I am aware of the relation ${\rm rank}(A+B)\leq {\rm rank}(A)+{\rm rank}(B)$, but I am curious of when the equality holds? What are the necessary assumptions for the matrices $A$ and $B$?
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$\begingroup$ Consider $$\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix} + \begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}$$ Think about the relationships between the component matrices' nullspaces and ranges. $\endgroup$– Ninad MunshiCommented May 22, 2023 at 8:06
1 Answer
Note that the rank is equal to the dimension of the column space (and row space).
$rank(A + B) = \dim col(A + B)$
$\leq \dim (col(A) + col(B))$
$\leq \dim col(A) + \dim col(B)$
$= rank(A) + rank(B)$
Let's think about the span. The column space of (A + B) should contain the column spaces of both A and B. Also, since the rank cannot exceed the amount of rows nor columns, $r \leq m \leq n,$ or $r \leq n \leq m.$ Where m, n represent the rows and columns.
Which means that if $\dim(A) + \dim(B) > n$ or $\dim(A) + \dim(B) > m$, there is certain dependence between the columns of A and B.
Finally
$rank(A + B) = \dim(col(A)) + \dim(col(B)) - \dim(col(A) \cap col(B)) = rank(A) + rank(B)$
If and only if
$\dim(col(A) \cap col(B)) = 0$
Indicating independence.
(i) The column space of (A + B) should contain A and B.
(ii) Vectors in column space of A should be independent from the vectors in column space of B.