Suppose that s is a subset of the positive real numbers r...

Suppose that S is a subset of the positive real numbers (R) whose members obey addition defined in the following way: A + B = √(AB) (for all A, B ∈ S) Show that + (using √) satisfies the addition properties for a vector space (which actually means that S is an Abelian group). You will have to find the "additive" identity (show this calculation explicitly; and no, it is not the real number zero, 0) as well as an expression for the "additive" inverse of a typical element of the set. Then choose any 5 numbers in S and calculate their specific inverses just for fun. Then demonstrate that + is associative (fun exercise), as well as commutative (easy). Finally, why is the operation closed? We all know from Euclidean geometry that two points define a unique linear function. By simple antidifferentiation principles, one needs three noncollinear points to define a unique quadratic function, and by induction, one needs n + 1 points to define a unique nth degree polynomial function. Suppose that you are given (1, 2), (-1, 6), and (2, 3). Using matrix methods for solving systems of equations, find the unique quadratic function that passes through these given points. No fair using your calculator's regression capabilities or matrix capabilities. Now suppose that you are given (1, 0), (-1, -24), (2, 0), and (4, 6). Using matrix methods for solving systems, find the unique cubic function that passes through these given points. No fair using your calculator's regression capabilities or matrix capabilities.

Transcript

00:01 The question says that we is a finite dimensional vector space with end of product defined on b by b and a linear transformation t is self -adjoined.

00:11 If it satisfies this condition, we need to prove that it is self -adjoint if and only if the matrix with respect to an orthonormal basis of v is symmetric.

00:25 So let us see how can we solve this question.

00:28 So what we need to do? let us see.

00:33 So we will start with the first one.

00:35 So i will consider an orthonormal basis first.

00:38 Okay, so let's be an element r e1, e2 and so on up to en.

00:46 Okay.

00:47 Be orthonormal, sorry, orthogonal basis, okay, of b with respect to b, okay.

00:59 So then i can say that t of ei from the definition i am defining p of ai to be equal to summation over j a of ij ej okay and then you see t of b of t of t of e i e k will be same as b of e i t of e k okay because we consider that the matrix it's sorry we consider that t is self -adjoint then it will satisfy this condition okay we have to show that the matrix is symmetric so here what i will get i will get b of now t of i will become this summation over j a i j j e j comma e k and here i will get v of e i summation over j a of kj e j okay so this is what i am getting here so this will implies what so this implies that a of i k is nothing but a of k i conjugate okay okay so this says that e i and e k sorry e j and e k are orthogonal if j is not equal to k okay so what does this mean this means that gb is self adjoined correct now we have b as orthogonal basis in which the matrix is self -adjoint okay, so now what i will do? gov, consider, b of t u comma b, which will be equal to b of u t b from the definition of self -aturn.

03:18 So this implies that you can be written as summation of i going from 1 to n, ui, ui, ei okay and we will be equal to summation over one to n viei correct then what i will get i will get t of u to be equal to what it will become summation over i from 1 to n of u i t of ei since t is linear transformation you should use the property we have used here so t is linear similarly here i will get p of b to be equal to summation over here it was i it is so j summation over j b j b of b j okay now what i will do i will consider b of t u b so b of tuv will be this now this is the converse part so this was forward part which we have proved that this is self adjoined and then we are considering it has to be self adjoined and we'll consider sorry in this part we need to prove that linear transformation is self -adjoint when the matrix is symmetric okay so this will become b of a summation over i u i i i i okay and this will be summation over j, uj, ej, correct.

05:05 So this can be written as double summation over i and j, ui, vj, v of, t of ei, ej, okay? so from here what i will say.

05:20 So this is my b of t ub, let's understand this is equation one.

05:24 So i will consider v of ugv.

05:27 Okay, so this will be nothing but b of summation u i .ei and here i will get summation b .j t ej okay so this is what i will get so if i solve i'm going to get double summation over i j u i j u i b j b of ei comma t e j okay so this is two okay, so we need to show that to show that one equals to two.

06:07 So that is to show, that is to show that v of t -e -i -e -j is same as b of e -e -i -t -e -j.

06:19 Okay, so this implies what, v of d -e -i -e -j, it is nothing but a -i -j...

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