Linear Algebra Answers

At linear programming for an unbounded region,maxima may exist or may not
z= ax+by and it has some constraints. and a feasible region that is unbounded. and M be the initial maxima. if ax+by>M . if the line ax+by=M does not overlap that previous feasible region, the initial maxima is the maxima. and if overlaps, then no maxima . why overlapping causes no maxima and without overlapping assures maxima?

Let V be the set of all functions that are twice differentiable in R and S={cosx,sinx,xcosx,xsinx}. a)Check that S is a linearly independent set over R.(Hint: Consider the equation a0cosx+a1sinx+a2xcosx+a3xsinx. Putx=0,π,π 2 ,π 4 ,etc.and solve for ai.) b) Let W=[S]and let T:V→V be the function defined by T(f(x))=d2 dx2(f(x))+2d dx(f(x)). Check that T is a linear transformation on V.
c)Check that T(W)⊂W.(7)d) Write down the matrix of T on W w.r.t. the basis S.(2)e)Is the matrix of the linear operator T non-singular? Justify your answer.

Which of the following statements are true and which are false? Justify your answer with a short proof or a counter example. i) ii) iii) iv) v) vi) vii) The function f:R→R define dbyf(x)=cosxis1-1. The operation∗defined by x∗y=log(xy)is a binary operation on S,where S is the set{x∈R|x>0}. The set{(x1.x2,...,xn)|x1,x2,...,xn∈R,x1=2x2+3}is a subspace of Rn. There is no 7×5 matrix of rank 6. If V and V are vector spaces and T:V→Visali near transformation,then whenever u1,u2,...,uk are linearly independent, Tu1,Tu2,...,Tuk are also linearly independent. If Visa vector space and T:V→Visali near operator with det(T)=0,then T is not diagonalisable. The degree of the minimal polynomial of a 3×3matrix is at most 2. viii) Forany 2×2 matrix A,Adj(At)=(Adj(A))t. ix) x) The only matrix which is both symmetric and skew-symmetric is the zero matrix. There is no co-ordinate transformation that transforms the quadratic form x2+y2+z2 to the quadratic form xz+yz.