e., every other columns $c_{k+1},\ldots,c_{n}$ can be written as a linear combination of $c_{1},\ldots,c_{k}$), $\tilde c_{1},\ldots,\tilde c_{k},\hat c_{1},\ldots,\hat c_{k}$ is the only independent set of column vectors in $B$. Therefore $Rank(B)=2k$.