Solutions

1) Put the vectors into a matrix and get the reduced row echelon form. If you put them in as columns, then---since elementary row operations don't change dependency among the columns---the columns of the original matrix with pivots are independent. If you put them in as rows, then---since elementary row operations don't change the span of the rows---the rows with pivots (in the rref) are a basis for the row space. In either case, the number of pivots gives you the dimension. Here, the answer is 3.

2) We form the matrix with columns Overscript[w_1, _],  Overscript[w_2, _], Overscript[w_3, _], Overscript[e_1, _], Overscript[e_2, _], Overscript[e_3, _], Overscript[e_4, _], and get the rref. The columns of this original matrix that have pivots (we use rref to find the pivots) are independent, and since the span of those columns is ^4, there will be 4 pivots.

RowReduce[({{1, 1, 2, 1, 0, 0, 0}, {2, 2, 2, 0, 1, 0, 0}, {2, 2, 2, 0, 0, 1, 0}, {1, 2, 1, 0, 0, 0, 1}})]//MatrixForm

( {{1, 0, 0, -1, 0, 3/2, -1}, {0, 1, 0, 0, 0, -1/2, 1}, {0, 0, 1, 1, 0, -1/2, 0}, {0, 0, 0, 0, 1, -1, 0}} )

The first, second, third, and fifth columns have pivots, so they are linearly independent. Since elementary row operations don't change dependency among the columns, we see that

Overscript[w_1, _] = ({{1}, {2}, {2}, {1}}), Overscript[w_2, _] = ({{1}, {2}, {2}, {2}}), Ov ... pt[w_3, _] = ({{2}, {2}, {2}, {1}}), Overscript[u, _] = ( {{0}, {1}, {0}, {0}} )

are independent. Four independent vectors in a space with dimension 4 (in this case ^4) are a basis.

3) M=P D P^(-1), where D is the diagonal matrix having the eigenvalues, and P is the matrix whose columns are the eigenvectors.

M = ({{1, 1, 2, 0}, {2, 2, 2, 1}, {2, 2, 2, 0}, {1, 2, 1, 0}}) . ({{2, 0, 0, 0}, {0, -3, 0, ... {0, 0, 0, 0}}) . Inverse[({{1, 1, 2, 0}, {2, 2, 2, 1}, {2, 2, 2, 0}, {1, 2, 1, 0}})]//MatrixForm

( {{6, 0, 1/2, -5}, {4, 0, 5, -10}, {4, 0, 5, -10}, {2, 0, 4, -8}} )

Created by Mathematica  (May 15, 2006) Valid XHTML 1.1!