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<math>f(a)=\frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z-a}\ dz.</math> | <math>f(a)=\frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z-a}\ dz.</math> | ||
| + | |||
| + | 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 | ||
| + | |||
| + | <PRE> | ||
| + | 1 1 a | ||
| + | 0 a-1 1-a | ||
| + | 0 1-a 1-a^2 | ||
| + | </PRE> | ||
| + | |||
| + | 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. | ||
Revision as of 10:48, 30 August 2010
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.
