Is the identity $dim(col(AB))=dim(row(A) \cap col(B))$ correct?

Question

I want to find an identity of dim(col(AB)) to gain some insight into the behavior of rank during matrix multiplication.

For an $m \times n$ matrix A and an $n \times p$ matrix B, is the identity $dim(col(AB))=dim(row(A) \cap col(B))$ correct? Here, col(M) and row(M) denotes the column and row space of matrix M.

The intuition behind:
See matrix A as a linear transformation from $R^n\to R^m$ ,while see matrix B as a set consists of $p$ column vectors in $R^n$.
On one hand, if a vector from col(B) lies in null(A), then it is mapped to zero and does not contribute to the dimension of AB.
On the other hand, for those vectors who lies in row(A), their relationship does not change and among these vectors the image of those linearly independent ones forms the base of col(AB).

Answer 1

No, it doesn't work. Consider for example \begin{align*} A &= \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} \\ B &= \begin{pmatrix} 1 & 0 \\ 2 & 0 \end{pmatrix}. \end{align*} We have $$AB = \begin{pmatrix} 3 & 0 \\ 0 & 0 \end{pmatrix},$$ which has rank $1$. On the other hand, the rowspace of $A$ and the rowspace of $B$ are the intersection of non-parallel lines $\operatorname{span}\{(1, 1)\}$ and $\operatorname{span}\{(1, 2)\}$ each passing through the origin. Thus, their intersection is $\{(0, 0)\}$, which has dimension $0$.