How to prove $\text{Rank}(AB)\leq \min(\text{Rank}(A), \text{Rank}(B))$?

Question

How to prove $\text{Rank}(AB)\leq \min(\text{Rank}(A), \text{Rank}(B))$?

Answer 1

I used a way to prove this, which I thought may not be the most concise way but it feels very intuitive to me. The matrix $AB$ is actually a matrix that consist the linear combination of $A$ with $B$ the multipliers. So it looks like... $$\boldsymbol{AB}=\begin{bmatrix} & & & \\ a_1 & a_2 & ... & a_n\\ & & & \end{bmatrix} \begin{bmatrix} & & & \\ b_1 & b_2 & ... & b_n\\ & & & \end{bmatrix} = \begin{bmatrix} & & & \\ \boldsymbol{A}b_1 & \boldsymbol{A}b_2 & ... & \boldsymbol{A}b_n\\ & & & \end{bmatrix}$$ Suppose if $B$ is singular, then when $B$, being the multipliers of $A$, will naturally obtain another singular matrix of $AB$. Similarly, if $B$ is non-singular, then $AB$ will be non-singular. Therefore, the $rank(AB) \leq rank(B)$.

Then now if $A$ is singular, then clearly, no matter what $B$ is, the $rank(AB)\leq rank(A)$. The $rank(AB)$ is immediately capped by the rank of $A$ unless the the rank of $B$ is even smaller.

Put these two ideas together, the rank of $AB$ must have been capped the rank of $A$ or $B$, which ever is smaller. Therefore, $rank(AB) \leq min(rank(A), rank(B))$.

Hope this helps you!

Answer 2

Hint: Show that rows of $AB$ are linear combinations of rows of $B$. Transpose this hint.

Answer 3

Surely vectors that are in the kernel of $B$ are also in the kernel of $AB$. Vectors that are in the kernel of $A^t$ are also in the kernel of $(AB)^t=B^tA^t$ therefore with the fact that Rank(A)=Rank($A^t$) and the knowledge that the rank gives you the size of the kernel of a Matrix you are done.

Answer 4

You know that a linear transformation cannot increase the dimension of its domain; i.e. If $T: V\rightarrow W$ is a linear transformation, $$\dim(T(V))\le \dim(V).$$