### Composition and Matrix Multiplication

Exercise

1) Prove that transformations **T** and **S** are ono-to-one ⇒
The composition `TS` is one-to-one.

Answer Let **T** and **S** be one-to-one. Then

**TS**(**x**) = `TS`(`x'`)
⇒ **S**(**x**) = **S**(`x'`)
⇒ **x** = `x'`

where the first ⇒ follows from **T** being one-to-one,
and the second ⇒ follows from **S** being one-to-one.
The whole implication means `TS` is one-to-one.

2) What about the converse? In other words,
if the composition `TS` is ono-to-one,
are the transformations **T** and **S** ono-to-one?

Answer Suppose `TS` is ono-to-one. Then we have

**S**(**x**) = **S**(`x'`)
⇒ **TS**(**x**) = `TS`(`x'`)
⇒ **x** = `x'`

where the first ⇒ is obtained by applying **T** to both sides,
and the second ⇒ follows from `TS` being one-to-one.
The whole implication means **S** is one-to-one.

On the other hand, **T** is not necessarily one-to-one. An example is given by
**S** = `ID`_{s}: student → integer,
**T** = `Absolute-Value`: integer → non-negative integer.
Note that student ID numbers are always positive.

3) Rephrase your conclusions in the first and second part in terms of matrix products.

Answer If solutions of `Ay` = **b** and
`Bx` = **c** are unique,
then the solution of `ABx` = **b** is unique.

Conversely, if the solution of `ABx` = **b** is unique, then solution of
`Bx` = **c** is unique. However, solution of `Ay` = **b**
may not be unique.

Alternatively: If all columns of **A** and **B** are pivot,
then all columns of `AB` are pivot. Conversely, if all columns of `AB` are pivot,
then all columns of **B** (but not necessarily **A**) are also pivot.