Originally Posted by

**Deveno** yes, dimension is the number of vectors is ANY basis.

the null space of L IS V. V itself is a subspace of R^{5}. we are given by the problem itself that ker(L) = V. we're supposed to find such an L.

ask yourself, is (1,0,0,0,0) in V? can we find a,b,c,d so that (1,0,0,0,0) = (a,a+b,a+b+c,a+b+c+d,a+b+c+d)? well if we can, then a = 1. but since a+b = 0, b = -1. since a+b+c = 0, and a+b = 0, c = 0. finally, d = 0 (similar reasoning).

that is:

(1,0,0,0,0) = (1,1,1,1,1) - (0,1,1,1,1).

if we make a matrix for L (in the standard basis), then since (1,0,0,0,0) is in V, the first column for L (which is L(1,0,0,0,0)) must be all 0's.

convince yourself that (0,1,0,0,0), (0,0,1,0,0) are both in V, and that (0,0,0,1,0) and (0,0,0,1,1) are NOT (hint: look at the last two coordinates).

you are mis-reading the question. the question says:

FIND a linear transformation L WITH kernel V. V is a 4-dimensional subspace of R^{5}. so L is going to take MOST of R^{5} to the 0-vector of R^{4}.

as you can see by the head-start i gave you, the matrix for L will take any vector (x,y,z,0,0) to 0. what do we need to happen, for L(0,0,0,1,1) = (0,0,0,0)? in particular if:

$\displaystyle L = \begin{bmatrix}0&0&0&l_1&?\\0&0&0&l_2&?\\0&0&0&l_3 &?\\0&0&0&l_4&? \end{bmatrix}$

what does that last column need to be?