Question

Consider an $\mathbf{ }\mathit{n}\mathbf{\times }\mathit{p}$matrix$\mathbf{ }\mathit{A}$and $\mathit{p}\mathbf{\times }\mathit{m}$matrix $\mathit{B}$. If ker ($\mathbf{ }\mathit{A}$) = im ($\mathit{B}$), what can you say about the product $\mathit{A}\mathit{B}$?

Short answer

The product $AB=0$.

Step-by-step solution

To define kernel of linear transformation

Thekernel of linear transformationis defined as follows:

The kernel of a linear transformation$T\left(\overrightarrow{x}\right)=A\overrightarrow{x}$ from${\mathrm{ℝ}}^m$ to${\mathrm{ℝ}}^n$ consists of all zeros of the transformation, i.e., the solutions of the equations $T\left(\overrightarrow{x}\right)=A\overrightarrow{x}=0$.

It is denoted by ker$\left(T\right)$ or ker$\left(A\right)$.

Let$A$ be an$n\times p$ matrix and$B$ be an$p\times m$ matrix.

Determine about the product 

Let $\overrightarrow{x}\in {\mathrm{ℝ}}^m$.

Therefore, $B\overrightarrow{x}\in im\left(B\right)=\ker \left(A\right)$.

This implies that

$\left(AB\right)\overrightarrow{x}=A\left(B\overrightarrow{x}\right)=0$.

Take $\overrightarrow{x}={\overrightarrow{e}}_i\in {\mathrm{ℝ}}^m$.

Thus, we get,

$AB{\overrightarrow{e}}_i=\overrightarrow{0}, i=1, 2, \dots , m$.

This implies that all columns of $AB$are 0.

Hence, the product $AB=0$.