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| − | = | + | = ''' ''' = |
| − | = | + | = '''Chpater3&4 MA351Spring2011''' = |
| − | = 3 | + | = ''' 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. | + | = '''3.1 11 Ax = 0 to find the span,very basic but very important<br>''' = |
| − | = | + | = = |
| − | = <br> = | + | = '''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.''' = |
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| + | = '''3.2 4 Fact 3.2.2.<br>''' = | ||
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| + | = '''3.2 12 Linearly dependent<br>''' = | ||
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| + | = '''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''' = | ||
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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>''' = | ||
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| + | = '''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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| + | = '''By Bingrou Zhou<br>''' = | ||
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| + | = = | ||
Latest revision as of 14:51, 30 April 2011
Contents
- 1
- 2 Chpater3&4 MA351Spring2011
- 3 This is the review for Chapter 3 and 4. The problems posted here are the one that I consider hard. Hope it can help.
- 4 3.1 11 Ax = 0 to find the span,very basic but very important
- 5
- 6 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.
- 7 3.2 4 Fact 3.2.2.
- 8 3.2 12 Linearly dependent
- 9 3.3 5 The first two vectors are non-redundant, but the third is a multiple of the first.
- 10 3.4 17 By inspection, we see that in order for x to be in V , x = 1v1 + 1v2 + 1~v3
- 11 4.1 11 Not a subspace: I3 is in rref, but the scalar multiple 2 I3 isn't.
- 12 4.2 53 Thus the kernel consists of all constant polynomials f(t) = a(when b = c = 0), and the nullity is 1.
- 13 By Bingrou Zhou
- 14
