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Linear Combinations, Spanning, Independence, , and

Learning goal: to recover these concepts in general vector spaces.

Let’s recover some central ideas from Rn and its subspaces for general vector spaces.

Definition: given a V, a subspace is any subset of V which is a vector space in its own right. (Note: subspaces must contain a zero vector, so must be non-empty.)

Lemma: to prove a nonempty subset is a subspace, you only have to show under and multiplication. The rest of the rules are inherited from the parent space.

Definition: given a of vectors in a vector space V, a linear combination of them is any finite sum of the form a1v1 + a2v2 + … + anvn for vectors v1, …,vn in the set.

We must have finite sums only because in general vector spaces we don’t have any idea of convergence. Worse, depending on how you define distance, some infinite series may converge in one definition and not in another. And worse, some sums may converge to things not in the vector space at all. For example, the infinite sum 1 + x + x2 / 2 + x3 / 6 + … of terms in the space of P adds to ex which is not in P.

Definition: given a subset S of vectors in vector space V, its span is the set of all linear combinations of vectors in S.

Lemma: the space of a set is always a subspace. (Note: the span of an empty set contains the empty sum, which adds to the zero vector.) For spans are always nonempty (contain at least the empty sum) and a easily seen to be closed under addition and .

So far everything has worked the same as it did for Rn. It continues to do so for a while:

Definition: a set S of vectors in V is linearly independent if only the trivial linear combination produces 0.

Hence, if you can find a non-trivial linear combination that adds to zero, you have a linearly dependent set.

Lemma: if a set is dependent, at least one of the vectors in it can be written as a linear combination of the others. For in the nontrivial combination that adds to zero, one of the scalar multiples is nonzero, so put that term on the other side of the equation, divide by the nonzero scalar, and you have solved for that vactor in terms of the others.

Definition: a basis is a linearly independent spanning set.

For example, all the stuff we have done before is just special cases. A more important example is that 1, x, x2, x3, … is a basis for P. For every nontrivial combination of these is some nonzero , and every polynomial is a finite combination of these. On the other hand, the space of sequences of real numbers is much harder to find a basis for. You might try (1, 0, 0, …), (0, 1, 0, 0, …), (0, 0, 1, 0, …), etc., but you will find that this does not span—not everything is a finite linear combination of these. It is equivalent to the axiom of choice as to whether this vector space even has a basis!

Lemma: given a basis for vector space V, every vector in V has a unique expression a a linear combination of basis vectors. For by spanning, it has at least one, and if it had at least two then subtracting one from the other gives a nontrivial combination of basis vectors that adds to zero, contradicting linear dependence.

Definition: a vector space is finite dimensional if it has any finite spanning set.

In view of the previous lemma, this means that it has at least one finite basis. For take the finite spanning set. If not already independent, some one of its elements may be written as a linear combination of the others. Toss it out. The remaining set of vectors still spans, because any use of the discarded vector can be replaced with a (finite) linear combination of the other vectors in the set. Keep this up, until eventually you end with a linearly independent set, or you have thrown out all the vectors (in which case, your vector space was the trivial space 0 consisting of only the zero vector, and whose basis is empty!).

Theorem: given a finite basis B for a (necessarily) finite dimensional vector space, any set of vector in V with more elements than B is linearly dependent.

Proof: (this is pretty much the same as it was before.) Let B have n vectors b1, b2, …, bn. Consider any n + 1 vectors v1, v2, …, vn, vn+1. Each of these may be written as a linear combination of the b’s: vj = a1jb1 + a2jb2 + … + anjbn. Now let M be the of aij., Note that since there are n + 1 v’s, this matrix is n × n + 1, so has more columns than rows. Let x = (x1, x2, …, xn+1) be any column vector of n + 1 scalars. Note that we can take the product Mx, using regularly defined , even if the scalar is something weird. Note that this gives an output vector of n entries. Call this vector y. If we take a linear combination of the b’s with weights given by y, we get the exact same vector as if we had taken a linear combination of the v’s with weights given by x. In essence, the matrix M expresses the v’s in terms of the b’s so you can turn a combination of v’s into a combination of b’s simply by multiplying the coefficients by the matrix M. Now since M has more columns than rows, standard row reduction will produce free variables and nontrivial null vectors. That is, we can find a vector z such that Mz = 0. But this means that z1v1 + … + zn+1vn+1 = 0b1 + 0b2 + … + 0bn = 0. Thus, there is a nontrivial linear combination of v’s that adds to zero, so this set is dependent.

Theorem: all bases for a finite dimensional vector space have the same number of vectors.

Proof: given one finite spanning set, we can find one finite basis. Then no larger set can be independent, and if any smaller set were a basis, then our first basis would be larger than a basis, and so dependent, which is a contradiction since we said it was a basis!

Corollary: in a finite dimensional vector space, any spanning set can be reduced to a basis and any linearly independent set can be extended to a basis.

Proof: Start with B = empty set, and repeat: if B isn’t yet a basis, choose any vector in the spanning set not in the span of B and add it to B. B remains independent because it was before, and since the new vector isn’t in the old span, no combo including it can be zero either. Since by finite dimensionality B can’t grow forever, this process will eventually stop with a basis. Going the other way, choose any spanning set, and start with B the given independent set, and repeat the process above.

Definition: the dimension of a finite dimensional vector space is the number of vectors in any basis.

And now the key theorem on which all of the rest of the theory builds:

Theorem: a linear function is completely determined by its action on a basis (note that this applies even if the spaces involved as domain and range aren’t finite dimensional!)

Proof: Let f be our function and x a vector. Let it be expressed as a1b1 + a2b2 + … + anbn. Then by f(x) = f(a1b1 + a2b2 + … + anbn) = a1f(b1) + … + anf(bn).

Notice how once we had a basis then good old column vectors and matrices made a swift reappearance? We shall explore that next.

Reading: 7.1 7.1: 5, 6, 13, 14, 15, 16, 18, 19