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[[Category:MA5272010Bell]]
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<span class="texhtml">''I''''n''''s''''e''''r''''t''''f''''o''''r''''m''''u''''l''''a''''h''''e''''r''''e''</span>
  
==Homework 1 collaboration area==
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== 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. --[[User:Bell|Steve Bell]] 12:11, 20 August 2010 (UTC)
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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. --[[User:Bell|Steve Bell]] 12:11, 20 August 2010 (UTC)  
  
Here is my favorite formula:
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Here is my favorite formula:  
  
<math>f(a)=\frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z-a}\ dz.</math>
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<math>f(a)=\frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z-a}\ dz.</math>  
  
Question from a student:
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Question from a student:  
  
I have a question about P.301 #33. I see in the back of the book that it is
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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?  
not a vector space. I don't understand why though. In the simplest form I
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would think an identity matrix would satisfy the requirements mentioned in
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the book for #33. Isn't an identity matrix a vector space?
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Answer from Bell:
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Answer from Bell:  
  
One of the key elements of being a vector space is
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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.)  
that the thing must be closed under addition. The
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set consisting of just the identity matrix is not
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a vector space because if I add the identity matrix
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to itself, I get a matrix with twos down the diagonal,
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and that isn't in the set. So it isn't closed under
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addition. (The set consisting of the ZERO matrix
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is a vector space, though.)
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Another question from the same student:
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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?
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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:
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Answer from Bell:  
  
Yes, a vector space must contain the zero vector
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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).  
(because a constant times any vector has to be in
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the space if the vector is, even if the constant
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is zero).
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The set of all linear combinations of a bunch of
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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.  
vectors is a vector space, and the zero vector is
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in there because one of the combinations involves
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taking all the constants to be zero.
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Question from a student:
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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.
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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:
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Answer from Bell:  
  
You'll need to do row reduction.
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You'll need to do row reduction. At some point you'll get  
At some point you'll get
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<pre> 1  1    a
 
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<PRE>
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1  1    a
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  0  a-1  1-a
 
  0  a-1  1-a
 
  0  1-a  1-a^2
 
  0  1-a  1-a^2
</PRE>
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</pre>  
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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
  
At this point, you'll need to consider the case  a=1.
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[ 1 1 1 ].
If  a=1, the matrix is all 1's and the row space is
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just the linear span of
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[ 1 1 1 ].
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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.  
  
If  a  is
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[Q: Only 1 more case? Don't we need to consider the case in which a is not equal to 1 and a is not equal to the value found from the third row after we continue row reduction? ]
not 1, then you can divide row two by  a-1 and continue
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doing row reduction.  I think you'll have one more case
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to consider when you try to deal with the third row
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after that.
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Question from student:
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Question from student:  
  
Also, what are you wanting for an answer to questions 22-25 (full proof, explanation, example, etc.)?
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Also, what are you wanting for an answer to questions 22-25 (full proof, explanation, example, etc.)?  
  
Anwer from Bell:
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Anwer from Bell:  
  
For problems 22-25, I'd want you to explain the answer
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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:  
so that someone who has just read 7.4 would understand
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you and believe what you say. For example, the answer
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to #25 might start:
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If the row vectors of a square matrix are linearly
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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.  
independent, then Rank(A) is equal to the number
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of rows. But Rank(A)=Rank(A^T). Etc.
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[[Category:MA5272010Bell]]

Revision as of 20:48, 30 August 2010

I'n's'e'r't'f'o'r'm'u'l'a'h'e'r'e

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.

[Q: Only 1 more case? Don't we need to consider the case in which a is not equal to 1 and a is not equal to the value found from the third row after we continue row reduction? ]

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.

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva