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Let Uu = (2, -1) and v = (2, 1). Show that the vector b = (h, k) is in [u, v] for all hand k. 11. Determine the subs subspace pace of V3 spanned by each of the following sets: (a) {(\, I, 1),(0, 1,2),(1,0,-1)} (b) {(2,1,0),(2,0,-I)} (c) {(I, 2,3), (1, 3, 5)}, (d) {(I, 2, 3), (1, 3, 5), (1,2, 4)}. 4 BASES AND DIMENSION In this section we shall assign a dimension to certain vector spaces, and dimension is based on the concept of a basis for the space. But basis is based on the concept of a linearly independent set.

Then W WII + W2 is finite-dimensional and dim(Jt; + W2) = dimJt; +dimW2 -dim(Jt; flW2) 52 Introductory Linear Algebra Propf: Let dim(W; nW2 )= k, dimW; = m, and dimW2 = n. =:> k 5, m and k 5, n . ';;;;W; =:> SI can be extended to form a basis for W WI. I• • Let S2 = {u p u 2 , ••• ,Uk , VI' V 2 , ••• 'Vm _ k } is a basis for W WI. ·, , ••• , w wn _k } is a basis for W2 • S4 = {up ... , Uk' uk ' vp Vp ... ••• ,, V vm_ k ' Wp wp ... , W,,-k} Our aim is to show that S4 is a basis for W WII + W2. First, we shall show that S4 is LI.

Thus W is a vector space, and hence a sllbspace subspace of V. 1, it is evident that W is a subspace of V if and only if (I) W is non-empty, that is, zero of V is the zero of W, (il) W is closed under vector addition, and (iil) (iii) W is closed under multiplication of vectors by scalars. The following theorem shows that we can simplify the properties of subspace still further. 2. A non-empty set W of a vector space V over the field F is a subspace of V if and only if alt all + v E W for all 1I, It, v in Wand for each a a in F.