Linear Algebra

110 Chapter Two. Vector SpacesThe verification is easy.1.5 Definition For any R n ⎛ ⎞ ⎛ ⎞ ⎛ ⎞1 0 00E n = 〈⎜ ⎟⎝.⎠ , 1⎜.⎟⎝.⎠ , . . . , 0⎜ ⎟⎝.⎠ 〉0 0 1is the standard (or natural) basis. We denote these vectors ⃗e 1 , . . . ,⃗e n .Calculus books refer to R 2 ’s standard basis vectors ⃗ı and ⃗j instead of ⃗e 1 and ⃗e 2 ,and they refer to R 3 ’s standard basis vectors ⃗ı, ⃗j, and ⃗k instead of ⃗e 1 , ⃗e 2 , and⃗e 3 . Note that ⃗e 1 means something different in a discussion of R 3 than it meansin a discussion of R 2 .1.6 Example Consider the space {a · cos θ + b · sin θ ∣ ∣ a, b ∈ R} of functions ofthe real variable θ. This is a natural basis.〈1 · cos θ + 0 · sin θ, 0 · cos θ + 1 · sin θ〉 = 〈cos θ, sin θ〉Another, more generic, basis is 〈cos θ − sin θ, 2 cos θ + 3 sin θ〉. Verification thatthese two are bases is Exercise 22.1.7 Example A natural basis for the vector space of cubic polynomials P 3 is〈1, x, x 2 , x 3 〉. Two other bases for this space are 〈x 3 , 3x 2 , 6x, 6〉 and 〈1, 1 + x, 1 +x + x 2 , 1 + x + x 2 + x 3 〉. Checking that these are linearly independent and spanthe space is easy.1.8 Example The trivial space {⃗0} has only one basis, the empty one 〈〉.1.9 Example The space of finite-degree polynomials has a basis with infinitelymany elements 〈1, x, x 2 , . . .〉.1.10 Example We have seen bases before. In the first chapter we described thesolution set of homogeneous systems such as this onex + y − w = 0z + w = 0by parametrizing.⎛ ⎞ ⎛ ⎞−1 11{ ⎜ ⎟⎝ 0⎠ y + 0⎜ ⎟⎝−1⎠ w ∣ y, w ∈ R}0 1Thus the vector space of solutions is the span of a two-element set. This twovectorset is also linearly independent; that is easy to check. Therefore thesolution set is a subspace of R 4 with a basis comprised of the above two elements.