what is single valued decomposition?
111130
Try theese sites\n\nhttp://public.lanl.gov/mewall/kluwer2002.html\nhttp://www.cs.cmu.edu/afs/cs/project/jair/pub/volume4/hogg96a-html/node11.html
325892
= 1-{1-[1-(0)]}\n= 1-{1-[1]}\n= 1-{0}\n= 1
119401
Yes, it's a function.  Each x-value has only 1 y-value.
469250
Discrete Case:\nE[X] = Sigma_{i=1}^n x_i*p(X = x_i) \n\nwhere Sigma_{i=1}^n p(X = x_i) = 1\n\nContinuous Case:\nE[X] = int(x*p(X = x)dx,-infty,infty)\n\nThink of it as a weighted *average*.
929836
v = b - u/k\nv - b = u/k\nk(v-b) = u
940114
I have no idea what physical domain is. If you're talking about function domains, it's all input values that give you a real output, as in all x values that give a y value.
224978
associative, commutative, identity, inverse
599592
At first I thought you were wanting the orthogonal matrices, but your example shows you are not. Orthogonal matrices are those whose inverse is their transpose. Yours are ones where the elements of the transpose are the inverse of the elements of the original.\n\nCertainly such a matrix has to have +-1 on the diagonal. Your matrix is singluar and any 2x2 matrix with your property and all 1 on the diagonal will be also. This means that one eigenvalue is 0 and the other is 2.\n\nI don't see anything special for 3x3 offhand, but I will look further.
796566
A function relates each of its inputs to exactly one output. That is for each value of x in the domain, there must be one and only one y.\n1) 2) are OK for each x in R there is exactly one y (and vice versa)\n3) Not OK.  Take y=-2 and y=2. Both have x=4\n4) OK. There are x-values that lead to the same y, but that is allowed.\n6) Not OK. Take x=2, y>4. There are an infinite number of y-values.
109963
The Identity Property
199213
You probably mean eigen-values. We say that x is an eigen value for the square matrix A if there is a vector v such that Av=xv. Eigen-values are usually found by solving det(A-xI)=0\nwhere I is the identity matrix.
92762
A = P + Prt\n\n==> P(1 + rt)\n\n==> P = A/(1 + rt)\n\n<done>
823012
Either one is possible. Consider y=x² (one critical value, which is not an inflection point) and y=x³ (one critical value, which IS an inflection point).
481388
The only way you'll be able to get a solution to 3 linear equations with 2 unknowns is if there's a dependence relation on the rows or equivalently the columns.  Otherwise, there will be no solution.\n\nI haven't checked the determinant of your matrix but if there are no dependence relations on the rows or columns of the matrix that you gave, Gaussian Elimination will give you a single solution.  Alternatively, for small matrices like the one you have you can use Cramer's Rule more efficiently.\n\nEither way, your coefficient matrix seemes to represent 3 equations and 3 unknowns so I would proceed with Gaussian Elimination.
216212
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.....
1154622
a variable quantity that can be resolved into components
725218
Modified Gram-Schmidt algorithm\n------------------- -----------------------\nA  is m x n,  with m ≥ n  s.t. rank(A) = n\n\n(a) For i = 1, ... , n \n`````````w_i = a_i\nEnd (a) For\n(b) For i = 1, ... , n\n`````````r_ii = ||w_i||\n`````````q_i = (w_i)/(r_ii)\n`````````(c) For j = i + 1, ... , n\n``````````````````r_ij = <q_i, w_j>\n``````````````````w_j = w_j - (r_ij)(q_i)\n`````````End (c) For\nEnd (b) For