A basis is a collection of vectors which consists of enough vectors to span the space, but few enough vectors that they remain linearly independent. It is the same as a minimal spanning set.
- it spans and
- it is linearly independent.
Fundamental Properties
- B1
- A non-zero set is a basis for iff it is a minimal spanning set for .
- B2
- Every non-zero vector space admits a basis.
- B3
- (finite case) If and are two bases for , then .
Proof of B1 Suppose is a basis for . Then it certainly spans . If it were not a minimal spanning set, it would mean there is a vector which is in the span of , which in turn would mean that could be written as a linear combination of vectors in . Thus the original set would not be vectorwise independent, which by the above exercise is a contradiction. Hence it must be minimal.
In the other direction, suppose it is a minimal spanning set. If it were not linearly independent, then by the same exercise it would not be vectorwise independent, so there would exist some vector which could be written as a linear combination of the other vectors in . Then , contradicting the fact that is minimal. Hence not only spans but must also be linearly independent.
Proof of B2 Given the first property, there are two different ways one can go about constructing a basis. We present both. Note: they do require familiarity with the operations of union and intersection for sets.
Method 1 Let . We can partially order by defining if . Every chain is clearly bounded below as a set by the empty set (even though the bound is not an element of ). Choose a totally ordered subset of ; by Zorn’s Lemma it contains a maximal element . Maximality with respect to this total ordering means is a minimal spanning set, hence a basis by B1.
Method 2 Let , and partially order by iff . Again, every chain has an upper bound (which again will not be in ). Choose a totally ordered subset of and let be a maximal element, implying it is a maximal linearly independent subset of , hence a basis.
The third property will require a bit more work; its proof will be deferred until later. It is stated above in the case that both and have a finite set of vectors, but it is true more generally. For an arbitrary set , let denote the number of elements of (this number, which can be infinite, is referred to as the cardinality of ). In full generality, property B3 states
B3 (general case) If is a non-zero vector space, and are two bases for , then .
Property B3 indicates that the dimension of a (non-zero) vector space doesn’t depend on the choice of basis for ; it only depends on . The dimension of a vector space is the single most important numerical invariant one can attach to that space.
Suppose is a subspace of , and we wish to find a basis. The first step is to determine a spanning set in the event it is not already given. Then forming the matrix we observe that (by construction) , the column space of . We can then use the method described in the section Spanning sets, row spaces, and column spaces to find a minimal spanning set from among the original vectors in . Thus we have an efficient algorithm for finding a basis for any vector space which is a subspace of (for some ).
An important consequence of the above is
- Proof
- We can assume, without loss of generality, that both and are non-zero vector spaces. Let be a basis for . As the set is linearly independent in , it is also linearly independent in , and so extends to a basis for where . Hence . And since the dimension is independent of the choice of basis, precisely when .
We will occasionally write as when we want to emphasize the fact that the vector lies in . Given the equivalence between bases and minimal spanning sets, we have
The basis is referred to as the standard basis for .
The above method works for finding a basis for a subspace of . In the event the finite-dimensional vector space is not naturally a subspace of for any this method still applies, but we need to do something else first. That is described in the next section.