1) Let S = {(-2,0,1),(5,-2,1),(11,7,-5),(-1,4,-2),(2,-3,1),}. Show that Span(S)=R^3, but S is not a basis for R^3. Find a subset of S that is a basis for R^3 and prove that your subset is actually a basis for R^3.

2) Let F(−∞,∞) represent the set of all real valued functions that are defined on (−∞,∞). Use the subspace theorem to show that the set of all differentiable functions on (−∞,∞) that satisfy f'(x) + 2f(x) = 0 is a subspace of F(−∞,∞).

3) Use the subspace theorem to show that {(0,t,s):t,s ∈ R} is a subspace of R^3 and graph the space in R^3. Also, state the dimension of the subspace.

4) suppose that {v1, v2, v3} is linearly independent subset of a vector space V with dim(V) = 4 and that v4 is not in span([v1, v2, v3}). Prove that {v1, v2, v3, v4} is a basis for V.

5) Show that ð‘¥!, ð‘¥! − ð‘¥!, ð‘¥! − ð‘¥! + ð‘¥!, ð‘¥! − ð‘¥! + ð‘¥! − ð‘¥, ð‘¥! − 1 is a basis for ð‘ƒ4.

6) Let A=[5 -3 1 -1 , 4 -2 2 2 , 1 2 8 18] which equal a matrix. Then find a basis for the null space of Na and dim(Na).

7) Let A1 = [1 0, 1 0], A2=[1 1, 0 0], A3=[1 0, 0 1], A4=[0 0, 1 0]

a) Show that {A1, A2, A3, A4} is a basis for M22.

b) find the coordinates of A = [6 2, 5 3] with respect to the basis in part (a)

8) Suppose that V is a vector space and dim(V) = 4 and W is a subspace of V. Prove directly without simple quoting a theorem that W must have finite dimension.

9) Let A= [1 2 1 0 -1 , 0 5 2 -1 -5 , 2 9 5 -1 -9] which is a matrix. Then find a basis for null space of A.

10)

**Part 1:** Show that if **ð‘‰! ð‘Žð‘›ð‘‘ ð‘‰! ð‘Žð‘Ÿð‘’ ð‘ ð‘¢ð‘ð‘ ð‘ð‘Žð‘ð‘’ð‘ ð‘œð‘“ ð‘Ž ð‘£ð‘’ð‘ð‘¡ð‘œð‘Ÿ ð‘‰, ð‘¡â„Žð‘’ð‘› ð‘‰! ∩ ð‘‰! ð‘–ð‘ ð‘Ž ð‘ ð‘¢ð‘ð‘ ð‘ð‘Žð‘ð‘’ ð‘œð‘“ ð‘‰.**

**Part 2:** Suppose that V1 is the subspace of R^3 given by V1={(2t-s, 3t, t+2s):t,s ∈ R} and V2 is the subspace of R^3 given by V2 = {(s,t,t):t,s ∈ R}. Then find a basis for V1 ∩ V2 and dim(V1 ∩ V2).

note: for this problem, you can assume that V1, V2 are subspace of R^3 and dont need to prove this fact.

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Subject | Mathematics |

Due By (Pacific Time) | 04/20/2016 11:00 am |

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