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#1 Help Me ! » Linear algebra question » 2007-04-30 14:19:56
- quackensack
- Replies: 0
If anyone could explain how the following problem on linear transformations is done, it would be greatly appreciated!
There is a theorem that assures us that if L: V->W is a linear transformation, then L(av_1 + bv_2) = aL(v_1)+bL(v_2), for all v1, v2 in V and all a, b in R. Prove that the converse of this statement is true by considering two cases: first a=b=1 and then b=0.
#2 Help Me ! » Maclaurin Series problem » 2007-04-25 07:10:57
- quackensack
- Replies: 1
If anyone could explain how the following problem is done, it would be greatly appreciated.
Write the Maclaurin series for cos(x^2). By antidifferentiating its first three (nonzero) terms, obtain an approximation for the integral of cos(x^2) dx from .6 to 1. Is this approximation larger than or smaller than the actual value of the integral of cos(x^2) from .6 to 1? How does the series itself tell you the answer to this question?
I've done the first part which is write out the Maclaurin series for cos(x^2):
cos(x^2) = 1 - x^4/2! +x^8/4! - x^12/6!...
And I've antidifferentiated the first three terms to get:
x - x^5/10 + x^9/216...
However, I'm confused about how to obtain an approximation for the integral of cos(x^2) dx from .6 to 1.
#3 Help Me ! » Linear algebra problem » 2007-04-19 09:06:58
- quackensack
- Replies: 3
This problem was confusing me so any explanations would be helpful and greatly appreciated!
Let V be a subspace of R^n with dim(v) = n - 1. (Such a subspace is called a hyperplane in R^n.) Prove that there is a nonzero x
R^n such that V = {v R^n|x v = 0}.#4 Help Me ! » Linear algebra problem » 2007-04-16 06:53:34
- quackensack
- Replies: 1
This problem was confusing me so any explanations would be helpful and greatly appreciated!
Examine S
R^5 defined byS = {x_1, x_2, x_3, x_4, x_5
R^5|x_2 = 0, x_3 + x_4 = x_5}.Verify that the subset S is, in fact, a subspace.
#5 Re: Help Me ! » Linear algebra problem » 2007-04-08 16:58:51
Thanks! I was wondering...part of the problem says to show that the intersection of W1 and W2 is nonempty. Did I already prove this by saying that ax and a+b are in the intersection, or is there more to it than that?
#6 Re: Help Me ! » Linear algebra problem » 2007-04-08 15:09:39
Just go through the properties one by one. Many you won't have to do. For example, being associative in V automatically means being associative in any subset of V, and the intersection of two subspaces is at least a subset.
So for example, to be a vector space, it must contain the 0 (I say "the" because you can prove that 0 is unique. i.e. there aren't two or more 0's). So we need to show that 0 is in the intersection. But part of being a subspace means that you have to contain the 0. So each subspace must contain the 0, and the intersection must therefore contain the 0.
Now lets try it for a + b. Let a and b be vectors in the intersection. This means that a is in W1 and a is in W2, and b is in W1 and b is in W2. So this means that a + b has to be in W1, as W1 must be closed. Same for W2. So a + b is in W1 and a + b is in W2. So a + b must be in the intersection.
All the rest are excruciatingly similar.
Thank you! 
So from what you said, I understand that I also have to prove it for scalar multiplication. So would it be right if I were to say:
Let a be a vector in the intersection. This means that a is in W1 and a is in W2. Therefore, a times a scalar x will be in W1 as well as W2 by the closure property of scalar multiplication. So a*x must be in the intersection.
#7 Help Me ! » Linear algebra problem » 2007-04-08 12:45:05
- quackensack
- Replies: 6
This proof was confusing me and I have a test next Thursday on this type of thing, so if anyone could explain how this is done, I would really appreciate it! Thanks in advance!
Let
be a vector space and let and be subspaces of . Prove that is a subspace of . (Do not forget to show that is nonempty.)#8 Help Me ! » Proof help » 2007-04-04 15:49:55
- quackensack
- Replies: 0
I was confused about how to do this so any explanations would be appreciated!
Let u = [u_1, u_2, u_3], v = [v_1, v_2, v_3], and w = [w_1, w_2, w_3] be three vectors in R^3. Show that S = {u, v, w} is linearly independent if and only if the determinant of
u_1 u_2 u_3
v_1 v_2 v_3
w_1 w_2 w_3
is not equal to 0.
(consider the transpose and the Independence Test Method)
#9 Help Me ! » linear algebra problem » 2007-04-03 06:29:00
- quackensack
- Replies: 1
If someone could explain how this is done, I would really appreciate it!
Show that the characteristic polynomial of a 2 x 2 matrix A is given by x^2 - (trace(A))x+|A|.
#10 Re: Help Me ! » Calc help » 2007-03-29 11:12:22
Thank you so much. I was wondering, what is the significance of the term -0.03FR in the F' equatiton and of the term -0.025FR in the R' equation? Meaning, what do they both represent?
#11 Help Me ! » Calc help » 2007-03-29 09:10:30
- quackensack
- Replies: 3
I'm given:
F' = 0.01F(10-F-3R)
R' = 0.0125R(8-R-2F)
F' = 0.1F(1 - F/10) - 0.03FR
R' = .1R (1 - R/8) - 0.025FR
This represents a mathematical model of competition, in which two species, wolves and tigers, compete for the same resources. F represents the wolf population, in hundreds of wolves, and R represents the tiger population in hundreds of tigers.
I have to find the four equilibrium points for the system. I'm a little confused about how to do this so if anyone could explain, I would REALLY appreciate it.
Also, I was wondering, what is the significance of the term -0.03FR in the F' equatiton and of the term -0.025FR in the R' equation?
Thanks in advance
#12 Help Me ! » Transition Matrix Help » 2007-03-22 03:55:09
- quackensack
- Replies: 0
I was given a transition matrix R and I have to find the steady state vector of that. I did the first step, which is to do R - the identity matrix. This is what I got (it's in table form just so you can tell the numbers apart more easily...it's a thumbnail so you can click it to make it larger and more readable).
Now, I know that in order to find a steady state vector I have to do this matrix multiplied by column vector [x1...x9] to get the column vector [0, 0, 0, 0, 0, 0, 0, 0, 0].
I'm just confused as to how to find the x1 through x9 because there are so many equations. Any help would be greatly appreciated!
#13 Re: Help Me ! » partial derivatives problem » 2007-03-15 15:55:29
would it just be h^0.725 * 6.64 * w^-.575 ?
(for w/ respect to w)
#14 Re: Help Me ! » partial derivatives problem » 2007-03-15 15:31:52
Sorry, I'm confused about how to do that.
#15 Help Me ! » partial derivatives problem » 2007-03-15 15:23:14
- quackensack
- Replies: 5
Not really sure how this is done so any help would be appreciated.
The function S(w, h) = (15.63w^0.425) * (h^.0725) gives the approximate surface area (in sq. inches) of a person h inches tall who weighs w pounds.
1) A boy grows taller while maintaining a weight of 130 pounds. At what rate is his surface area increasing when he's 5'4" tall?
2) Now at his full height of 5'9", he goes to the gym and bulks up. At what rate is his body surface increasing when he reaches 140 pounds?
#16 Help Me ! » partial derivatives problem » 2007-03-15 12:55:17
- quackensack
- Replies: 1
Not really sure how this is done so any help would be appreciated.
The function S(w, h) = (15.63w^0.425) * (h^.0725) gives the approximate surface area (in sq. inches) of a person h inches tall who weighs w pounds.
1) A boy grows taller while maintaining a weight of 130 pounds. At what rate is his surface area increasing when he's 5'4" tall?
2) Now at his full height of 5'9", he goes to the gym and bulks up. At what rate is his body surface increasing when he reaches 140 pounds?
#17 Re: Help Me ! » Proof » 2007-03-11 08:44:56
Using the fact that AX=0 has the trivial solution won't help you, as your teacher pointed out.
What you need to use is that it has just the trivial solution. That means that no other matrix X will fit the equation, and you should be able to use that to help you.Also, keep in mind that it is an 'if and only if' proof, which means that you need to prove that if A has rank n then AX=0 has no non-trivial solutions, and that if AX=0 has no non-trivial solutions then A has rank n.
Okay, but where I'm stuck is that I'm not sure how to prove that if A has rank n, then AX=0 has no non-trivial solutions.
#18 Re: Help Me ! » Proof » 2007-03-11 02:14:10
i talked to my teacher this morning, and he said that the everything I wrote starting with "The inverse of the identity matrix is itself..." is correct. but he also said that AX=0 always has the trivial solution, but A isn't necessarily row-equivalent to the identity matrix.
i am confused...how do i prove that it is in this case? we haven't learned the invertible matrix theorem, so i'm not sure how else to prove it.
#19 Re: Help Me ! » Proof » 2007-03-10 15:52:43
Can't one just say:
If AX=0 has the trivial solution, then A is row equivalent to the identity matrix:
(1 0 0)
(0 1 0)
(0 0 1)
The inverse of the identity matrix is itself. And thus it is nonsingular.
Thus, A has rank n.
Because the identity matrix is equal to A, then A has rank n.
#20 Re: Help Me ! » Proof » 2007-03-10 15:31:18
I'm sorry - I'm a little stuck because I haven't learned about linear dependence, kernel, or morphism in my class. Is it possible to show what you mean using a matrix? If it isn't, that's okay. I just think that it would be easier for me to understand it visually.
#21 Re: Help Me ! » Proof » 2007-03-10 14:24:28
Thanks for all your input, guys!
I was wondering - is there a way to do it without using the nullity theorem? I haven't about the concept of linear dependence in my class (as far as I know, that is).
A friend told me that I can do this proof by putting a matrix A in row-reduced echelon form. If doing this yields a matrix with at least one non-zero entry in each row, then the only solution to AX=0 is trivial. This , in turn, is the definition of rank n if A is an n by n matrix.
I guess where I'm stuck is...how does one row-reduce the matrix A without knowing what it is?
#22 Help Me ! » Proof » 2007-03-10 12:57:30
- quackensack
- Replies: 16
hey i take linear algebra, and i have a quiz coming up on monday so i'm studying proofs. i was wondering if someone could show me how to do the following:
suppose that A is an n by n matrix. prove that the rank of A is n if and only if the equation AX=0 has just the trivial solution.
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