Answers to some Final Review questions.

1) the inhomogeneous solution set is equal to the homogeneous solution
set (ie the null space of the matrix)
plus a constant vector. This constant is any solution to the
inhomogeneous system.

2) basis is the one vector (3,-1,2). dim W = 1.

3) check : lin indep ? (yes). Does B span W? (ie is each vector in W in
span B ? yes since the two vectors comprising W are in span B). So, yes.

4) no since not lin indep. (even though do span W).

5) yes since the coordinate matrix
(1 0 1)
(1 1 0)
(0 1 1)
has all 3 pivots.

6) 8-3 = 5.

7) No since always have 2 or more free vars. So solution set is a constant
plus a null-space

8) If a linear combination of v_1 ... v_n gives the zero vector,
then the only possible set of weights is the trivial set with zero weight
for each.

9) A linear transformation T:V->W is an isomorphism when it is both one-to-one
and onto W. In this case it is also invertible.

10) No free vars and 3 pivots (dim row space) means exactly 3 columns.
In which case, there are b for which not consistent.

11) Eigenvectors from distinct eigenspaces are linearly independent
(proved in class).
So, the matrix has n lin. indep. eigenvectors, so is diagonalizable. 

12) k=0, since otherwise the eigenspace only has dim 1, not enough
to equal the multiplicity 2 of the eigenvalue lambda=3.

13) It is the sum of squares, each of which never negative.

14) The set of all vectors which are orthogonal to every vector in W.

15) In the direction of the eigenvector corresponding to the eigenvalue
of largest magnitude. Because whe you take eigenvalues to the power k,
as k -> infinity, the largest magnitude eigenvalue dominates over the
other weights.

16) The product A^T A must be invertible, which is true if A has 
lin indep columns.