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

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

## marked as duplicate by Marc van Leeuwen linear-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 3 '14 at 14:11

• The rank of a matrix $A$ which maps $V$ to $W$ is the same as the dimension of the image of $V$, which will be some subspace $U$. $AB$ is the composite of two linear maps. Try tracking the image and its dimension through subspaces as you do one after the other. – Mark Bennet Jul 2 '11 at 7:14
• math.stackexchange.com/questions/978/… – user92843 Jul 2 '11 at 12:08

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!

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

Since we have $\text{Col }AB \subseteq \text{Col }A$ and $\text{Row }AB \subseteq \text{Row }B$, therefore $\text{Rank }AB \leq \text{Rank }A$ and $\text{Rank }AB \leq \text{Rank }B$, then the result follows.

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

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.