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echelon form. For #4 on page 301, you'll | echelon form. For #4 on page 301, you'll | ||
need to consider various cases: case a=0, | need to consider various cases: case a=0, | ||
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| + | |||
case b=0, etc. | case b=0, etc. | ||
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[[Category:MA5272010Bell]] | [[Category:MA5272010Bell]] | ||
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| + | Question regarding Question #4: | ||
| + | I have the matrix reduced to [a-b b-a 0; b a c]. If I say c=0, then [a-b b-a 0; b a 0]. This can be reduced to [0 b^2-a^2 0; 1 a/b 0]. This doesn't make sense because the null space would be span((a^2-b^2 1 a/b)^T (0 0 1)^T) and the nullity would be 1. However, since the rank is 2, rank+nullity does NOT equal # columns. Please explain. | ||
Revision as of 07:13, 2 September 2010
Homework 2 work collaboration area
Question from student:
If we find that a matrix has a nullity value such as 1, then there has to be a null space correct?
Answer from Bell:
A matrix with a nullity of one has a one dimensional null space. (The zero vector is always in the null space of a matrix, so it never happens that a matrix does not have a null space.)
Question from student:
Regarding # 4 on page 301: it doesn't seem that the original matrix can be put in row reduced echelon form, which is a requirement, correct?
Answer from Bell:
Any matrix can be put into reduced row echelon form. For #4 on page 301, you'll need to consider various cases: case a=0,
case b=0, etc.
Question from student:
About #4 on p. 301, I've made it to [a b c; b-a a-b 0] and plugged a=1, b=2, c=3 arbitrarily. Then I reduced it to [1 2 3; 0 -3 -3]. I'm not even sure whether plugging in random values was the right idea, but I'm stuck here. How do I proceed from here?
Answer from Bell:
Plugging in values is never the way to go. That's like doing an experiment in science. You'd have to plug in lots of random values if you were doing science, but you'd miss the key points in math. You'll need to consider various cases in this problem, e.g., the case a=0, the case b=0, etc.
Question regarding Question #4: I have the matrix reduced to [a-b b-a 0; b a c]. If I say c=0, then [a-b b-a 0; b a 0]. This can be reduced to [0 b^2-a^2 0; 1 a/b 0]. This doesn't make sense because the null space would be span((a^2-b^2 1 a/b)^T (0 0 1)^T) and the nullity would be 1. However, since the rank is 2, rank+nullity does NOT equal # columns. Please explain.
