Proving rank of $AB$ is at most equal to rank of $B$

Question

$A=m\times n$ matrix. $B = n\times p$ matrix. Prove that the rank of of the product $AB$ is at most equal to the rank of $B$.

Current state of my work:

(1) First idea: show that the kernel of $B$, $k(B)$, is contained in $k(AB)$. Let $x$ denote an arbitrary element in $k(B)$. Then $ABx = A(Bx) = $A(0) = 0, where 0 is the zero element. Thus, every element of $k(B)$ is in $k(AB)$. Thus the kernel of $B$ is contained in $k(AB)$.

(2) Second idea: Explain why the nullity of $B$ is at most the nullity of $AB$. This follows from (1). If nullity of $B >$ the nullity of $AB$, then there would be some elements of $k(B)$ not expressed in $k(AB)$. But this can't be so since (1) shows that every element of $k(B) \in k(AB)$.

(3) Last idea: Use the rank-nullity theorem. This is where I get stuck. The only thing I can come up with is Dim(domain(AB)) = k(AB) + rank(AB). $k(AB) \geq k(B)$. So $Dim(domain(AB)) \geq k(B) + rank(AB)$.

So I'm stuck. Also, feedback on (1), (2) is appreciated.

Answer 1

Identifying a matrix with its corresponding linear map, the rank of a map $A:\mathbb{R}^n \to \mathbb{R}^m$ is just $\dim (\operatorname{im} A)$. The required statement follows immediately. The rank-nullity theorem allows you make the same argument in terms of $\dim (\ker A)$ rather than $\dim (\operatorname{im} A)$, which looks like the argument in your post.

Answer 2

Consider the block matrix multiplication. Let $ b_1, b_2, \ldots, b_n $ - rows of the matrix $B$. Multiply $A$ by $B$. $$ \begin{bmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22} & \ldots & a_{2n} \\ \ldots & \ldots & \ldots & \ldots \\ a_{m1} & a_{m2} & \ldots & a_{mn} \\ \end{bmatrix} \begin{bmatrix} b_1 \\ b_2 \\ \ldots \\ b_n \\ \end{bmatrix} $$

Obviously, in this case rows of $ AB $ will be linear combination of the rows of the matrix $ B $ with coefficients of $A$. Hence dimention of the linear hull spanned by the rows $ AB $ will be less or equal dimention of the linear hull spanned by the rows $ B $. Hence $ rankAB \le rankB $

Answer 3

we will use the fact that the row vector $a^\top B$ is a linear combination of the rows of $B$ weighted by the components of $a.$ let us write the matrix $A$ as made up of the rows $\pmatrix{a_1^\top\\a_2^\top\\\vdots\\a_m^\top}.$ then the $i$-th row of $AB$ is $a_i^\top B$ which is a linear combination of the rows of $B.$ that is every row of $AB$ is a linear combination of the rows of $A.$ that implies the that a basis of the row space of $B$ spans the row space of $AB.$ that is equivalent to $$rank(AB) \le rank(B). $$

Answer 4

Any $m\times n$ matrix $A$ with $m$ rows and $n$ columns represents a linear transformation, which acts on vectors from $\mathbb{R}^n$ and produces vectors in $\mathbb{R}^m$. Similarly, given a $n\times k$ matrix $B$, we can view it as a linear transformation from $\mathbb{R}^k$ to $\mathbb{R}^n$. Then the product $AB$ stands for the superposition of linear transformations $A$ and $B$, which is a linear transformation itself.

More explicitly,
$$ A:\mathbb{R}^{n}\to\mathbb{R}^{m}, \quad B:\mathbb{R}^{k}\to\mathbb{R}^{n} \implies(A\cdot B) :\mathbb{R}^{k} \to \mathbb{R}^{m} $$

To illustrate, assume that vector $v \in \mathbb{R}^{k}$, then $$ w = A\cdot B\cdot\!\!\underbrace{v}_{\in\mathbb{R}^{k}} = A\cdot\!\underbrace{B\,v}_{\in \mathbb{R}^{n}} = \underbrace{ABv}_{\in\mathbb{R}^{m}} $$ Now, rank of a matrix is the number of linear independent vectors needed to span the set of all possible outputs of this matrix, i.e. the dimension of its image: $$ A:\mathbb{R}^{n}\to\mathbb{R}^{m} \implies \operatorname{rank}(A) = \dim \big(\operatorname{im}(A)\big) = \dim \big\lbrace w\in \mathbb{R}^{m} \, \big| \ w = Au \ \text{ for some } \ u \in \mathbb{R}^{n} \big\rbrace \\ B:\mathbb{R}^{k}\to\mathbb{R}^{n} \implies \operatorname{rank}(B) = \dim \big(\operatorname{im}(B)\big) = \dim \big\lbrace u\in \mathbb{R}^{n} \, \big| \ u = Bv \ \text{ for some } \ v \in \mathbb{R}^{k} \big\rbrace $$ Now, in superposition $(A\cdot B)$ matrix $A$ acts on the image of $B$, therefore the resulting image cannot have dimensionality more that $\operatorname{rank}(B)$ because of linearity of $A$. Thus, we conclude that $$\operatorname{rank}(AB)\leq \operatorname{rank}(B)$$ Q.E.D.

Answer 5

Since the rows of $AB$ are linear combinations of rows of $B$, we have $$ rank AB\le rank B $$