| Line 21: | Line 21: | ||
Any matrix can be put into reduced row | Any matrix can be put into reduced row | ||
echelon form. For #4 on page 301, you'll | echelon form. For #4 on page 301, you'll | ||
| − | need to consider various cases | + | need to consider various cases. I've added |
| + | some hints at the end of | ||
| − | + | [http://www.math.purdue.edu/~bell/MA527/Lectures/lec09-03.pdf Lecture 9 notes] | |
| − | + | ||
Question from student: | Question from student: | ||
| Line 34: | Line 34: | ||
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. | 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 from student regarding p. 301, #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. | + | |
| + | 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. | ||
| + | |||
| + | Answer from Bell: | ||
| + | |||
| + | [0 b^2-a^2 0; 1 a/b 0] | ||
| + | |||
| + | does not appear to be in row echelon form to me. If b^2-a^2 is not zero, | ||
| + | then you can row reduce what you have to | ||
| + | |||
| + | <PRE> | ||
| + | 1 a/b 0 | ||
| + | 0 1 0 | ||
| + | </PRE> | ||
| + | |||
| + | which further reduces to | ||
| + | |||
| + | <PRE> | ||
| + | 1 0 0 | ||
| + | 0 1 0 | ||
| + | </PRE> | ||
| + | |||
| + | Two bound variables plus one free variable. | ||
| + | |||
| + | Rank = 2, nullity =1. A basis for the null space | ||
| + | is | ||
| + | |||
| + | [0 0 1]. | ||
| + | |||
| + | It all adds up. | ||
| + | |||
| + | |||
[[2010 MA 527 Bell|Back to the MA 527 start page]] | [[2010 MA 527 Bell|Back to the MA 527 start page]] | ||
Revision as of 10:01, 3 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. I've added some hints at the end of
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 from student regarding p. 301, #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.
Answer from Bell:
[0 b^2-a^2 0; 1 a/b 0]
does not appear to be in row echelon form to me. If b^2-a^2 is not zero, then you can row reduce what you have to
1 a/b 0 0 1 0
which further reduces to
1 0 0 0 1 0
Two bound variables plus one free variable.
Rank = 2, nullity =1. A basis for the null space is
[0 0 1].
It all adds up.
