Consider the set of vectors S={(01),(10),(23),(47)}.
What is the span of S? Span(S)=R2.
Is S a basis? No, because it is a linearly dependent set – the vectors are in
R2 but there are four of them, and 4>2.
Is {(01),(23)} (S with two vectors removed) a basis? Yes, because it is linearly independent and spans the entire space of R2.
Theorem.
- Let S={v1,...,vp} be a set of vectors in V and let H=Span(S). If one of the vectors in S, say vk, is a linear combination of the remaining vectors in S, then the set formed by removing vk from S also spans H.
- If H={0}, then some subset of S is a basis for H.
(Note that the basis produced by this method will be different depending on which vectors you remove.)
For a matrix A, a basis for NulA is formed from the solution set of Ax=0, and a basis for ColA is formed from pivot columns of A.
Ex.
ABasis for ColAx1x3⎝⎜⎜⎜⎜⎛x1x2x3x4x5⎠⎟⎟⎟⎟⎞Basis for NulA=⎝⎛−3126−2−4−125138−7−1−4⎠⎞→⎝⎛100−200010−1203−20⎠⎞={⎝⎛100⎠⎞,⎝⎛010⎠⎞}x2, x4, x5 are free=2x2+x4−3x5=−2x4+2x5=x2⎝⎜⎜⎜⎜⎛21000⎠⎟⎟⎟⎟⎞+x4⎝⎜⎜⎜⎜⎛10−210⎠⎟⎟⎟⎟⎞+x5⎝⎜⎜⎜⎜⎛−30201⎠⎟⎟⎟⎟⎞={⎝⎜⎜⎜⎜⎛21000⎠⎟⎟⎟⎟⎞,⎝⎜⎜⎜⎜⎛10−210⎠⎟⎟⎟⎟⎞,⎝⎜⎜⎜⎜⎛−30201⎠⎟⎟⎟⎟⎞}
Theorem. Let B={b1,...,bn} be a basis for the vector space V. Given any x in V, there are unique c1,...,cn such that x=c1b1+...+cnbn.
This means that once a basis is found for a vector space, we can uniquely identify every vector in that space by the list of scalars c1,...,cn associated with that vector. We call these the coordinates of x in B, denoted by [x]B.
Ex. Let B={(13),(40)}, and let [x]B=(711).
x=7(13)+11(40)=(5121)