Advanced Calculus fi..

Advanced Calculus, Fifth EditionBy a subspuce of V" we mean a collection W of vectors of V" such that(i) W contains 0, and(ii) if u, v are in W, then so is each linear combination nu + bv.Let W be a subspace and let W contain vectors besides 0. We then seek setsof linearly independent vectors in W. Each such set contains at most n vectors butmay in fact always contain fewer than n. We let k be the largest integer such that Wcontains k linearly independent vectors, and call k the dimension of W. When Wcontains 0 alone, we assign the dimension 0 to W.Whcn k = 11, W necessarily coincides with Vn. For then W contains linearlyindependent vectors ul , . . . , u,,, and by repeatedly applying (ii) we conclude that Wcontains each linear combination of u,. . . . , u,,. Since ul, . . . . u, form a basis forvn, W = V".For k = I . W consists of all scalar multiples of a single vector u,, that is, allw = slul. If, say. n = 3. then the corresponding points P such that 3 = wlultrace out a line L through 0 in 3-dimensional space (Fig. 1.18a).If k = 2, then W contains two linearly independent vectors ul . u7 and, by (ii),contains all the linear combinationsThese linear combinations exhaust W. since any vector not expressible in this formwould have to be linearly independent of ul , uz (Rule (c) of Section 1.14). If againn = 3, then the corresponding points P such that 3 = w fill a plane in space, asin Fig. 1.1 8b.In general, we see that if W has dimension k, then W consists of all linearcombinationswhere ul, . . . . ur are linearly independent vectors of W. We term ul. . . . , uk a basisof W. There are many choices of basis for a given W, but all havc the same numberof vectors (Problem 4, which follows below).Geometrically, the subspaces of V" correspond to the point 0 (k = O), and lines,planes. and "hyperplanes" through 0, in E", and to E" itself (k = n).Figure 1.18suhspace of V '.(a) One-dimensional subspace of V '. (b) Two-dimensional