Theorem 1 `v`_{1}, `v`_{2}, ...,
**v**_{k}` _{}`,

Proof In case `k` = 2,
we assume `v`_{1}, `v`_{2}, `v`_{3} are linearly independent. Then we have the following implication

`c`_{1}`v`_{1}
+ `c`_{2}`v`_{2}
= **0** ⇒
`c`_{1}`v`_{1}
+ `c`_{2}`v`_{2}
+ 0`v`_{3}
= **0** ⇒
`c`_{1} = `c`_{2} = 0,

where the second ⇒ follows from the linear independence of `v`_{1},
`v`_{2}, `v`_{3} (take `c`_{3} = 0
in the definition). The implication means exactly that `v`_{1},
`v`_{2} are linearly independent.

The proof for general `k` is similar.

Theorem 2 `v`_{1}, `v`_{2}, ..., **v**_{k}` _{}` are linearly independent and

Proof In case `k` = 3,
we assume `v`_{1}, `v`_{2}, `v`_{3} are linearly independent and **v**_{4} ∉ `span`{`v`_{1}, `v`_{2}, `v`_{3}}.

Assume `v`_{1}, `v`_{2}, `v`_{3}, `v`_{4} are linearly dependent, then we have

`c`_{1}`v`_{1}
+ `c`_{2}`v`_{2}
+ `c`_{3}`v`_{3}
+ `c`_{4}`v`_{4}
= **0**,

with at least one of the coefficients `c`_{1}, `c`_{2}, `c`_{3}, `c`_{4} being nonzero. If `c`_{4} = 0, then at least one of the coefficients `c`_{1}, `c`_{2}, `c`_{3} is nonzero, and the equality above becomes `c`_{1}`v`_{1}
+ `c`_{2}`v`_{2}
+ `c`_{3}`v`_{3}
= **0**. This means that `v`_{1}, `v`_{2}, `v`_{3} are linearly dependent. The contradiction shows that we must have `c`_{4} ≠ 0. Thus we may divide `c`_{4} and get

`v`_{4} = (-`c`_{1}/`c`_{4})`v`_{1}
+ (-`c`_{2}/`c`_{4})`v`_{2}
+ (-`c`_{3}/`c`_{4})`v`_{3} ∈ `span`{`v`_{1}, `v`_{2}, `v`_{3}},

again contradicting with the assumption.

Therefore we conclude that `v`_{1}, `v`_{2}, `v`_{3}, `v`_{4} must be linearly independent. The proof for general `k` is similar.

Theorem 3 The three operations do not change linear independence.

Proof We only consider special case of the operations on three vectors. The general case is similar.

For the first operation, we will prove

`v`_{1}, `v`_{2}, `v`_{3}
linearly independent ⇔
`v`_{1}, `v`_{2} + `r v`

Assuming left side holds, we have

`c`_{1}`v`_{1} +
`c`_{2}(`v`_{2} + `r v`

⇒ (

⇒

⇒

The whole implication means exactly the right side. Assume the right side holds. Then

`c`_{1}`v`_{1} +
`c`_{2}`v`_{2} +
`c`_{3}`v`_{3}
= 0

⇒
(`c`_{1} - `rc`_{2})`v`_{1} +
`c`_{2}(`v`_{2} + `r v`

⇒

⇒

The whole implication means exactly the left side.

For the second operation, we will prove

`v`_{1}, `v`_{2}, `v`_{3}
linearly independent ⇔
`v`_{2}, **v**_{1}, `v`_{3}
linearly independent.

Assume left side holds. Then

`c`_{1}`v`_{2} +
`c`_{2}`v`_{1} +
`c`_{3}`v`_{3}
= 0

⇒ `c`_{2}`v`_{1} +
`c`_{1}`v`_{2} +
`c`_{3}`v`_{3}
= 0

⇒ `c`_{2} = 0,
`c`_{1} = 0,
`c`_{3} = 0
(`v`_{1}, `v`_{2}, `v`_{3}
linearly independent)

⇒ `c`_{1} = 0,
`c`_{2} = 0,
`c`_{3} = 0.

The whole implication means exactly the right side. The implication in the other direction can be proved similarly.

For the third operation, we will prove that for `d` ≠ 0,

`v`_{1}, `v`_{2}, `v`_{3}
linearly independent ⇔
`d v`

Assume left side holds. Then

`c`_{1}(`d v`

⇒

⇒

The whole implication means exactly the right side. The implication in the other direction can be proved similarly.