d. Find a basis for the column space of of A. If necessary, enter a1 for a, etc., or enter coordinate vectors of the form <1,2,3> or <1,2,3,4,5>. Enter your answer as a comma separated list of vectors. A basis for the column space of A is { } e. The dimension of the null space of A is , and the null space of A is a subspace of f. If x1 = (5, 2, –1, 1,0), then Ax1 = .Is x1 in the null space of A? choose v g. If x2 = (-4, –2, –1,0, 1), then Ax2 = Is x2 in the null space of A? choose v h. If xg = 3x2 – 4x1 = , then Axg = Is xg in the null space of A? choose v

There are four vectors (1,a,b,c), (1,1,1,1), (1,2,3,1), (3,4,5,3). If the vector space (subspace) consisting of the linear combination of these vectors is T, find the condition that the number of vectors constituting the basis of T will be 3.

Suppose that V is a that has subspaces of U and W. Furthermore, suppose that {u¹, u2} is a basis for U, that {1, 2} is a basis for W and that the only vector that U and W have in common is the zero vector 0. Show that {u¹, u², w¹, w²}. After you get done, note where you used the fact that the only vector common to both U and W is 0. (Without this condition the vectors {u¹, u², w², w2} don't need to be linearly independent, so if you didn't use this condition you definitely made a mistake somewhere.)

Please provide a solution using the answer key I have attached below.