Difference between revisions of "HWK2MA527Fall2010" - Rhea

Revision as of 11:52, 6 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 about this at the end of

Lecture 9 notes

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.

Question from student regarding p. 329, #1:

Can you provide some direction for starting problems 1, 3, 4, & 9 in Problem Set 7.9? I am haveing a very difficult time with starting these.

Answer/Question from another student regarding p. 329, #1:

I found that, if setting v1 = 1, I can find that v2 = 1 and v3 = -1 to solve the one given equation. Then, we can go do our tests regarding addition and multiplication to determine if its a vector space. Prof. Bell, can you please confirm this or provide other help? Thanks.

Question from student regarding HW #3: HW 3 Question: Do to scheduled work travel next week I am trying to get ahead on HW #3. I keep getting stuck on the Eigenvalues. When the directions state to find the spectrum, is it asking for the spectrum of the original matrix or the spectrum of the symmetric, skew-symmetric, orthoganal, hermitian, skew-hermitian, or unitary matrix? I have worked the problems trying to get the spectrum of the original matrix and I am not having much succcess. Thank you.

HWK #3,4,7 on page 329 question: We would want to use variables in these matrices correct? Other wise certain components would indicate a vector space and some would not. Am I correct in stating this?

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