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= Chpater3&4 MA351Spring2011 =
  
 
=    This is the review for Chapter 3 and 4. The problems posted here are the one that I consider hard. Hope it can help.  =
 
=    This is the review for Chapter 3 and 4. The problems posted here are the one that I consider hard. Hope it can help.  =
  
= 3.1 11 Ax = 0 to find the span,very basic but very important<br> =
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= 3.1 11 Ax = 0 to find the span,very basic but very important<br> =
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= 3.115 By Fact 3.1.3, the image of A is the span of the columns of A, any two of these vectors span all of R2 already.  =
 
= 3.115 By Fact 3.1.3, the image of A is the span of the columns of A, any two of these vectors span all of R2 already.  =
  
= 3.2 4 Fact 3.2.2.<br> =
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= 3.2 4 Fact 3.2.2.<br> =
  
= 3.2 12 Linearly dependent<br> =
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= 3.2 12 Linearly dependent<br> =
  
 
= 3.3 5 The first two vectors are non-redundant, but the third is a multiple of the first.  =
 
= 3.3 5 The first two vectors are non-redundant, but the third is a multiple of the first.  =
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= 3.4 17 By inspection, we see that in order for x to be in V , x = 1v1 + 1v2 + 1~v3  =
 
= 3.4 17 By inspection, we see that in order for x to be in V , x = 1v1 + 1v2 + 1~v3  =
  
= <br>4.1 11 Not a subspace: I3 is in rref, but the scalar multiple 2 I3 isn't.<br> <br> =
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= <br>4.1 11 Not a subspace: I3 is in rref, but the scalar multiple 2 I3 isn't.<br> <br> =
  
 
= 4.2 53 Thus the kernel consists of all constant polynomials f(t) = a(when b = c = 0), and the nullity is 1.  =
 
= 4.2 53 Thus the kernel consists of all constant polynomials f(t) = a(when b = c = 0), and the nullity is 1.  =
  
 
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Revision as of 15:42, 30 April 2011

 

Chpater3&4 MA351Spring2011

   This is the review for Chapter 3 and 4. The problems posted here are the one that I consider hard. Hope it can help.

3.1 11 Ax = 0 to find the span,very basic but very important

3.115 By Fact 3.1.3, the image of A is the span of the columns of A, any two of these vectors span all of R2 already.

3.2 4 Fact 3.2.2.

3.2 12 Linearly dependent

3.3 5 The first two vectors are non-redundant, but the third is a multiple of the first.

3.4 17 By inspection, we see that in order for x to be in V , x = 1v1 + 1v2 + 1~v3


4.1 11 Not a subspace: I3 is in rref, but the scalar multiple 2 I3 isn't.

4.2 53 Thus the kernel consists of all constant polynomials f(t) = a(when b = c = 0), and the nullity is 1.


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