in a vector space, if and only if or is the zero vector
Proof. Let us denote by and by the zero and unit elements in respectively. Similarly, we denote by the zero vector in . Suppose . Then, by axiom 8 (http://planetmath.org/VectorSpace), we have that
for all . By axiom 6 (http://planetmath.org/VectorSpace), there is an element in that cancels . Adding this element to both yields . Next, suppose that . We claim that for all . This follows from the previous claim if , so let us assume that . Then exists, and axiom 7 (http://planetmath.org/VectorSpace) implies that
holds for all . Then using axiom 3 (http://planetmath.org/VectorSpace), we have that
for all . Thus satisfies the axiom for the zero vector, and for all .
For the other direction, suppose and . Then, using axiom 3 (http://planetmath.org/VectorSpace), we have that
On the other hand, suppose and . If , then the above calculation for is again valid whence
which is a contradiction, so .
This result with proof can be found in , page 6.
- 1 W. Greub, Linear Algebra, Springer-Verlag, Fourth edition, 1975.
|Title||in a vector space, if and only if or is the zero vector|
|Date of creation||2013-03-22 13:37:34|
|Last modified on||2013-03-22 13:37:34|
|Last modified by||aoh45 (5079)|