Linear Algebra

118 Chapter Two. Vector Spaces2.10 Corollary No linearly independent set can have a size greater than thedimension of the enclosing space.Proof The proof of Theorem 2.5 never uses that D spans the space, only thatit is linearly independent.QED2.11 Example Recall the subspace diagram from the prior section showing thesubspaces of R 3 . Each subspace shown is described with a minimal spanning set,for which we now have the term ‘basis’. The whole space has a basis with threemembers, the plane subspaces have bases with two members, the line subspaceshave bases with one member, and the trivial subspace has a basis with zeromembers. When we saw that diagram we could not show that these are R 3 ’sonly subspaces. We can show it now. The prior corollary proves that the onlysubspaces of R 3 are either three-, two-, one-, or zero-dimensional. Therefore, thediagram indicates all of the subspaces. There are no subspaces somehow, say,between lines and planes.2.12 Corollary Any linearly independent set can be expanded to make a basis.Proof If a linearly independent set is not already a basis then it must not spanthe space. Adding to the set a vector that is not in the span will preserve linearindependence. Keep adding until the resulting set does span the space, whichthe prior corollary shows will happen after only a finite number of steps. QED2.13 Corollary Any spanning set can be shrunk to a basis.Proof Call the spanning set S. If S is empty then it is already a basis (thespace must be a trivial space). If S = {⃗0} then it can be shrunk to the emptybasis, thereby making it linearly independent, without changing its span.Otherwise, S contains a vector ⃗s 1 with ⃗s 1 ≠ ⃗0 and we can form a basisB 1 = 〈⃗s 1 〉. If [B 1 ] = [S] then we are done. If not then there is a ⃗s 2 ∈ [S] suchthat ⃗s 2 ∉ [B 1 ]. Let B 2 = 〈⃗s 1 , ⃗s 2 〉; if [B 2 ] = [S] then we are done.We can repeat this process until the spans are equal, which must happen inat most finitely many steps.QED2.14 Corollary In an n-dimensional space, a set composed of n vectors is linearlyindependent if and only if it spans the space.Proof First we will show that a subset with n vectors is linearly independent ifand only if it is a basis. The ‘if’ is trivially true — bases are linearly independent.‘Only if’ holds because a linearly independent set can be expanded to a basis,but a basis has n elements, so this expansion is actually the set that we beganwith.To finish, we will show that any subset with n vectors spans the space if andonly if it is a basis. Again, ‘if’ is trivial. ‘Only if’ holds because any spanningset can be shrunk to a basis, but a basis has n elements and so this shrunkenset is just the one we started with.QED