### Rank and Nullity

##### Properties of rank and nullity

Theroem 1 `rank`**A** = `rank`**A**^{T}.

Proof The equality can be derived by comparing the two methods for computing a basis of the column space.

`rank`**A**

= number of pivots in the column echelon form of `A` (second method of computing a basis of `col`**A**)

= number of pivots in the row echelon form of `A`^{T} (column op on `A` = row op on `A`^{T})

= `rank`**A**^{T}. (first method of computing a basis of `col`**A**^{T})

Theorem 2 For an `m` by `n` matrix `A`, `rank`**A** ≤ `m` and `n`. Moreover

`rank`**A** = `m` ⇔ Columns of `A` span **R**^{m} ⇔ `Ax` = `b` has solutions for all `b`.
`rank`**A** = `n` ⇔ Columns of `A` are linearly independent ⇔ Solution of `Ax` = `b` is unique.

Proof First,

`rank`**A** = number of pivot columns of `A` ≤ number of all columns of `A`.

Then,

`rank`**A** = `rank`**A**^{T} ≤ number of all columns of `A`^{T} = number of all rows of `A`.

Moreover,

`rank`**A** = `m`

⇔ `dim`(`col`**A**) = `dim`**R**^{m} (definition of rank)

⇔ `col`**A** = **R**^{m} (take `H` = `col`**A** and `V` = **R**^{m} in this property)

⇔ The column vectors span **R**^{m}.

`rank`**A** = `n`

⇔ `dim`(`nul`**A**) = `nullity`**A** = `n` - `rank`**A** = 0 (equality: `rank` + `nullity` = number of columns)

⇔ `nul`**A** ={**0**}

⇔ `Ax` = **0** has only the trivial solution (definition of `nul`**A**)

⇔ The column vectors are linearly independent.