### Chapter Two, Sections 2.3 - 2.8

You need to know the definitions and theorems, with these exceptions (about isomorphisms): Definition 2.13, Theorems 2.13, 2.14, 2.15, Corollary 2.6.

Procedure 1. (page 116). To test whether the vectors v1, v2, ..., vk are linearly independent or linearly dependent:

• Solve the equation x1 v1 + x2 v2 + ... + xk vk = 0.
Note: the vectors end up as columns in a matrix.
If the only solution is all zeros, then the vectors are linearly independent.
If there is a nonzero solution, then the vectors are linearly dependent.

Procedure 2. (page 114). To check that the vectors v1, v2, ..., vk span the subspace W:

• Show that for every vector b in W there is a soluton to x1v1 + x2v2 + ... + xkvk = b.

Procedure 3. (page 131). To find a basis for the subspace span { v1, v2, ..., vk } by deleting vectors:

1. Construct the matrix whose columns are the coordinate vectors for the v's
2. Row reduce
3. Keep the vectors whose column contains a leading 1
The advantage of this procedure is that the answer consists of some of the vectors in the original set.

Procedure 4. (page 156). To find the transition matrix PS<-T:

1. Construct the matrix A whose columns are the coordinate vectors for the basis S
2. Construct the matrix B whose columns are the coordinate vectors for the basis T
3. Row reduce the matrix [ A | B ] to the form [ I | P ]
4. The matrix P is the transition matrix
The purpose of the procedure is to allow a change of coordinates [ v ]S = PS<-T [ v ]T .

Procedure 5. (page 143). To find a basis for the solution space of the system A x = 0 :

1. Row reduce A
2. Identify the independent variables in the solution
3. In turn, let one of these variables be 1, and all others be 0
4. The corresponding solution vectors form a basis
```

```

Procedure 6. To find a simplified basis for the subspace span { v1, v2, ..., vk } :

1. Construct the matrix whose rows are the coordinate vectors for the v's
2. Row reduce
3. The nonzero rows form a basis
The advantage of this procedure is that the vectors in the basis have lots of zeros, so they are in a useful form.

### Chapter Three, Sections 3.3, 3.4

You need to know the definitions and theorems, with these exceptions: Theorem 3.2; Theorem 3.8.

Procedure 1. (the Gram-Schmidt Process, page 215). Let find an orthonormal basis for an m-dimensional subspace of an inner product space V:
Given a basis S = { u1, u2, . . . um } for W, first find an orthogonal basis T* = { v1, v2, . . . vm } as follows:

1. v1 = u1
2. v2 = u2 - [(u2 , v1) / (v1 , v1) ] v1
3. v3 = u3 - [(u3 , v1) / (v1 , v1) ] v1 - [(u3 , v2) / (v2 , v2) ] v2
4. etc

Finally, T = { w1, w2, . . . wm } where wi is found by dividing vi its length