Question

Consider an $\mathbf{ }\mathit{n}\mathbf{\times }\mathit{p}$matrix $\mathbf{ }\mathit{A}$and$\mathbf{ }\mathit{p}\mathbf{\times }\mathit{m}$matrix $\mathit{B}$.

1. What is the relationship between ker($\mathit{A}\mathit{B}$) and ker role="math" localid="1664188035115" $\mathit{B}$)? Are they always equal? Is one of them always contained in the other?
2. What is the relationship between im($\mathit{A}$) and im (role="math" localid="1664187975738" $\mathit{A}\mathit{B}$)?

Short answer

1. $\ker \left(B\right)\subseteq \ker \left(AB\right)$
   
2. $im\left(AB\right)\subseteq im\left(A\right)$
   

Step-by-step solution

Define kernel of linear transformation

Thekernel of linear transformation is 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.

(a) Find relationship between kernels

Let$\overrightarrow{x}\in \ker \left(B\right).$

Therefore, the equation is:

$B\overrightarrow{x}=0$ …… (1)

Now,

$$\begin{gathered} \left(AB\right)\overrightarrow{x}=A\left(B\overrightarrow{x}\right) \\ =A·0 \\ =0 \cdots \cdots \left(\because by \left(1\right)\right) \end{gathered}$$

$\therefore \left(AB\right)\overrightarrow{x}=0$

Thus, $\overrightarrow{x}\in \ker \left(AB\right).$

Therefore, $\ker \left(B\right)\subseteq \ker \left(AB\right)$.

Hence, by Exercise 37, they do not need to be equal.

(b) Find relationship between images

Consider $\overrightarrow{z}\in im\left(AB\right)$.

This implies that there exists a $\overrightarrow{y}$such that

$$\begin{gathered} \left(AB\right)\overrightarrow{y}=\overrightarrow{z} \\ \Rightarrow A\left(B\overrightarrow{y}\right)=\overrightarrow{z} \\ \Rightarrow \overrightarrow{z}\in im\left(A\right) \end{gathered}$$

Thus, $\mathrm{im}\left(\mathrm{AB}\right)\subseteq \mathrm{im}\left(\mathrm{A}\right)$.

Hence, by Exercise 37, they do not need to be equal.