### Inverse

##### Criterion for invertibility of a matrix

Theorem For a square matrix **A**, the following are equivalent

**A** is invertible
**AB** = **I** for some matrix **B**
**BA** = **I** for some matrix **B**
**Ax** = **b** has solution for any **b**
- Solution of
**Ax** = **b** is unique
- All rows of
**A** are pivot
- All columns of
**A** are pivot
**A** can be row operated to become **I**

Proof By definition of invertibility, the 1st property
implies the 2nd and the 3rd. Then by this
result, the 2nd property implies
the 4th, and the 3rd implies the 5th. The 4th property
means
that all rows of **A** are pivot. The 5th property
means
that all columns of **A** are pivot. For a square matrix, either case
implies
that the reduced row echelon form of **A** is the identity matrix **I**.
This completes the proof for the following implications.

1 ⇒ 2 ⇒ 4 ⇒ 6 ⇒ 8

1 ⇒ 3⇒ 5 ⇒ 7 ⇒ 8

Thus it remains to show that the 8th property implies the 1st. The idea is to show that our
method for computing the inverse, which was based on
the assumption that the 8th property holds, indeed gives us the inverse matrix.

The method is to use row operations (presumably the same ones that change **A**
to **I**) to change [`A I`] to [`I B`]. By taking columns
1, 2, ..., `n`, `n` + `i` only, we see that [**A e**i] can be
row operated to become [`I b``i`], where **e**i is the
`i`-th standard basis vector, and **BI** is the `i`-th column of **B**.
By thinking of [**A e**i] as the augmented matrix of the system
`Ax` = **e**i, the solution of the system is the same as
the solution of `Ix` = **BI** Since `Ix` = **BI**
clearly has **x** = **BI** as the unique solution, we have `Ab``i` = **e**`i`. Then we
conclude

`AB` =
[`Ab`_{1}, `Ab`_{2}, ..., **Ab**_{n}] =
[**e**_{1} **e**_{2} ... **e**_{n}] =
**I**.

To show **B** is really the inverse of **A**, we still need to establish
`BA` = **I**. The following argument shows that this is a consequence of
`AB` = **I** and **A** is a square matrix.

**AB** = **I**

⇒ `Ax` = **b** has solution for any **b**
(2 ⇒ 4)

⇒ Solution of `Ax` = **b** is unique
(by basic principle of linear algebra)

**AB** = **I**

⇒ **A**(`BA`) = (`AB`)**A** = `IA`
= **A** = `AI`

⇒ **A**(`i`-th column of `BA`) = **Ae**_{i}
(by this equality)

⇒ `i`-th column of `BA` = **e**_{i}
(by solution of `Ax` = **b** unique)

⇒ `BA` = [**e**_{1} **e**_{2} ... **e**_{n}]
= **I**.