MHB Amy's question at Yahoo Answers (Self adjoint operator)

  • Thread starter Thread starter Fernando Revilla
  • Start date Start date
  • Tags Tags
    Operator
AI Thread Summary
The discussion focuses on proving that the linear transformation T defined on the vector space V = P2(C) is self-adjoint. The inner product is defined as <a_0 + a_1x + a_2x^2, b_0 + b_1x + b_2x^2> = a0b0 + a1b1 + a2b2, with the b's being conjugates. To show T is self-adjoint, it is necessary to demonstrate that <T(p), q> = <p, T(q)> for all p, q in V. The calculations reveal that both sides of the equation yield the same result, confirming that T is indeed self-adjoint. This conclusion is essential for understanding the properties of linear transformations in functional analysis.
Fernando Revilla
Gold Member
MHB
Messages
631
Reaction score
0
Here is the question:

Let V = P2(C) with inner product

< a_0+a_1x+a_2x^2 , b_0+b_1x+b_2x^2 > = a0b0 + a1b1 + a2b2 (with the b's being conjugates)

Show that T:V--->V define by T(a_0+a_1x+a_2x^2) = - ia_2 - a_1x + ia_0x^2

is self adjoint

Here is a link to the question:

Show a linear transformation is self adjoint? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
Mathematics news on Phys.org
Hello Amy,

For all $p,q\in V=P_2(\mathbb{C})$ we need to prove $$<T(p),q>=<p,T(q)>\quad\mbox{ (definition of self adjoint operator)}$$ Denote $p(x)=a_0+a_1x+a_2x^2$ and $q(x)= b_0+b_1x+b_2x^2$. Then, $$<T(p),q>=< - ia_2 - a_1x + ia_0x^2, b_0+b_1x+b_2x^2>=\\-ia_2\overline{b_0}-a_1\overline{b_1}+ia_0\overline{b_2}$$ $$<p,T(q)>=<a_0+a_1x+a_2x^2,- ib_2 - b_1x + ib_0x^2>=\\a_0\overline{(-ib_2)}+a_1\overline{(-b_1)}+a_2\overline{(ib_0)}=ia_0\overline{b_2}-a_1\overline{b_1}-ia_2\overline{b_0}$$ That is, or all $p,q\in V=P_2(\mathbb{C})$ we have proven $<T(p),q>=<p,T(q)>$ as a consequence, $T$ is self adjoint.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top