Rank of the Sum of Matrices

Let \(A\) and \(B\) be two matrices of rank \(m\) and \(n\) respectively. Let \(\{C_{a_k}\}_{k=1}^m\) and \(\{C_{b_k}\}_{k=1}^n\) denote the sets of basis for the column space of matrix \(A\) and \(B\) respectively. Each column of \(A\) is, thus, a linear combination of the elements in the set \(\{C_{a_k}\}_{k=1}^m\)and each column of \(B\) is a linear combination of the elements in the set \(\{C_{b_k}\}_{k=1}^n\). From this it follows that each column in \(A+B\) is a linear combination of the elements in the combined set \(\{C_{a_k}\}_{k=1}^m \cup \{C_{b_k}\}_{k=1}^n\). Let \(rank(C_A \cap C_B )\) denote the rank of the intersection of the column spaces of \(A\) and \(B\). Note that this is, informally, equal to the number of elements common to the sets \(\{C_{b_k}\}_{k=1}^n\) and \(\{C_{b_k}\}_{k=1}^n\). Then:

$$ rank(A+B) \leqslant rank(A) + rank(B) - rank(C_A \cap C_B) $$

Similarly, letting \(R_A\) and \(R_B\) denote the row space of \(A\) and \(B\) respectively leads to:

$$ rank(A+B) \leqslant rank(A) + rank(B) - rank(R_A \cap R_B) $$

Now, because the rank due to the row space must be equal to the rank due to the column space, we have:

$$ rank(A+B) \leqslant rank(A) + rank(B) - max(rank(C_A \cap C_B),rank(R_A \cap R_B)) $$