Are C and R^2 isomorphic as vector spaces?
396588
If one considers C as a 2-dimensional vector space over R (this is the complex plane representation of complex numbers), then it is trivially isomorphic to R^2 (the real plane) by mapping the complex number (a, b) to the real pair (a, b). The vector space structure is trivially preserved (with respect to real-linear morphisms). If it was not isomorphic, there would be no complex plane representation and we'd be stuck working with an exotic 1-dimensional structure. See http://mathworld.wolfram.com/ComplexPlane.html and http://en.wikipedia.org/wiki/Complex_numbers#Real_vector_space .
869774
That's kind of a tall order, mentioning every single definition and theorem related to vector spaces and subspaces.  I suggest you ask your professor for help on this one and read the textbook if you need a thorough explanation.\n\nBut anyway, the main idea is: suppose you have a subset W of a vector space V.  Then W is a subspace of V if W itself is a vector space when using the operations inherited from V.\n\nThen there's this theorem which basically says the same thing:\n\nLet V be a vector space and W a subset of V.  Then W is a vector subspace if and only if\n\n1) W is nonempty,\n2) for every x in W and scalar c, cw is in W, and\n3) for every x, y in W, x + y is also in W.\n\nExamples:\n\nA) Let P3 be the polynomials of degree at most 3.  Then P2 (which is the set of polynomials of degree at most 2) is a vector subspace of P3.\n\nB) Let P3 be as above, and let W = { f in P3 : f(7)=0 }.  Then W is a subspace of P3.\n\nC) Consider R^4 = { (x,y,z,w) : x,y,z,w in R }.  Let U = { (x,y,z,w) in R^4 : x + y = 0, z = w }.  Then U is a subspace of R^4.\n\nNonexamples:\n\nA) Consider R^2 = { (x,y) : x,y in R } and D = { (x,y) : x^2 + y^2 <= 1 }.  So R^2 is the real plane and D is the disk of radius 1 centered at the origin.  D is not a subspace of R^2 since it is not closed under addition or scalar multiplication.  For instance (0,1) is in D, but (0,1) + (0,1) = (0,2) is not.  Also, if |c| > 1, then c (0,1) is not in D as well.\n\nB) Consider P3 as above.  Then the subset S = { f in P3 : f(7) = 1 } is not a subspace.  I'll let you figure out why.\n\n\nYou can verify that all these are subspaces or not using the theorem above.  I hope this helps you out.  Cheers!
20441
Linear algebra deals with linear vector spaces. Operations on linear vector spaces are limited to additon of two vectors and multiplication of a vector by a constant. Matrix algebra is a subset of linear algebra.\n\nIf X, Y, and Z are vectors, aX + bY = Z has meaning\nbut aX^2 + bY^2 = Z does not.
1048436
As a^2 is contained in F(a), clearly F(a^2) is contained in F(a). So we need to show the reverse inclusion.\n\nWe have the chain of fields F < F(a^2) <= F(a).\n\n(where I use <= and < for subset and proper subset containment) \n\nand hence [F(a) : F] = [F(a) : F(a^2)][F(a^2) : F]. Set n = [F(a) : F] and m = [F(a^2) : F]. The preceeding equation implies that m divides n. It will suffice to that m = n. Let f(X) = X^m + c_{m-1}X^{m-1} + ... + c_1X + c_0 denote the minimal polynomial of a^2 over F. Then f(X^2) = X^{2m} + c_{m-1}X^{2(m-1)} + ... + c_1X^2 + c_0 has a as a root. This implies that the minimal polynomial of a has degree less than or equal to 2m. So we get m divides n and n <= 2m. The only ways for this to occur are when m=n or when 2m = n. But n is odd, so n does not equal 2m. Hence m = n. This completes the proof.
1027338
(Remark: the fact that this is supposed to work with any field K is a hint that our matrices - elements of GLn(K) - can't be too complicated!  In fact we can get away with using only 0s and 1s.)\n\nThe key is to think about the elements of G, call them g1, g2, ..., gn, also as permutations of the set { g1, g2, ..., gn } itself:  any element gi permutes { g1, g2, ..., gn } to { g1 gi, g2 gi, ..., gn gi } (verify that that really is a permutation of the set!).\n\nThis is the key to showing that every finite group of order n is isomorphic to a subgroup of Sn, the symmetric group of order n!.  But Sn itself is isomorphic to a subgroup of GLn(K), namely the subgroup of "permutation matrices", which have exactly one 1 in every row and column and 0s everywhere else.
204094
ok, i think i have an idea for this one. \nSketch of a proof:\nSuppose that a is an element of order k. Also suppose that f is an isomorphism. Then we know that f(a*a*a...k times)=f(a)*f(a)*f(a)*...k times (homomorphism definition). We know that f(a)^k is the identity because isomorphisms take the identity to the identity. Now see that k must be the first number with the quality f(a)^k=e. This comes as a result of the isomorphism having a kernel with one element. If we suppose for the sake of contradiction that f(a)^m=e with m<k, then we would have f(a*a*a*a...m times)=e, so that means that f(a^m)=e at the same time as f(a^k)=e with a^m not equal a^k, this is a contradiction.\n\nQED...ish
139989
Here is a web site that gives a nice example that might be helpful. Unfortunately, I am not an expert here so i can't give a quick answer, even if one exists. I suspect that one might have a group that is isomorphic to a Galois group, and not know it.
277316
let |f|=k (some constant).let u,v be the real & img parts of 'f'.\n            the we have |f|=u^2+v^2, u^2+v^2=k^2\n      differentiating partially,wrt to x&y ,and \n\n    USING CAUCHY-REI EQ'S FROM TWO RESULTING EQ'S \n\nu finally get     (u^2+v^2)(du/dx)=0.\n                 &  (u^2+v^2)(du/dy)=0. \nfrom above two eq's   we have \n                     k^2=u^2+v^2=0  implies u=0&v=0\n\n                  so 'f' is constant.
302837
Yes, you can, but the representation of the vectors is going to be position dependent. So you won't be able to add a vector at the origin and another vector at another point. You would first have to 'parallel transport' one vector to wherever the other is and then add them. This type of thing is usually done in more generality in differential geometry
260979
Assume the vectors (x,y,z) and (u,v,w) are elements of this set and check that they are closed under the vector space operations (that under those operations, the resultant elements also satisfy the equation).\nAs a matter of fact, as a subset of R^3, it is not a subspace as it does not contain the 0 vector. However, it is a 2-dimensional vector space if you give it its own coordinate system. It is properly an affine subspace of R^3.
336113
S3 is isomorphic to the dihedral group of order 6, which consists of maps over the equilateral triangle with unique sides. There are only two basis elements, the flip over a central axis, and a single rotation. All of the other 4 maps can be written as products of these two actions. Since a homomorphism is completely defined by its action on the basis, we have 2! automorphisms (ways of mapping the basis into itself).
460866
Not true. It is true for continuous homomorphisms. In that case a homomorphism is of the form f(x)=ax for some constant a. THen a^n=1, so a=+-1.
419013
This is a cute one. Suppose A is the linear operator corresponding to the matrix A. \nThe equation implies A^2 - A = 0 which implies (A - I)A = 0 where I is the identity of the image of A. A could be the 0 operator; you have not said otherwise. It doesn't matter, because 0 is trivially diagonalizable. Suppose A is not the 0 operator (You can find an easy example in operators over R^2). Then A - I = 0 which implies A is the identity for some basis of its image which is the same as saying the matrix of A is diagonalizable.\nNote that the above is an equation of operators, and as was said, the I is not necessarily the identity of the original vector space. For example, a projection onto the x-axis also satisfies the above conditions and is diagonalizable. The identity in that case is over R.\nThis is cute because it has the analogue x^2 = x which has the only solutions 0 (the 0 operator) and 1 (the identity over R) over the reals. Over higher dimensions, this simple algebraic fact tells us a lot more.
615433
The notion of a Vector space, a mathematical system of a certain kind, has developed into one of the richest concepts in both pure and applied mathematics.\n\nThe concept of vector space pervades mathematics, engineering and the sciences. The questioner is surely familiar with vectors in the form of directed line segment or columns of numbers studied in calculus and mechanics. \n\nThe scope of a general notion of vector is much broader than these examples. For example, functions, matrices and the operation of differentiation will turn out to be vectors.\n\nVector spaces turn up in a wide variety of studies. For example, vector space arise naturally in the study of solutions of systems of equations, geometry in 3-space, solutions of differential equations, finite elements, discrete and continuous Fourier transforms, quantum mechanics, approximation theory and the study of a wide variety of other mathematical and scientific disciplines
837082
Hint: Let C=X&#92;O and let f(x)=dist(x,C). The other way is trivial.
814015
Let us say you have a certain coordinate system C with the basis vectors:\nex= (1,0) and ey=(0,1)\n. ey↑\n. . .C →\n. . . . ex\nand you want to express points according to another coordinate system C' with the basis vectors\ne'x = (-1,0), and e'y =( 0,-1), the coordinates of the new basis vectors is according to C\n. ex'0(-1,0)\n←C'\n. . .↓ey'=(0,-1)\n\nThe relationship between the (x,y) in C and (x',y') in C' can be expressed by a Matrix mapping.\n. . . . . . .(-1 o) ..(x)\n(x',y') =  (0, -1) *(y) \n\nremember it is matrix multiplication. This was of cause a very simple example\n\nif we take a point (5,-1) in C its coordinates will be:\n\n. . . . . . . . . . . . . . (5)\n. . . (x',y') = (-1,0)*(-1) = (-5,1)\n. . . . . . . . .(0 ,-1)\n\naccording to the new basis ex',ey' in C'\nI hope its might have helped you \nthe matrix I have use is of cause \n\nE= (-1, 0)\n. . .(0 ,-1)\n\nI this simple situation I could also have used this very simple matrix = (-1,-1)\nand then the mapping would have looked like this:\n\n. . . . . .(-1)\n(x',y' )=(-1)*(x.y)\n\nAgain remember its matrix multiplication.
182274
...found first counterexample to Dowker's conjecture that normality in topological spaces is preserved under cross product with the closed unit interval of the reals.\n\n...assuming the continuum hypothesis proved that the box topology on the product of countably many copies of the reals is paracompact.