Difference between revisions of "REVIEWMA527Fall2010" - Rhea

Revision as of 13:37, 28 September 2010

Talk about the Review Problems for Exam 1 here

Here are some

SOLUTIONS

to the Review Problems.

Question: I'm still a bit confused about the definition for the dimension of a matrix. I understand that the dimension of A (or nullity, depending on what you're looking for) = (# variables/# columns) - (rank of A). To me, this would mean that the dimension is not generally equal to the rank.

However, I am confused by p. 300 of the book, which says that the dimension of V is "the maximum number of linearly independent vectors in V," which is the same definition of rank, given on p. 297. Furthermore, p. 301 says that the "row space and column space of matrix A have the same dimension, equal to rank A." I don't know what I am missing here; what is it that am I misunderstanding about the way this is written?

Reply from Bell: A matrix doesn't have a dimension. But it does have a rank, which is the dimension of the row space. That theorem says that the dimension of the column space is equal to the dimension of the row space. The nullity is the dimension of the null space, i.e., the dimension of the space of solutions x to Ax=0.

In general, the dimension of a vector space is equal to the number of vectors in a basis for the space, which is the same as the maximum number of linearly independent vectors in the space.

Question continued: But what I don't understand is that if dimension = rank, then how is it also equal to (#variables) - (rank)? If I have a vector space after row-reducing that looks like this

1 7 9 10 | 0
0 0 0 0  | 0
0 0 0 0  | 0
0 0 0 0  | 0

I would say that the rank is 1, and that the dimension is 4 - 1 = 3, not 1, because there are 3 free variables. I'm really sorry, but I'm still confused.

Reply from Bell: Yes, the rank is 1, and the null space is spanned by 3 linearly independent vectors (and that number is equal to the number of free variables).

The rank plus the number of free variables alwasy adds up to the number of variables, 4 in this case. It all adds up.

Rank + Nullity = # of columns in A.

is the same as saying

(# bound variables) + (# free variables) = (# of variables)

When you say "dimension = rank" you are using the word dimension to refer to the dimension of the vector space given by the linear span of the rows of A, i.e., the row space of A.

Question: For problem 2d of the review exam, you had the null space as being spanned by

[-1 -1 1 0]^T

and

[-4/5 -8/5 0 1]^T.

Why isn't the null space [-3 -1 1 0]^T and [-4 -8/5 0 1]^T since the matrix is in row echelon form?

Answer: To get the basis vectors, let one of the free variables be equal to one and the other (or others) equal to zero. Then you do back substitution from the row reduced equations to solve for the bound variables. Make a column vector out of the results to get a basis vector for the null space. Repeat for the other free variables to get the whole basis.

I'm just really confused because when I do this problem, I first set x3=0 and then x4 = 1, then vice-versa. The basis I get is [11 -8 0 5]T and [0 -1 1 0]T. What am I doing wrong here?

I am also confused on part 2B. I am getting a column basis of [1 0 4]T, [0 1 -1]. What am I doing wrong?

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