show that all elements of original diagonal of symetric positive definit matrix are positive(in details)?
100903
If A is an n-by-n symmetric positive definite matrix with real entries, that means ((v transpose)Av) > 0 for nonzero real n-by-1 column vectors v.\n\nConsider the 3x3 case with A equal to\n\n[ a b c ]\n[ b d e ]\n[ c e f ]\n\nLet u, v, and w be the nonzero column vectors\n\n[ 1 ]\n[ 0 ]\n[ 0 ]\n\n[ 0 ]\n[ 1 ]\n[ 0 ]\n\n[ 0 ]\n[ 0 ]\n[ 1 ]\n\nrespectively.  Then\n\n((u transpose)Au) = a > 0\n((v transpose)Av) = d > 0\n((w transpose)Aw) = f > 0\n\nCan you see how to generalize this for other values of n besides three and/or for matrices and vectors with complex entries?
481076
Since you already know what vectors are, take two vectors from this plane as a basis for all vectors in the plane. The easiest basis is just to take two sides from the triangle formed by the 3 given points. \nFrom your study of linear algebra, you may remember that one way of checking whether a set of n vectors was linearly independent was to take the n-volume of their parallelotope (their determinant). If the n-volume was 0, then obviously the set of vectors makes less than an n-dimensional parallelotope, so they can't span an n-dimensional space. We also define a vector v to be orthogonal to a vector w if their inner product is 0.\nThus, we are looking for the vector w = (w1, w2, w3) such that the determinant (u, v, w) is non-zero and u*w=0 and v*w=0. If we expand the determinant above along the unknown vector w, choosing the n-volume to be the same as the area of the parallelegram spanned by u and v and expand the dot products, we find we have system of 3 linear equations in 3 variables. The solution vector of this system is called cross product of u and v, the formula for which you can find in the source below. It is a vector perpendicular to u and v that has as magnitude the area of the parallelegram spanned by u and v.
735897
Let s be in S.  Since f is a homomorphism, f(r)=f(1)*f(r), thus \ns*f(r)=s*f(1)*f(r).  We rearrange and factor to get\n[s-s*f(1)]*f(r)=0.  The hypotheses imply that f(r) is not a zero divisor, so we conclude that s=s*f(1).  Similarly using f(r)=f(r)*f(1) we get s=f(1)*s, thus f(1) is an identity element.
331428
Can you show that T has a real eigenvalue? If so, let v be an eigenvector corresponding to this eigenvalue. Then, the line tv for all t in Reals is invariant under T :)
433823
Revolving about the x-axis yields a cone whose tip is at the origin and opens in the positive x direction.  Taking discrete x slices (dx) yields a disk whose volume is given by pi*(kx)^2 dx.  Integrating from 0 to h gives us (where "S" is used as the integration symbol):\n\nV = S pi*(kx)^2 dx\n= pi * k^2 S x^2 dx\n= (pi * k^2) * (x^3)/3\n\nEvaluating from 0 to h gives us\n= (pi * k^2) * (h^3)/3 - (pi * k^2) * (0^3)/3\n= (pi * k^2 * h^3)/3
744408
Say a[i,j] are the elements of A. The definition of the determinant is:\n\ndetA = Sum over P: (-1)^P * a[1,P(1)] * a[2,P(2)]*...* a[n,P(n)]\n\nwhere the sum is over all permutations P of (1,2,...,n) and (-1)^P is -1,+1 for odd and even permutations respectively. (P(i) is the image of i under P).\n\nNow let's say (for example) that matrix B is created multiplying the second row by a constant c and adding it to the first row:\n\na[1,i] --> a[1,i]+c*a[2,i]\n\nThen the determinant of this new matrix will be:\n\ndetB = Sum over P:\n(-1)^P * (a[1,P(1)]+c*a[2,P(1)]) * a[2,P(2)]*...* a[n,P(n)]\n\n= det A +\nc * Sum over P: (-1)^P * a[2,P(1)] * a[2,P(2)]*...* a[n,P(n)]\n\nbut the last sum is zero, because for each permutation P there is exactly one other permutation differing only by a transposition T of (1,2)\n\nP' = PT , P = P'T\n\nand P and P' have opposite parity so the two terms generated by each of these n!/2 pairs cancel each other --> detB = detA
102340
I assume you mean invertible, not convertible.\n\nIt depends on how detailed you want to get. If you do not want to dig in extremely deeply, you can just do it by construction. \n\n1. A symmetric matrix M has an orthogonal decomposition\n\nM=O'DO\n\nwhere O is an orthogonal matrix, O'O=I, the identity, and D is a diagonal matrix with eigenvalues of M down the diagonal.\n\n2. If M is positive definite, then the diagonal elements of D are >0\n\n3. Produce the matrix \n\nA=O'(D^(-1))O\n\n\n4. Verify that A is the inverse of M\n\ni.e., \n\nA*M=M*A=I
139869
[ 0 0 ]\n[ 0 0 ] has determinate 0, thus it is not invertible.  All nxn matrices are diagnalizable, but the zero matrix is already in diagnol form, having zeros everywhere except the main diagnol.  It is just has diagnol entries (0, 0).  Thus I would say it is not non-diagnol.
1017458
symetric properety...\n\n\nok heres why.....\n\n(pretend the equal signs are parallel line segments)\n\n===========================\n\n\ndo you see how you can split them dont the middle, thats called the symetric property
130487
it's delta-y divided by delta-x\n\n(5 - (-1))/((-1) - (-3))
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.
773251
The Wikipedia entry discusses the linear algebraic proof of the existence of the SVD factorization (if that is what you want)\n\nhttp://en.wikipedia.org/wiki/Singular_value_decomposition\n\nA few other easy-to-understand links:\n\nFrom netlib:\nhttp://www.netlib.org/lapack/lug/node53.html\n\nFrom Numerical Recipes:\nhttp://www.nrbook.com/a/bookfpdf/f2-6.pdf
724685
r-s = t\nadd s to both sides\nr = t+s
844003
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.
585433
The Jacobian determinant for the suggested transformation is:\n\n|d(u,v,w)/d(x,y,z)| = |y x 0 , z 0 x , 0 z y| = -2xyz\n==>\ndudvdw = dxdydz(-2xyz)\n==>\ndxdydz = -dudvdw/(2xyz) = -dudvdw/(2*sqrt(uvw))\n\n(The sign just depends on the order of variables and is not significant.) All variables are positive. The volume is the integral:\n\n&#92;int_1^9 du &#92;int_1^4 dv &#92;int_1^16 dw 1/(2*sqrt(uvw)\n=\n(1/2)\n* &#92;int_1^9 du/sqrt(u) \n* &#92;int_1^4 dv/sqrt(v)\n* &#92;int_1^16 dw/sqrt(w)\n=(1/2) * 2^3 * (3-1)*(2-1)*(4-1) = 24
823237
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
255796
a^4 is already positive; c^0 is 1, and b^-7 goes in the denominator so\n\n(a^4)/(b^7)