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 ,  , , , , , , 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 , there will be 4 pivots.

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

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

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

 Created by Mathematica  (May 15, 2006)