### Inverse

Exercise

1) Prove that `Ax` = **b** has solutions for any **b** ⇔
There is a matrix **X**, such that `AX` = **I**.

Answer Assume `Ax` = **b** has solutions for any **b**.
By taking **b** to be vectors in the standard basis
**e**_{1}, **e**_{2}, ..., **e**_{n},
we get vectors **x**_{1}, **x**_{2}, ..., **x**_{n},
satisfying **Ax**_{i} = **e**_{i}.
Thus if we take **X** =
[**x**_{1} **x**_{2} ... **x**_{n}],
then by this formula, `AX` =
[`Ax`_{1} `Ax`_{2} ... **Ax**_{n}] =
[**e**_{1} **e**_{2} ... **e**_{n}] =
**I**.

Conversely, assume `AX` = **I**. Then for any **b**,
we have **A**(`Xb`) = `Ib` = **b**.
In other words, **x** = `Xb` is a solution of `Ax` = **b**.

2) Prove that solution of `Ax` = **b** is unique ⇔
There is a matrix **X**, such that `XA` = **I**.

Answer Assume **A** is an `m` by `n` matrix such that
the solution of `Ax` = **b** is unique. Then all columns of **A**
are pivot, which means that the reduced row echelon form of **A** is

**J** = [**e**_{1} **e**_{2} ... **e**_{n}] = [ |
**I**_{n} |
] |

**O** |

where **e**_{1}, **e**_{2}, ..., **e**_{n}
are the first `n` vectors in the standard basis of **R**^{m}
(recall
that the uniqueness implies `n` ≤ `m`). In other words,
we can use row operations to change **A** to become **J**.
The same row operations will change [**I**_{n} **A**] to [**Y J**],
where **Y** is some `n` by `n` matrix. Then the reversed row operations change
[**Y J**] =
[**Y** **e**_{1} **e**_{2} ... **e**_{n}]
to [**I**_{n} **A**] =
[**I**_{n} **a**_{1} **a**_{2} ... **a**_{n}].
Consequently, [**Y** **e**_{i}] can be row operated to become
[**I**_{n} **a**_{i}]. In other words, the systems
`Yx` = **e**_{i} and
**I**_{n}**x** = **a**_{i} have the same solutions.
Since **x** = **AI** is the unique solution of
**I**_{n}**x** = **AI**, we conclude that
**Ya**_{i} = **e**_{i}. Thus

**YA** =
[`Ya`_{1} `Ya`_{2} ... **Ya**_{n}] =
[**e**_{1} **e**_{2} ... **e**_{n}] =
**J**

and if we take **X** = [**I**_{n} **O**]**Y** (i.e., **X** is the first `n` rows of **Y**), then

**XA** = [**I**_{n} **O**]`YA` =
[**I**_{n} **O**]**J** = **I**_{n}.

Conversely, assume `XA` = **I**. Then `Ax` = **0** implies
`XAx` = `X`0 = **0**. Since `XA` = **I**,
we conclude that **x** = **0**. The whole implication is that the homogeneous system
`Ax` = **0** has only the trivial solution.

3) Rephrase the first and second parts in terms of linear transformations.

Answer A linear transformation **T** is onto ⇔
There is a linear transformation **S**, such that `TS` = `id`.

A linear transformation **T** is one-to-one ⇔ There is a linear transformation
**S**, such that `ST` = `id`.

We remark that the statements also hold for general transformations.