How to prove that if A and B are matrices, then the rank of AB <= the rank of B

Question

So far I know that when matrices A and B are multiplied, with B on the right, the result, AB, is a linear combination of the columns of A, but I'm not sure what to do with this.

Answer 1

$$rank(AB)=rank((AB)^T)=rank(B^TA^T)\leq rank(B^T)=rank(B)$$

Answer 2

If you know the rank-nullity theorem, it's easy :

$\forall x \in \text{Ker} B, ABx = A(Bx) = A(0) = 0$

So $x\in \text{Ker} AB$, hence $\text{Ker} B \subset \text{Ker} AB$, that imply that

$\text{dim Ker} B \leq \text{dim Ker} AB$ , and by the rank nullity theorem, you get that $\text{rank} B \geq \text{rank} AB$

Answer 3

The image of $B$ is a subspace of dimension $\mathrm{rank}(B)$. Left multiplication by $A$ transforms this into a new subspace, which is the image of $AB$ having dimension $\mathrm{rank}(AB)$; this linear transformation cannot increase the dimension of the subspace.

Put in other words, $\mathrm{rank}(B)$ is the dimension of the space generated by the rows of $B$, and $\mathrm{rank}(AB)$ is the dimension of the space generated by the rows of $AB$. The rows of $AB$ are linear combinations of the rows of $B$, so the dimension of this space is restricted by $\mathrm{rank}(B)$ (the columns of $AB$ are also linear combinations of the columns of $A$ as you mention, but this fact is less helpful here).

Answer 4

Consider the linear maps associated to $A$ and $B$ in canonical bases, respectively ($K$ denotes the base field): \begin{align*} f\colon K^p\to K^q\\ g\colon K^n\to K^p \end{align*} $\DeclareMathOperator\rk{rank}\DeclareMathOperator\img{Im}$The rank of a matrix is the dimension of the image of the associated linear map. Thus $$\rk(AB)=\rk(fg),\quad \rk A=\rk f,\quad \rk B=\rk g,$$ and $\;\rk(fg)=\dim f(g(K^n))=\dim(\img(f\,\vert_{\img g})\le \dim(\img g)=\rk B$.

Answer 5

If $AB$ has rank $r$ then there are vectors $v_1,...,v_r$ such that $ABv_1,...,AB v_r$ are linearly independent.

Hence $Bv_1,...,B v_r$ must be linearly independent (or an immediate contradiction), and so $B$ must have rank $\ge r$.

Answer 6

Let $A$ be $m \times n$ and $B$ be $n \times p$ both with entries in the field $F$. Consider the maps: $L_{A}: F^{n} \rightarrow F^{m}: L_{A}(X) = AX$ and $L_{B}: F^{p} \rightarrow F^{n}: L_{B}(X) = BX$.

Then $L_{A} \circ L_{B}: F^{p} \rightarrow F^{m}$ satisfies $Y \in \text{Image } L_{A} \circ L_{B} \implies L_{A}(L_{B}(X)) = Y \implies Y \in \text{Image } L_{A} $. It follows that $\text{Image } L_{A} \circ L_{B} \subset \text{Image } L_{A}$. The inequality in question then follows.