Determine the basis of the following subspace of R3.\nall vectors of the form (a,b,c) where b=a+c.\nThe answer is {(0,1,1)(1,1,0)}. HOW DO YOU GET THIS ANSWER ??
125303
R³ basis:{(1,0,0), (0,1,0), (0,0,1)}\n\n=> (a,a+c,c)\n\n= (a,0,0) + (0,a+c,0) + (0,0,c)\n\n= a*(1,0,0) + (a+c)*(0,1,0) + c*(0,0,1)\n\n= a*(1,0,0) + a*(0,1,0) + c*(0,1,0) + c*(0,0,1)\n\n= a*((1,0,0) + (0,1,0)) + c*((0,1,0) + (0,0,1))\n\n= a*(1,1,0) + c*(0,1,1)\n\n=> {(1,1,0), (0,1,1)} is a basis of all vectors of the form (a,a+c,c)
550930
Since you know elements of the center commute with *everything*, try a couple of possibilities. Suppose [a b // c d] is in the center. Multiplying that by [1 0 // 1 1] on each side and equating them shows you that b=0 and a=d. Multiplying by [ 1 1 // 0 1] on each side and equating them shows you that c=0 and a=d. So all you're left with is multiples of the identity, which obviously work.
433746
There is several way to solve the problem, but I think it is easies, to find the parametric representation, and then use, that the scalar-product of the orientations-vectors is zero, if the line are perpendicular, and the orientations-vectors is linear depended if parallel.\n\nL1:  y = -½x-3 => (x,y)= (1,-½)*t+(0,-3)\n4x-2y+5 = 0 <=> y=2x+5/2=>\nL2: (x,y) = (1,2)*t + (0+5/2)\n\nbecause (1,-½)*(1,2)=1-1=0 the lines are perpendicular
589622
It's too cumbersome to rewrite it all here, but check out http://www.euclideanspace.com/maths/algebra/vectors/angleBetween/index.htm\n\nThat's 2D, but the formula is the same for 3D.
510108
slider is correct.  \n\nRemember the distributive property?\na(b+c) = ab + ac\n\nThis is using it in reverse.  \n\nIn part A, notice that there is an a in each term, so you can factor out the a's.  \n3a² + 2a = (3a)*a + 2 * a = a(3a + 2).\n\nIn part B, there aren't any letters in common but you can factor out numbers too.  \n12d²+6 = 6*2d² + 6*1 = 6(2d²+1)\n\nThe other parts use similar reasoning.
555908
These vectors are expressed in terms of polar coordinates (r, theta).  To add them, you'll first need to express them in terms of rectangular (also known as Cartesian) coordinates, (x,y), using the transformations:\n\nx = r * cos(theta)\ny = r * sin(theta)\n\n(You can't just add the angles as suggested by the first answerer.  This is simply wrong.)\n\nYour first vector then has Cartesian representation:\n\n(10, 135 degrees) => (10*cos(135), 10*sin(135)) = (-7.071, +7.071)\n\nand the second has Cartesian representation:\n\n(5, 45 degrees) => (5*cos(45), 5*sin(45)) = (3.536, 3.536)\n\nNow, to add the vectors together, simply add the x components and y components of the two vectors:\n\n(-7.071+3.536, 7.071+3.536) = (-3.536, 10.607)\n\nAnother way to think about this is to recognize that:\ncos(135) = -cos(45)\nsin(135) = sin(45)\n\nThis means we can add the x and y coordinates while still expressed in terms of trig functions:\n\n\n(10*cos(135), 10*sin(135)) + (5*cos(45), 5*sin(45)) = \n(-10*cos(45), 10*sin(45)) + (5*cos(45), 5*sin(45)) = \n(-5*cos(45), 15*sin(45))
449214
HI: I tried to answer your email but it said that your email address has not been confirmed and so the email didn't go through.\n\nalpha and beta are scalers for the vector space V.  This problem is actually so easy that people don't believe what they've done.  You really just need to show that Span S behaves like V.  \n\nSpan S must be a vector space itself.  Two elements from the span is \nu=a1*v1 + a2*v2...ak*vk\nand \nw=b1*v1 + b2*v2...bk*vk\n\nshow that alpha*u+beta*w is in Span S where alpha, beta are scalers for your vector space.\n\nalpha*u+beta*w = (alpha*a1+beta*b1)*v1 + ....\nis a linear combination of the vectors in Span S, thus it's in Span S.  \n\nSo, Span S is not empty and is closed, thus its a subspace.
554392
Firstly, there is a formula. However, its quite easy to forget - so why not learn how to actually come up with it?\nSuppose you are given coordinates (a,b) in the first system. That really corresponds to the point a[2,1] + b[2,0]. Writing that in matrix form, we have: (// denoting next line)\n[2 2 // 1 0][a // b]\nNow, we want to express it as c[-2,-1]+d[0,1]. Writing that in matrix form we have:\n[-2 0 // -1 1][c // d].\nSo how do we work out [c // d]? Just multiply on the left by the inverse of that second matrix.\nSo it would be [-2 0 // -1 1]^(-1) * [2 2 // 1 0].\nThat turns out to be [-1 -1 // 0 -1].
153521
Let the two sides be formed from three points, (a,b), (c,d), (e,f). So: Side A is (a,b) to (c,d) and Side B is (c,d) to (e,f).\n\nMidpoint of A: ([a+c]/2, [b+d]/2)\nMidpoint of B: ([c+e]/2, [d+f]/2)\nThe vector formed is:\n([c+e]/2 - [a+c]/2, [d+f]/2 - [b+d]/2) =\n([e-a]/2, [f-b]/2)\nThe third side is formed from (e,f) to (a,b), or\n([e-a], [f-b])\nIt's easy to see then that they are parallel.
461901
This is impossible.  For suppose that S={v_1,...,v_k} is a linearly dependent subset of R2.  Then there are scalars r1, ..., rk, not all zero, such that\n\nr1 v_1 + ... + rk v_k =0\n\nIf T: R2 -> R2 is a linear transformation, then\n\n0 = T(0) = T(r1 v_1 + ... + rk v_k)\n\n             = r1 T(v_1) + ... + rk T(v_k)\n\nTherefore, the set\n\nT(S) = {T(v_1), ... , T(v_k)}\n\nis linearly dependent, since one of the ri's is nonzero.
198209
Assume there is a pair a_i, b_j with a non zero product.\nWrite fg= (a_i x^i + other terms) (b_j x^j + other terms) =\na_i b_j x^(i+j) is non zero.\n\nContradiction.
219910
r u 4 r3al ?!! \n\nwhat mapping ? what equasion?  deckard maybe ? \n\nmore info please.
328490
It has dimension 2.\nI will not bother to prove that this subset is indeed a subspace except to note that (0,0,0) is an element.\nThis is most readily seen by constructing a basis.\nI propose that {(1,1,0),(0,1,1)} is a basis for this space.\nLet (a,b,c) be an element of this space. Then (a,b,c)=(a,a+c,c) =  a*(1,1,0) + c*(0,1,1)\nThis our proposed basis Spans this subspace.\nNow I think we can clearly see that our proposed basis is linearly independant since:\nIf:   a*(1,1,0) + c*(0,1,1) = (0,0,0)\nthen: 0 = a, 0 = a+c and 0 = c by equating the components.\nThus we have a linearly independant spanning set.\nThus we have a basis!\nSince our basis is of magnitude 2, our space must be of finite dimension 2.
220859
Let a(P,Q) be the direction angle of the vector going from point P to Q measured counterclockwise from AB (so that a(A,B)=0). Then the direction angles of the bisectors going through B and C respectively are:\n\na(B,C)/2\n(a(B,C)-pi+a(A,C))/2\n\nand the angle between the two (A1)\n\nA1 = a(B,C)/2 - (a(B,C)-pi+a(A,C))/2 = pi/2 - a(A,C)/2\n\nIn terms of the original angle at A (which is a(A,C)) this is:\n\nA1 = pi/2 - A/2\n\nThat is, the recursion relation is:\n\nA_(n+1) = pi/2 - A_n/2\n\nSubstituting A_n = pi/3 + X_n, gives:\n\nX_(n+1) = - (1/2)*X_n\n\nSo clearly X_n = (-1/2)^n * Constant\n-->\nA_n = pi/3 + (-1/2)^n * Constant\n\nSince A_0 = A , the constant is: A-pi/3\n\n--> answer 4\n\nTo guess the answer without doing any work, just note that A_0 must be A, so only 3 and 4 are eligible. It's easy to see A_n can't be a constant just by sketching a triangle with a very obtuse angle at A.
518462
If R := I - BA (:= means "equals by def.") then, by rearranging (BA = I - R, so A-1 B-1 = (I - R)-1), we get \n\nA-1 = (I - R)-1 B\n\nand the matrix (I - R)-1 can be expanded in a geometric series (just as, for numbers, 1/(1 - r) = I + r + r^2 +...), that is, \n\n(I - R)-1 = I + R + R^2 + ...\n\nbecause B approximates A-1, so R is close to 0. (More precisely, the expansion holds if each eigenvalue of R is less than 1 in absolute value.) Putting the two together, and now writing the inverse properly as A^(-1) instead of A-1 (which might be mistaken for A - I), we get \n\nA^(-1) = (I + R + R^2 + R^3 + ...)B\n\nas was required. This formula is exact, so this is not actually an iterative process, by the way (an iterative process would consist in getting successive approximations to A-1 by repeating some transformation again and again).\nPls award Best Answer :))
390543
Let [a, b, c] denote the vector <4a+3b, 0, a+b+c, c-2a>. Vectors of this form are a subset of R^4. In order to show that they form a subspace, we must show that they are closed under linear combinations. It suffices to show that they are closed under vector addition and scalar multiplication. [a, b, c] + [a', b', c'] = <4a+3b, 0, a+b+c, c-2a> + <4a'+3b', 0, a'+b'+c', c'-2a'> = <4(a+a') + 3(b+b'), 0, (a+a') + (b+b') + (c+c'), (c+c') - 2(a+a')> = [a+a', b+b', c+c'], so the set is closed under addition. Similarly, k[a, b, c] = k<4a+3b, 0, a+b+c, c-2a> = <4ka+3kb, 0, ka+kb+kc, kc-2ka> = [ka, kb, kc], so the set is closed under scalar multiplication.\n\nThe spanning set is {[1, 0, 0], [0, 1, 0], [0, 0, 1]}. Writing this out in full: {<4, 0,1, -2>, <3, 0, 1, 0>, <0, 0, 1, 1>}
87689
Yes, but you have to be more clear what you want in general.\n\nSuppose we are working in n-dimensional Euclidean space. Suppose also that you have a hyperplane going through the origin. If you take n-1 independent vectors in that hyperplane and do a determinant construction similar to that for the cross product, you will get a vector normal to your hyperplane. But you will need n-1 *independent* vectors and you will have to do a nxn determinant (with the top row consisting of the unit tangent vectors along the axes).\n\nIf you don't have a full hyperplane, you will be wanting a basis for the normal subspace. Other than the previously mentioned Graham-Schmidt process, I don't know a good way to do that. Hmmm.....