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Consider a matrix

$$ X = \begin{bmatrix} A & B \\ 0 & D \end{bmatrix} $$

where the number of rows of each matrix is greater than that of columns. I know that $$\mbox{rank}(X) \ge \mbox{rank}(A) + \mbox{rank}(D)$$ holds. Then, can we say that $X$ has full column rank when both $A$ and $D$ have full column ranks?

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  • $\begingroup$ Let $A: n_1 \times p_1$, $B: n_1 \times p_2$ and $D: n_2 \times p_2$ where $n_1>p_1$, $n_1 >p_2$ and $n_2 > p_2$ with $p=p_1+p_2$, $n=n_1+n_2$ and $n>p$. My conjecture is that if $A$ and $D$ have full column rank such that $rank(A)=p_1$ and $rank(D)=p_2$, we have $rank(A)+rank(D)=p_1+p_2 \le rank(X) \le p$, and hence, $rank(X)=p$ holds if both $A$ and $D$ have full column rank. I'm wondering if this is correct. $\endgroup$
    – user0131
    Feb 16, 2022 at 23:37
  • $\begingroup$ This is correct and easy to show: If the columns of $A$ and those of $D$ are linearly independent, then so are those of $X$. $\endgroup$
    – Helmut
    Feb 17, 2022 at 11:07
  • $\begingroup$ Many thanks for your confirmation! $\endgroup$
    – user0131
    Feb 18, 2022 at 0:58

1 Answer 1

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Yes in the case where $A$ and $D$ are both square. Here we have $\det(X) = \det(A)\det(D)\neq 0$, so $X$ is non-singular. Thus $X$ has full-rank.

If $A$ and $D$ are rectangular, then this is not true. For instance, let $$X = \begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 1 \end{bmatrix}.$$

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  • $\begingroup$ Many thanks. But I think, in your example, $A$ and/or $D$ is not full rank... $\endgroup$
    – user0131
    Feb 16, 2022 at 3:32
  • $\begingroup$ It is. You take $A$ to be the column vector $(1;0)$. $\endgroup$
    – Magi
    Feb 16, 2022 at 4:31
  • $\begingroup$ Sorry, I'm considering the case where both $A$ and $D$ have more rows than columns. $\endgroup$
    – user0131
    Feb 16, 2022 at 5:40
  • $\begingroup$ Here $A$ is the column vector $\begin{bmatrix} 1\\ 0\end{bmatrix}$. It has more rows than columns. $\endgroup$
    – Magi
    Feb 16, 2022 at 6:46
  • $\begingroup$ Thanks. But $D=[0,1]$ is $1 \times 2$ vector, which I'm not assuming. $\endgroup$
    – user0131
    Feb 16, 2022 at 7:06

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