Homework 2 collaboration area
Here is the Homework 2 collaboration area. Since HWK 2 is due the Wednesday after Labor Day, I won't have a chance to answer questions on Monday like usual. I will answer any and all questions here on the Rhea on Tuesday with help from Eun Young Park. - Steve Bell
Questions from a student :
When finding a basis, does it always have to be fully reduced? For example, if you have a basis [4 -2 6] does it need to be reduced to [2 -1 3] or is either answer acceptable? Jones947
Answer from Eun Young :
No, it doesn't need to be reduced. If { [4 -2 6] } is a basis for some vector space V, then { [2 -1 3] } is also a basis for V and vice versa. If v belongs to span{[4 -2 6 ]}, v = c[4 -2 6] = 2c [2 -1 3 ] for some c. Hence, v belongs to span{[2 -1 3]}. The opposite direction is same. So, span{[4 -2 6]}= span{[2 -1 3]}. A basis for a vector space is not unique but a dimension of a vector space is unique.
Question from a student:
From what I understand, a basis is a set of vectors that can be used to create any vector in the span. So for example, if the basis is [1 0] [0 1], then the span could be [1 0] [0 1] [2 2] [2 0]. Is that correct?
Answer from Steve Bell:
The span is ALL vectors you get by taking linear combinations. Hence, the span is x*[1,0] + y*[0,1] = [x,y] as x and y range over all possible values, i.e., the span is R^2.
Question from a student:
I also have a question about how to find the basis. Should I be reducing the matrix to RREF? Or leave it when I get it to REF? I do not think it should matter either way, as the basis can still make up any vector in the set even if I use RREF. My answer is different from the back of the book but I think that is typical for linear algebra problems.
Answer from Steve Bell:
I tried to explain in class why the RREF matrix is better. It is easier to read off information from it and it has the virtue of being unique. But the plain old RE form does give you the answers with a little extra work sometimes(like when you need to do back substitution to get the general solution to a system). You are right that it is not uniquely determined, so you might get the right answer, even if it looks different than the one given in the back of the book.
Question from a student:
is my logic correct here.... to find the row space basis of a vector space you REF the matrix A and any non zero rows of the REF form the basis vectors for the row space. The column space is the corresponding columns of the matrix A that have pivots in the REF. Is this correct?
Remark from Steve Bell:
Yes, that's how you get a basis for the row space. The book expects you to put the columns of the matrix in as rows -- i.e., take the transpose, and repeat the process, being sure to change the row vectors you get as the non-zero rows of the row echelon matrix back into column vectors.
However, you are correct that another way to find a basis for the column space is to row reduce the matrix to row echelon form. The original columns of the matrix corresponding to the first non-zero entry in each of the non-zero rows in the row echelon matrix form a basis for the column space. This fact does not appear to be mentioned in the text.
Question from a student:
Re: Pg 300 #4 - It doesn't appear to me that N! multiplications are required to evaluate determinates of Nth order. 2x2 requires 2 multiplications, ok. 3x3 is 3 scalars multiplied by 3 (2x2)'s, so that is 9 multiplications. A 4x4 is 4 scalars mult by 4 (3x3)'s, which would be 40 multiplications. I must be missing something.....
Remark from Steve Bell:
Let's change this problem to read: Show that computing an nxn determinant by iterated expansion along row 1 takes at least n! multiplications. I just want you to think about this problem and come up with something reasonable. Use induction and define the determinant by expanding along the first row as in class. The purpose of the problem is to make you realize how nasty determinants are when computed via brute force.
I'm still getting questions about problems 32 and 34 on p. 287, so I am putting a little additional blurb about them here. Steve Bell 17:57, 1 September 2013 (UTC)
Question from the Aussie student who does not have a long weekend...:
Re Pg 318 Q6 All functions y(x)=acos2x+bsin2x with arbitrary constants a and b. Steve I think that I have proved that this is indeed a vector space by showing that all the axioms for vector addition and multiplication hold true. I did this by considering two functions f(x)=acos2x+bsin2x and g(x)=ccos2x+dsin2x. The problem that I am having is finding the dimension and basis for this space. I think that the problem is definitely me because the function space is a bit abstract for me but I considered the zero function acos2x+bsin2x = 0; if it is indeed a space it must contain the zero function. If I can prove that this is only zero when a and b equal zero then the basis for the function space must be cos2x and sin2x; and dimension is then 2? Or am I on the wrong track?
Answer from Steve Bell 21:44, 1 September 2013 (UTC):
G'Day James. You are definitely on the right track. In my lecture on Friday, I dealt with the two functions exp(x) and exp(−x)>. You can do exactly the same thing with cos(2x) and sin(2x). Keep in mind that sine over cosine is equal to tangent, and the tangent function is not a constant function. You can find my lecture notes from Friday at
G'Day Steve - thanks for that I just wanted to make sure that I was on the right track. Cheers James.
Jim Fink again here on pg 300 #4. Makes much more sense saying "greater than N! ... " multiplications, thanks. I figured out the number of mults req'd (say, Y) is a recursive formula; Y = n (Yn-1+1), where n = order of the matrix, which does make Y proportional to, but greater than n!. Recognizing that this may be getting a bit off topic, and I am obsessing about this problem a bit, but is there a way to write this recursive formula entirely in terms of n?
Remark from Steve Bell:
Jim, I don't know if that recursive function can be expressed in a clean way. It is traditional to say that the number of multiplications needed to compute a determinant by brute force is ON THE ORDER OF n!. That means that there are constants A and B so that
$ A n!\le\text{(number multiplications)}\le B n! $
