Discover Linear Algebra

Proposition 18.5.5 states that every finite spanning set can be reduced to a linearly independent spanning set. But that's exactly what a basis is — a linearly independent spanning set.

1. 1. Suppose \(\basisfont{B}\) is a basis for a vector space. That is, suppose \(\basisfont{B}\) is a linearly independent spanning set. If we remove even one vector \(\uvec{v}\) from \(\basisfont{B}\text{,}\) the remaining vectors cannot still form a spanning set for the space. Because if they could, then \(\uvec{v}\text{,}\) being a vector in that vector space, could be expressed as a linear combination of some number of the remaining vectors in \(\basisfont{B}\text{.}\) In other words, some vector in \(\basisfont{B}\) would be expressible as a linear combination of others, which would violate the assumption that \(\basisfont{B}\) is linearly independent.
   2. Suppose \(S\) is a spanning set for a vector space, and that \(S\) contains a finite number of vectors. Further suppose that \(S\) is minimal in the sense that no proper subcollection of \(S\) also forms a spanning set for the space. We would like to prove that this forces \(S\) to be a basis. We already assuming one-half of the definition of basis, so we only need to consider the other half: must \(S\) also be linearly independent? If it were linearly dependent instead, then it could be reduced to subcollection that forms a linearly independent spanning set (Proposition 18.5.5). But \(S\) doesn't have any subcollections that form spanning sets for the vector space, let alone any linearly independent ones. So \(S\) cannot be linearly dependent, forcing it to be linearly independent, as required.
2. 1. Suppose \(\basisfont{B}\) is a basis for a vector space, and that \(\basisfont{B}\) contains a finite number of vectors. Then by definition of basis, \(\basisfont{B}\) is a spanning set for the vector space, and so any collection of vectors that contains more vectors than \(\basisfont{B}\) must be linearly dependent (Lemma 18.5.7). In particular, if some collection of vectors contains \(\basisfont{B}\) as a proper subcollection, then that larger collection must be linearly dependent.
   2. Suppose \(S\) is a linearly independent collection of vectors in a vector space, and that \(S\) is maximal in the sense that no other linearly independent collection of vectors can contain \(S\) as a proper subcollection. We would like to prove that this forces \(S\) to be a basis for the space. We already assuming one-half of the definition of basis, so we only need to consider the other half: must \(S\) also be a spanning set for the space? If it were not a spanning set, then \(\Span S\) would merely be a proper subspace, and there would be other vectors in the full vector space that are not in that subspace. Let \(\uvec{v}\) be one such vector. Then Proposition 18.5.6 tells us that the collection of vectors containing both \(\uvec{v}\) and every vector in \(S\) must be linearly independent. But this is not possible, since this new, “larger” linearly independent collection would contain \(S\) as a proper subcollection, and we have assumed that \(S\) is a maximal linearly independent set of vectors. So \(S\) must also be a spanning set for the vector space, as required.

We will prove that two different linear combination expressions involving basis vectors must compute to two different vectors, which will imply that one single vector in the vector space cannot have two different expressions as linear combinations of basis vectors. So suppose we have two different linear combination expressions involving basis vectors. Let \(\uvec{v}_1,\uvec{v}_2,\dotsc,\uvec{v}_m\) be a complete list of the basis vectors involved in both expressions. By attaching a zero coefficient to missing vectors, we can assume that both linear combination expressions involve all of these basis vectors. Let \(a_1,a_2,\dotsc,a_m\) represent the corresponding coefficients in one of these linear combination expressions, and let \(b_1,b_2,\dotsc,b_m\) represent the corresponding coefficients in the other. Note that we must have at least one instance of \(a_j\neq b_j\) in these collections of coefficients, because we have assumed that these linear combination expressions are different. Now, these two linear combination expressions compute to two vectors in the vector space,

We would like to prove that \(\uvec{v}\neq\uvec{w}\text{.}\) Equivalently, we would like to prove that \(\uvec{v}-\uvec{w}\neq\zerovec\text{.}\) By collecting like terms, this difference vector can also be expressed as a linear combination as

Since we have at least one instance of \(a_j\neq b_j\text{,}\) we have at least one nonzero coefficient in the expression above, and so the linear combination above is nontrivial. And our basis vectors are linearly independent, so a nontrivial linear combination of basis vectors cannot equal the zero vector. Therefore, \(\uvec{v}-\uvec{w}\neq\zerovec\) as desired.