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

#### Fernando Revilla

##### Well-known member
MHB Math Helper
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.

#### Fernando Revilla

##### Well-known member
MHB Math Helper
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.