### Composition and Matrix Multiplication

Exercise

1) Prove that transformations **T** and **S** are onto ⇒
The composition `TS` is onto.

Answer For any **b**, by **T** onto,
we can find **y**, such that **T**(**y**) = **b**.
For this **y**, by **S** onto, we can find **x**,
such that **S**(**x**) = **y**. Then we have

**TS**(**x**) = **T**(**S**(**x**))
= **T**(**y**) = **b**.

Thus any **b** is the image of some element under the transformation `TS`.
This shows `TS` is onto.

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

Answer If `TS` is onto, then for any **b**,
we have **x**, such that **T**(**S**(**x**)) = **b**.
In particular, **b** is the image of **S**(**x**) under the transformation
**T**. The observation tells us **T** must be onto.

On the other hand, **S** is not necessarily onto. An example is given by
**S** = `ID`_{s}: student → integer,
**T** = `Last-Digit`: integer → {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.

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

Answer If `Ay` = **b** and
`Bx` = **c** has solutions for any **b** and **c**,
then `ABx` = **b** has solution for any **b**.

Conversely, if `ABx` = **b** has solutions for any **b**,
then `Ay` = **b** has solutions for any **b**.
However, `Bx` = **c** may not have solution for any **c**.

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