Let Z be the set of all vectors fo the form \(\displaystyle \begin{bmatrix} 2a - 3b - e \\ 0 \\ a + c +2d \\ -3b + c + d + e \end{bmatrix}\), where \(\displaystyle a, b, c, d, e \ \epsilon \ R\). Z is a subspace of \(\displaystyle R^4\).

a) Find a set that spans Z.

b) Is the set from part (a) a basis for Z? Please explain.

c) Does Z have any two dimensional subspaces? If so give an example of such a subspace. If not, please explain why not.

Attempted Answers:

For part (a) I got: \(\displaystyle \{(2, 0, 1, 0), (-3, 0, 0, -3), (0, 0, 1, 1) \}

\).This is the column space of W. Correct?

Part (b) seems like Yes but I don't know the reason. Part (c) I have no clue.

For part (a), I would recognize that this vector is any combination of the vectors (2,0,1,0),(-3,0,0,-3),(0,0,1,1),(0,0,2,1),(-1,0,0,1). This is the easiest way to determine a set that spans Z. To be sure that your (a) is correct, you must recognize that the vectors (0,0,2,1) and (-1,0,0,1) are linear combinations of the other three, which I think you do.

For part (b), using the above set, we would need to verify that the five vectors are linearly independent (and it is fairly easy to see upon inspection that they are not).

Notice that

\(\displaystyle \frac{1}{2}(2,0,1,0) + \frac{1}{3}(-3,0,0,-3) + \frac{3}{2}(0,0,1,1) = (0,0,2,1)\)

and

\(\displaystyle \frac{-2}{3}(2,0,1,0) - \frac{1}{9}(-3,0,0,-3) + \frac{2}{3}(0,0,1,1) = (-1,0,0,1)\)

Therefore, since these vectors are linear combinations of the other three, they are linearly dependent. Removing the two vectors (-1,0,0,1) and (0,0,2,1) yields a list of three linearly independent vectors, {(2,0,1,0),(0,0,1,1),(-3,0,0,-3)} which span Z. Thus, the vectors form a basis.