The rank of an $m \times n$ matrix $M$ is the dimension of the range $\calR(M)$ of the matrix $M$.
The range of the matrix $M$ is
\[ \calR(M)=\{\mathbf{y} \in \R^m \mid \mathbf{y}=M\mathbf{x} \text{ for some } \mathbf{x} \in \R^n\}.\]

Proof.

(a) $\rk(AB) \leq \rk(A)$.

Recall that the rank of a matrix $M$ is the dimension of the range $\calR(M)$ of the matrix $M$.
So we have
\[\rk(AB)=\dim(\calR(AB)), \quad \rk(A)=\dim(\calR(A)).\]

In general, if a vector space $V$ is a subset of a vector space $W$, then we have
\[\dim(V) \leq \dim(W).\]
Thus, it suffices to show that the vector space $\calR(AB)$ is a subset of the vector space $\calR(A)$.

Consider any vector $\mathbf{y} \in \calR(AB)$. Then there exists a vector $\mathbf{x}\in R^{l}$ such that $\mathbf{y}=(AB)\mathbf{x}$ by the definition of the range.
Let $\mathbf{z}=B\mathbf{x} \in \R^n$.

Then we have
\[\mathbf{y}=A(B\mathbf{x})=A\mathbf{z}\]
and thus the vector $\mathbf{y}$ is in $\calR(A)$. Thus $\calR(AB)$ is a subset of $\calR(A)$ and we have
\[\rk(AB)=\dim(\calR(AB)) \leq \dim(\calR(A))=\rk(A)\]
as required.

(b) If the matrix $B$ is nonsingular, then $\rk(AB)=\rk(A)$.

Since the matrix $B$ is nonsingular, it is invertible. Thus the inverse matrix $B^{-1}$ exists. We apply part (a) with the matrices $AB$ and $B^{-1}$, instead of $A$ and $B$. Then we have
\[\rk((AB)B^{-1}) \leq \rk(AB)\]
from (a).

Combining this with the result of (a), we have
\[\rk(A)=\rk((AB)B^{-1}) \leq \rk(AB) \leq \rk(A).\]
Therefore all the inequalities are in fact equalities, and hence we have
\[\rk(AB)=\rk(A).\]

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