1. Class 26: Question 1 1.An orthogonal basis for A 2.An orthogonal basis for the column space of A 3.An orthogonal basis for the row space of A 4.An orthogonal basis for the null space of A

2. Class 26: Answer 1: (B) The given vectors are n linear independent columns of a matrix A so they form a basis for the column space of A. Starting with this basis, Gram-Schmidt Orthogonalization produces an orthogonal basis which spans the same space the original collection of vectors does. Thus the answer is (B), it will span the column space of A

3. Class 26: Question 2

4. Class 26: Answer 2: (B) The way to do this problem is to do one step of Gram- Schmidt. Therefore the new pair (-2,1,0) and (3/5,6/5,1) is an orthogonal set of vectors formed from (-2,1,0) and (1,1,1)

5. Class 26: Question 3 1. TRUE 2. FALSE TRUE or FALSE:

6. Class 26: Answer 3: (A) Although this second pair of vectors is not formed by using Gram-Schmidt, it is still an orthogonal pair. The question is does span{(2,-1,0),(1,1,1)} equal span{2,-1,0),(3,6,5)}? In other words are they bases for the same space? Is (3,6,5) a linear combination of (-2,1,0) and (1,1,1)? Yes! 5(1,1,1)+(-2,1,0)=(3,6,5). Another way to see if the spaces are the same is to see if their cross-product is the same, i.e. they have the same normal vector, and thus they must lie in the same plane

7. Class 26: Question 4 TRUE or FALSE: The Gram-Schmidt Orthogonalization process can be used to construct an orthonormal set of vectors from an arbitrary set of vectors. 1. TRUE. 2. FALSE.

8. Class 26: Answer 4: (B) FALSE! The set of vectors that Gram-Schmidt works on is not any arbitrary set of vectors, but a linearly independent set. Although technically Gram-Schmidt produces an orthogonal set, one can always continue the process to the end to produce an orthonormal set.

9. Class 25: Question 5 1. TRUE. 2. FALSE.

10. Class 25: Answer 5 (B) There are multiple bases for R 2 so there must be multiple orthogonal bases for R 2. All you need to do is pick two vectors that are not scalar multiples of each other, are orthogonal and then normalize them to produce multiple orthonormal bases in R 2. Here are some more

11. Class 25: Question 6 Let Q be a square matrix with orthonormal columns. TRUE or FALSE: Q -1 =Q T. 1. TRUE. 2. FALSE.

12. Class 25: Answer 6: A If Q is a square matrix with orthonormal columns then Q T Q must equal the identity matrix. This is clear because if you multiply a matrix by it’s transpose then each element of the product consists of dot products of the rows of the matrix with it’s own columns. Since the matrix has orthonormal columns those dot products will produce either 1 or 0, which will result in the identity matrix. The product has 1 along the diagonal because that is when a particular column vector is being dotted with itself. Thus since Q T Q=I that must mean that Q -1 =Q T.

13. Class 25: Question 7

14. Class 25: Answer 7 (A)

15. Class 25: Question 8

16. Class 25: Answer 8: A The key thing to understand about this question is that it is the same question as the one before! In other words, to make an orthogonal projection onto a subspace one needs to have a basis for that subspace. What is a basis for the subspace corresponding to the line y=x/2? The line corresponds to span{(2,1)}. Thus (2,1) is a basis for the subspace. Thus the orthogonal projection of (-3,1) onto (2,1) is (2,1).

17. Class 25: Question 9

18. Class 25: Answer 9: D The line named l is y=3x. This is the subspace span{(1,3)}. If z is the projection of b onto this subspace it will be some scalar multiple of (1,3), i.e. (c,3c). However, the more important part of the problem is the interpretation of b-z which equals b-proj l (b). Recall that this new vector b-z will be orthogonal to l by definition. There’s no way b-z could be a point on l and orthogonal to it. Thus only two of the statements are true.