### Vector Space

##### 4. Proofs in vector space

Logically, the further development of the theory of vector spaces will depend entirely on the eight properties. In other words, all the proofs will trace their reasons back to the properties. Since writing all such details down is rather tedious and may not contribute much to the understanding of linear algebra, we will only present here a sample of how strict logic works. We will not be concerned too much with such details later on.

We will prove the following simple and "apparently" obvious fact.

cu = 0c = 0 or u = 0.

The key problem here is that the statement has to be justified by using the eight properties (and the usual properties about numbers).

Proof We break the statement into three parts.

(1) 0u = 0. (note that the plain 0 on the left means the number zero, and the bold 0 on the right means the vector zero)

0u
= 0u + 0 (property 3: u + 0 = u)
= 0u + (u + (- u)) (property 4: u + (- u) = 0)
= (0u + u) + (- u) (property 2: (u + v) + w = u + (v + w))
= (0u + 1u) + (- u) (property 8: 1u = u)
= (0 + 1)u + (- u) (property 6: (c + d)u = cu + du)
= 1u + (- u) (property of numbers: 0 + 1 = 1)
= u + (- u) (property 8: 1u = u)
= 0. (property 4: u + (- u) = 0)

We emphasis that the simple property takes five of the eight properties to prove.

(2) c0 = 0.

0
= c0 + (- c0) (property 4: u + (- u) = 0)
= c(0 + 0) + (- c0) (property 3: u + 0 = u)
= (c0 + c0) + (- c0) (property 5: c(u + v) = cu + cv)
= c0 + (c0 + (- c0)) (property 2: (u + v) + w = u + (v + w))
= c0 + 0 (property 4: u + (- u) = 0)
= c0. (property 3: u + 0 = u)

(3) cu = 0c = 0 or u = 0

The statement is equivalent to cu = 0 and c ≠ 0 ⇒ u = 0, which is what we will actually prove.

u
= 1u (property 8: 1u = u)
= (c-1c) u (c ≠ 0)
= c-1(cu) (property 7: (cd)u = c(du))
= c-10 (assumption: cu = 0)
= 0. (2nd part of proof)