Consider the following set of vectors in R6
S={[-9 7 -8 3 0 -5], [1 -7 3 2 -8 -8], [-6 -14 1 9 -23 -29], [11-21 14 1 -16 -11], [8 16 -8 8 10 1], [17 -7 13 -8 8 18]
(a) (2 points) Demonstrate that S is not a basis for R6.
(b) (4 points) Let H = Span S. Find a basis for H and determineits dimension.
(c) (2 points) Determine whether v= [1,1,1,?1,?1,?1] ^T belongsin H or not.
(d) (6 points) Find a basis for R6 consists of the basis vectorsof H found in part (b) and some additional linearly independentvectors.
(e) (4 points)Suppose A is a 6Ă—6 matrix and T(x) =Ax. Show thatT(H), the set of images of vectors of H, is a subspace of R6.
(f) (3 points) Show that dimT(H)?dimH
(g) (4 points) Suppose A is invertible. Show that dimT(H) =dimH.
(h) (5 points)Suppose K is a 4 dimensional subspace of R6. Showthat Hand K must have a nonzero vector in common.Hint: Start withbases for the two subspaces. If H andKonly have the trivial vectorin common, then what’s a basis for the subspace H+K?
