# What is the rank of this matrix

$Q$ is a $n_Q \times m_Q$ matrix and has known rank $r_Q$

$P$ is a $n_P \times m_P$ matrix and has known rank $r_P$

Let $q_k$ and $p_k$ be the k-th kolumn of $Q$ and $P$ resp.

## My goal

Find an expression of rank $R$ in function of properties of Q and P.

## The matrix R

$$R = \left[ \begin{array}{@{}c|c|c@{}} Q & Q & Q & \cdots\\ p_1\ \ p_1 \cdots & p_2\ \ p_2 \cdots & p_3\ \ p_3 \cdots \end{array} \right]$$

$$\hat{R} = \left[ \begin{array}{@{}c|c@{}} q_1\ \ q_1 \cdots & q_2\ \ q_2 \cdots & q_3\ \ q_3 \cdots \\ P & P & P & \cdots \end{array} \right]$$

It can be easily seen that $\text{rank}(R) = \text{rank}(\hat{R})$ because it is just a rearrangement of the columns.

## Results so far

Based on intuition and experimental results I believe that there exists an upper bound: $\text{rank}(R) \le \text{rank}(Q) + \text{rank}(P)$

I have no proof of this though.

## Update

Based on the suggestions the problem is reduced to the following problem: I substracted the first column from every other column.

$$\text{rank}(R) = \text{rank}( \left[ \begin{array}{@{}c|c|c@{}} Q \\ p_1\ \ p_1 \cdots \end{array} \right]) + \text{rank}(R) - 1$$

• I think that if you made some simple subtractions you will get: $$$$\left[\begin{array}{@{}c|c|c@{}} Q & 0_Q & 0_Q & \cdots\\ p_1\ \ p_1 \cdots & p_2-p_1\ p_3-p_1 \cdots & 0\ \ 0 \cdots \end{array} \right]$$$$ – Elaqqad Mar 2 '15 at 20:44
• Interesting suggestion. I'll look into it! – tgoossens Mar 2 '15 at 20:51
• I think your result is wrong but your idea lead me to the solution I believe. I'll post my results soon – tgoossens Mar 2 '15 at 21:02
• You can identify the columns of your matrix $R$ with the vectors $\left(q_i, p_j\right)$ in the direct sum $F^{n_Q} \oplus F^{n_P}$ of vector spaces (where $F$ is your ground field). The rank of $R$ is thus the dimension of the span of these vectors. But this span is a subspace of the subspace $\operatorname{Im} Q \oplus \operatorname{Im} P$ (where $\operatorname{Im}$ means "image"), and thus has rank $\leq \operatorname{rank} Q + \operatorname{rank} P$. Whether this inequality becomes an equality generally depends on the actual matrices, not just their ranks. – darij grinberg Mar 2 '15 at 21:11