<![CDATA[Physics Forums - Linear & Abstract Algebra]]> http://www.physicsforums.com Vector spaces and linear transformations. Groups and other algebraic structures along with Number Theory. en Tue, 18 Jun 2013 23:54:45 GMT vBulletin 60 http://www.physicsforums.com/images/physicsforums/misc/rss.jpg <![CDATA[Physics Forums - Linear & Abstract Algebra]]> http://www.physicsforums.com Problem A1 from the 2009 Putnam Exam http://www.physicsforums.com/showthread.php?t=697507&goto=newpost Mon, 17 Jun 2013 20:01:04 GMT Here is the problem: Let f be a real-valued function on the plane such that for every square ABCD in the plane, f(A) + f(B) + f(C) +f(D) = 0.... Here is the problem:

Let f be a real-valued function on the plane such that
for every square ABCD in the plane, f(A) + f(B) +
f(C) +f(D) = 0. Does it follow that f(P) = 0 for all
points P in the plane?



Below is my solution:

Create a 3x3 set of boxes (where each box has equal lengths as all of the other boxes) like so:

. . . .
. . . .
. . . .
. . . .

The periods represent vertices. Now, denote the value of the function at each vertice by a(1),...,a(16).

Now, since the sum of any four vertices making a square is zero, we get 18 unique equations (each containing a linear relationship between a unique quadruple of vertices):

9 (each representing a unit box)
+
4 (each representing a box of side length 2 units)
+
1 (this equation relates the vertices of the entire 3x3 square)
+
4 (each representing a box of side length sqrt(2) units (the diagonal squares)


Writing these into a matrix, we get an 18x16 matrix mapping a vector whose coordinates are the values of the function at each vertex) to the zero vector. Since we have 18 homogeneous equations and 16 variables, all of the variables must be zero. From this we can clearly conclude that the value of the function is zero for everywhere on the plane.

Is this solution correct? I can't see any errors, I just thought it was an interesting answer (it was different from the official answer, which I've posted below) and wasn't sure if I was missing anything.

OFFICIAL ANSWER:

Yes, it does follow. Let P be any point in the plane. Let
ABCD be any square with center P. Let E; F; G; H
be the midpoints of the segments AB; BC; CD; DA,
respectively. The function f must satisfy the equations
0 = f(A) + f(B) + f(C) + f(D)
0 = f(E) + f(F) + f(G) + f(H)
0 = f(A) + f(E) + f(P) + f(H)
0 = f(B) + f(F) + f(P) + f(E)
0 = f(C) + f(G) + f(P) + f(F)
0 = f(D) + f(H) + f(P) + f(G):
If we add the last four equations, then subtract the first
equation and twice the second equation, we obtain 0 =
4f(P), whence f(P) = 0. ]]> hello95 http://www.physicsforums.com/showthread.php?t=697507 Solving for a Lie Algebra in General http://www.physicsforums.com/showthread.php?t=697480&goto=newpost Mon, 17 Jun 2013 17:15:32 GMT So, I just went through the derivation of the Lie algebra for SO(n). in order to do so, we considered ##b^{-1}ab##, and related it to... So,

I just went through the derivation of the Lie algebra for SO(n). in order to do so, we considered ##b^{-1}ab##, and related it to ##U\left(b^{-1}ab\right)##, and since we have a group homomorphism, ##U^{-1}\left(b\right)U\left(a\right)U\left(b\right)##, all of which correspond to the whole similarity matrix thing. By careful choice of element representation, one is able to massage a commutator structure, then it all reduces down. What is the deal with the choice of similarity? Is this why mathematicians always say in passing how important similarity transformations are for physicists?

cheers ]]> pdxautodidact http://www.physicsforums.com/showthread.php?t=697480 Ax=Bx for all x implies A=B http://www.physicsforums.com/showthread.php?t=697412&goto=newpost Mon, 17 Jun 2013 06:21:47 GMT Suppose A and B are matrices of the same size, and x is a column vector such that the matrix products Ax and Bx are defined. Suppose that Ax=Bx... Suppose A and B are matrices of the same size, and x is a column vector such that the matrix products Ax and Bx are defined.

Suppose that Ax=Bx for all x. Then is it true that A=B?

I know that this is true and I can prove it using the idea of transformation matrices, and viewing Ax and Bx each as linear transformations and showing that those two transformations are equivalent, but I was curious if this can be proved without appealing to the notion of a linear transformation.

Tips?

BiP ]]> Bipolarity http://www.physicsforums.com/showthread.php?t=697412 Mazur-Ulam theorem (bijective isometries are affine maps) http://www.physicsforums.com/showthread.php?t=697380&goto=newpost Mon, 17 Jun 2013 01:49:48 GMT I've been studying the proof of the Mazur-Ulam theorem in the pdf linked to at the end of this Wikipedia article. I'm struggling with some details in that pdf.

Theorem: Let E and F be arbitrary normed spaces. If ##f:E\to F## is a bijective isometry, then f is an affine map.

Some of the things I don't get:

1. Their definition of "affine map". How does ##f(tx+(1-t)y)=tf(x)+(1-t)f(y)## for all ##x,y\in X## and all ##t\in[0,1]## imply that f-f(0) is linear? I don't see how to deal with (f-f(0))(ax) where ##a\in\mathbb R## is arbitrary.

2. The claim that ##\psi## is an isometry. ##\psi:E\to E## is defined by ##\psi(x)=2z-x## for all ##x\in E##. This map sends an arbitrary point in E to the point that's "on the opposite side of z", i.e. the point y such that y=z+(z-x). They claim that ##\psi## is an isometry, but consider e.g. ##E=\mathbb R##, z=2, x=1. We have ##\psi(1)=2\cdot 2-1=3##, but ##\|\psi(1)\|=3\neq 1=\|1\|##.

3. If ##\psi## isn't an isometry, then I don't see a reason to think that the map the author denotes by g* should be an isometry either. It's defined by ##g^*=\psi\circ g^{-1}\circ\psi\circ g##, where g is a bijective isometry. The step ##\|g^*(z)-z\|\leq\lambda## relies on ##g^*## being an isometry.

4. All they're proving is that for all ##a,b\in E##, we have ##f\left(\frac{a+b}{2}\right)=\frac 1 2 f(a)+\frac 1 2 f(b)##. It's not obvious that this implies that f is affine. ]]> Fredrik http://www.physicsforums.com/showthread.php?t=697380 Angle between vectors http://www.physicsforums.com/showthread.php?t=697321&goto=newpost Sun, 16 Jun 2013 19:19:09 GMT I have some question guys i have four points in the x,y plane in cartesian coordinates. A (Ax,Ay) B(Bx,By) C(Cx,Cy) D(Dx,Dy) I have some question guys

i have four points in the x,y plane in cartesian coordinates.

A (Ax,Ay)
B(Bx,By)

C(Cx,Cy)
D(Dx,Dy)

A and B is vector G
C and D is vector F

I would like to know what is the equation to get the angle between those two vectors (G,F) . and what are the limitations of this equation.

thank you ]]> dudu3060 http://www.physicsforums.com/showthread.php?t=697321 inverse transformation matrix entry bounds http://www.physicsforums.com/showthread.php?t=697014&goto=newpost Fri, 14 Jun 2013 18:21:34 GMT I have sets of 2d vectors to be transformed by an augmented matrix A that performs an affine transform. Matrix A can have values that differ at most... I have sets of 2d vectors to be transformed by an augmented matrix A that performs an affine transform.
Matrix A can have values that differ at most |d| from the identity matrix, to limit the transformation, meaning that the min/max bounds for A are [itex] I_3 \pm dI_3[/itex]

The problem is that i'd lke to have bounds for the inverse as well, expressed as a function of d, so that if i know that the transformation matrix is bound by d, that the matrix of the inverse transformation is bound by f(d).
I thought the same bounds would apply, but they don't.
Is there a way to find them? ]]> atrus_ovis http://www.physicsforums.com/showthread.php?t=697014 Explanation of exponential operator proof http://www.physicsforums.com/showthread.php?t=696914&goto=newpost Fri, 14 Jun 2013 01:26:18 GMT Can someone please explain the below proof in more detail?
http://i4.photobucket.com/albums/y11...psb4f8f1f9.jpg

The part in particular which is confusing me is http://i4.photobucket.com/albums/y11...psf444f5b1.jpg

Thanks in advance! ]]> gkirkland http://www.physicsforums.com/showthread.php?t=696914 Tensor notation and common operations http://www.physicsforums.com/showthread.php?t=696814&goto=newpost Thu, 13 Jun 2013 15:03:23 GMT Hello,
sorry for my english..
I have more doubts about the tensor notation.
If i have a (1,1)-tensor A[itex]\mu[/itex] [itex]\nu[/itex]
what does it correspond A[itex]\nu[/itex] [itex]\mu[/itex], A [itex]\nu[/itex][itex]\mu[/itex]??

And what does it correspond in index notation A-1 and A+ ??
And if i have a (p,q)-tensor how does it appear in index notation the inverse tensor and hermitian conjugate tensor? ]]> oliveriandrea http://www.physicsforums.com/showthread.php?t=696814 Family of sets without maximal element http://www.physicsforums.com/showthread.php?t=696748&goto=newpost Thu, 13 Jun 2013 03:18:20 GMT I have begun to learn about maximal elements from a linear algebraic perspective (maximal linearly independent subsets of vector spaces). I have a... I have begun to learn about maximal elements from a linear algebraic perspective (maximal linearly independent subsets of vector spaces). I have a few questions of which I have been able to get few insights online:

1) Does every family of sets have a maximal element? How can I make a family of sets that does not have a maximal element? I have to obviously make the hypothesis of Zorn's lemma fail, but I can't quite see how to do that.

2) Does every chain of sets have a maximal element? It seems that a chain of sets necessarily satisfies the criteria for Zorn's lemma but I am not sure.

Thanks!

BiP ]]> Bipolarity http://www.physicsforums.com/showthread.php?t=696748 The definition of an algebra http://www.physicsforums.com/showthread.php?t=696663&goto=newpost Wed, 12 Jun 2013 19:02:34 GMT "Let R be a commutative ring. We say that M is an algebra over R, or that M is an R-algebra if M is an R-module that is also a ring (not necessarily commutative), and the ring and module operations are compatible, i.e., [tex]r(xy) = (rx)y = x(ry)[/tex] for all [itex]x, y \in M[/itex] and [itex]r \in R[/itex]."

I'm not really sure why the second equality is true, because it implies commutativity and the definition tells us that an R-module is not necessarily commutative, right?

Thank you in advance ]]> Artusartos http://www.physicsforums.com/showthread.php?t=696663 Need resources for solving nonlinear matrix equations http://www.physicsforums.com/showthread.php?t=696595&goto=newpost Wed, 12 Jun 2013 10:53:43 GMT In my work I've encountered equations of the type:

(Ax).*(Bx) + Cx = d

Where A,B and C are non-unitary square matrices, x and d column vectors and .* denote component-wise multiplication.

I have a few books which discuss nonlinear matrix equations, but not of this kind. Any suggestions? ]]> Kosh Naranek http://www.physicsforums.com/showthread.php?t=696595 <![CDATA[zeta(3) and Euler's formula]]> http://www.physicsforums.com/showthread.php?t=696507&goto=newpost Tue, 11 Jun 2013 19:33:59 GMT Hi everyone. I'm trying to understand the step where they wrote

1/2 ∏1/(1+p^-3) =1/2 Ʃ(-1)^ord(k)/k^3

How can I see this? I know the Euler product formula, but it has a negative sign before the p^-3, where here we have a + sign.

Thanks for the help.
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]]> camilus http://www.physicsforums.com/showthread.php?t=696507 Invertible transformations http://www.physicsforums.com/showthread.php?t=696397&goto=newpost Tue, 11 Jun 2013 02:50:43 GMT For a linear transformation to be invertible, is it a requirement that the domain and codomain be the same vector space, or merely that they have the... For a linear transformation to be invertible, is it a requirement that the domain and codomain be the same vector space, or merely that they have the same dimension? My intuition tells me they merely need the same dimension but someone can correct me please?

BiP ]]> Bipolarity http://www.physicsforums.com/showthread.php?t=696397 Uniqueness of linear maps http://www.physicsforums.com/showthread.php?t=696331&goto=newpost Mon, 10 Jun 2013 17:19:32 GMT Friedberg proves the following theorem: Let V and W be vector spaces over a common field F, and suppose that V is finite-dimensional with a basis... Friedberg proves the following theorem:
Let V and W be vector spaces over a common field F, and suppose that V is finite-dimensional with a basis [itex] \{ x_{1}...x_{n} \}. [/itex] For any vectors [itex] y_{1}...y_{n} [/itex] in W, there exists exactly one linear transformation [itex] T: V → W [/itex] such that [itex] T(x_{i}) = y_{i} [/itex] for i = 1,...,n.

He uses this theorem to assert the following:
Suppose that V and W are finite-dimensional vector spaces with ordered bases [itex] = \{ x_{1}...x_{n} \} [/itex] and [itex] = \{ y_{1}...y_{m} \} [/itex], respectively. Let [itex] T: V →W [/itex] be a linear map. Then there exist scalars [itex] a_{ij} \in F \mbox{ (i = 1,..., m and j = 1,...,n) } [/itex] such that

[tex] T(x_{j}) = \sum^{m}_{i=1}a_{ij}y_{i} \mbox{ for } 1 \leq j \leq n [/tex]

He doesn't really prove the assertion he makes regarding the matrix representation, and it is not obvious to me. Obviously the first theorem (regarding the action of linear maps upon bases) is necessary to show that the matrix representation exists and is the unique linear map satisfying the action of the linear map upon bases. The problem is that in the first theorem, he uses n vectors from W. In the second theorem, the basis he uses for W has m vectors which may or may not be equal to n. If it is not equal to n, why is he allowed to use the theorem?

I apologize if I'm not clear enough. Please let me know which part is not clear and I will clarify further.

BiP ]]> Bipolarity http://www.physicsforums.com/showthread.php?t=696331 Finding the inverse of matrices larger than 2x2 http://www.physicsforums.com/showthread.php?t=696145&goto=newpost Sun, 09 Jun 2013 15:06:50 GMT Is there a way without using the algorithm to find A-1 of a square matrix greater than 2x2?

The question we are given in the books is:

[-25 -9 -27]
[536 185 537]
[154 52 143]

We are asked to find A-1 of the second and third column without computing the first column.

(Sorry about the format...I couldn't figure out how to use the brackets on this one.)

Thanks. ]]> dwn http://www.physicsforums.com/showthread.php?t=696145