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
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 problem 1 in Problem Set 7.9? I am haveing a very difficult time with starting these.
Answer:
Think of the single equation in three unknowns as a matrix equation:
Av=0
where A is a 1x3 matrix and v is a column 3 vector. When you think of it this way, you realize that the problem is simply to show that the null space of a certain matrix is a vector space (and this is done near the top of page 305 -- you have to show that the sum of two solutions is a solution and that a constant times a solution is a solution). That one line matrix is already in row echelon form. Hence v_1 is a bound variable and v_2 and v_3 are free variables. When you solve the system you'll see that the dimension of the null space is two and you'll get two basis vectors for it by the usual method.
Question from a student regarding p. 329, #4:
How do you find the dimension if the problem is not explicitly stated? For example, a 3x3 matrix could have dimension 1, 2, or 3, right?
Answer:
Write the typical 2x2 matrix:
a b c d
There are four numbers there. I bet we are going to get a 4 dimensional vector space. It is not hard to come up with a basis if you've got the idea of unit vectors in the back of your mind:
[a b] [1 0] [0 1] [0 0] [0 0] [ ] = a [ ] + b [ ] + c [ ] + d [ ] [c d] [0 0] [0 0] [1 0] [0 1]
Sure enough, those four "unit" matrices are linearly independent and they form a basis for the vector space of 2x2 matrices.
Question: I'm confused about the book answer regarding a 2 x 3 matrix with all positive entries. Shouldn't that always make a vector space? If you add it to itself it's still a 2 x 3 and if you multiply by a scalar it's still a 2 x 3. The book says no but that is confusing to me. Anybody else puzzled about this?
Answer: Note that it explicitly states "positive" entries. If you multiply by a negative scalar, you are no longer in the same space.
