0
$\begingroup$

I forgot how to prove the following inequality. Could anybody help?

Let $A, B, C$ be matrices of appropriate dimensions: $rank\left( \left[\begin{array}{rr} A & 0\\ C & B \end{array}\right]\right)\ge rank\left( \left[\begin{array}{rr} A & 0\\ 0 & B \end{array}\right]\right).$

  • 0
    @ehsanmo: does rank-nullity imply this directly? If so, please post an answer with such reasoning. I do not see this argument.2011-04-19

3 Answers 3

1

If you apply the elementary row operations that bring the right-hand matrix into row echelon form to both matrices, the result is the same on both sides except in the $C$ block, which may have additional non-zero entries. Then all rows that are linearly independent on the right-hand side are also linearly independent on the left-hand side, and possibly more.

  • 0
    I think it would just be somewhat difficult to write down the detailed proof. I guess we should separately bring $A$ and $B$ to row echelon form, rather than bringing the whole matrix to its row echelon form, to give the argument, correct?2011-04-15
0

Let $A \in M_{p\times q}(\mathbb K)$, $C \in M_{(n-p)\times q}(\mathbb K)$ and $B \in M_{(n-p)\times (n-q)}(\mathbb K)$. We may decompose $\mathbb K^n=\mathbb K^p\times 0 \oplus 0 \times \mathbb K^(n-p)$ and consider $\{a_1,\dots,a_q\} \subset \mathbb K^p\times 0$ the set of columms of $A$, $\{b_1,\dots,b_q\} \subset 0 \times \mathbb K^{n-p}$ the set of columms of $B$ and $\{c_1,\dots,c_q\} \subset 0 \times \mathbb K^{n-p}$ the set of columms of $C$.

So, it suffices to show that given an LI subset $\{a_{i_1},\dots,a_{i_s},b_{j_1},\dots,b_{j_r}\}$ of $\{a_1,\dots,a_q,b_1,\dots,b_{n-q}\}$ we have that $\{a_{i_1}+c_{i_1},\dots,a_{i_s}+c_{i_s},b_{j_1},\dots,b_{j_r}\}$ is a LI subset of $\{a_1+c_1,\dots,a_q+c_q,b_1,\dots,b_{n-q}\}.$ In fact, if $\alpha_{i_1}(a_{i_1}+c_{i_1})+\dots+\alpha_{i_s}(a_{i_s}+c_{i_s})+\beta_{j_1}b_{j_1},\dots,\beta_{j_r}b_{j_r} = 0$ for some $\alpha$s and $\beta$s \in $\mathbb K$, we must have that $\mathbb K^p \times 0 \ni \alpha_{i_1}a_{i_1} +\dots+\alpha_{i_s}a_{i_s} = 0$ and $0 \times \mathbb K^{n-p} \ni \alpha_{i_1}c_{i_1}+\dots+\alpha_{i_s}c_{i_s}+\beta_{j_1}b_{j_1},\dots,\beta_{j_r}b_{j_r} = 0.$ By the linear independence of $\{a_{i_1},\dots,a_{i_s}\} \subset \{a_{i_1},\dots,a_{i_s},b_{j_1},\dots,b_{j_r}\}$ we must have that $\alpha_{i_1}=\dots=\alpha_{i_s}=0$ and, hence, $\beta_{j_1}b_{j_1}+\dots+\beta_{j_r}b_{j_r} = 0.$ Again, $\{b_{j_1},\dots,b_{j_r}\}\subset \{a_{i_1},\dots,a_{i_s},b_{j_1},\dots,b_{j_r}\}$ implies that $\beta_{j_1}=\dots=\beta_{j_r}=0.$ Therefore, we conclude that $\{a_{i_1}+c_{i_1},\dots,a_{i_s}+c_{i_s},b_{j_1},\dots,b_{j_r}\}$ is a LI set.

0

Let c'_1,..., c'_n be the columns of

$\left( \left[\begin{array}{rr} A & 0\\ C & B \end{array}\right]\right)$

and $c_1,..., c_n$ be the colums of

$\left( \left[\begin{array}{rr} A & 0\\ 0 & B \end{array}\right]\right)$

If $c_{i_1},..., c_{i_j}$ are linearly independent and

k_1c'_{i_1}+...+ k_jc'_{i_j} =0 \,.

then, it follows immediately by writing explicitely the columns that $k_1=..=k_j=0$.

Hence if we pick a basis in the column space of $\left( \left[\begin{array}{rr} A & 0\\ 0 & B \end{array}\right]\right)$, the correcponding colums are linearly independent in $\left( \left[\begin{array}{rr} A & 0\\ C & B \end{array}\right]\right)$, which proves the inequality.