Revision as of 09:48, 30 August 2010 by Bell (Talk | contribs)


Homework 1 collaboration area

Feel free to toss around ideas here. Feel free to form teams to toss around ideas. Feel free to create your own workspace for your own team. --Steve Bell 12:11, 20 August 2010 (UTC)

Here is my favorite formula:

$ f(a)=\frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z-a}\ dz. $

Question from a student:

I have a question about P.301 #33. I see in the back of the book that it is not a vector space. I don't understand why though. In the simplest form I would think an identity matrix would satisfy the requirements mentioned in the book for #33. Isn't an identity matrix a vector space?

Answer from Bell:

One of the key elements of being a vector space is that the thing must be closed under addition. The set consisting of just the identity matrix is not a vector space because if I add the identity matrix to itself, I get a matrix with twos down the diagonal, and that isn't in the set. So it isn't closed under addition. (The set consisting of the ZERO matrix is a vector space, though.)

Another question from the same student:

Does a set of vectors have to contain the 0 vector to be a vector space? If that is true, then any set of vectors that are linearly independent would not be a vector space?

Answer from Bell:

Yes, a vector space must contain the zero vector (because a constant times any vector has to be in the space if the vector is, even if the constant is zero).

The set of all linear combinations of a bunch of vectors is a vector space, and the zero vector is in there because one of the combinations involves taking all the constants to be zero.

Question from a student:

How do I start problem 6 (reducing the matrix) on page 301? I understand that the row and column space will be the same, but I'm not sure how to get it to reduce to row-echelon form.

Answer from Bell:

You'll need to do row reduction. At some point you'll get

 1  1    a
 0  a-1  1-a
 0  1-a  1-a^2

At this point, you'll need to consider the case a=1. If a=1, the matrix is all 1's and the row space is just the linear span of

[ 1 1 1 ].

If a is not 1, then you can divide row two by a-1 and continue doing row reduction. I think you'll have one more case to consider when you try to deal with the third row after that.

Question from student:

Also, what are you wanting for an answer to questions 22-25 (full proof, explanation, example, etc.)?

Anwer from Bell:

For problems 22-25, I'd want you to explain the answer so that someone who has just read 7.4 would understand you and believe what you say. For example, the answer to #25 might start:

If the row vectors of a square matrix are linearly independent, then Rank(A) is equal to the number of rows. But Rank(A)=Rank(A^T). Etc.

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