Show that the bracket $⟨p(z), q(z)⟩ = p(0)\overline{q(0)}+ p(2)\overline{q(2)}$ defines an inner product on V .

Question

Given that V is a the vector space of polynomials $f(z) = az + b$ for complex a, b. Show that the bracket $⟨p(z), q(z)⟩ = p(0)\overline{q(0)}+ p(2)\overline{q(2)}$ defines an inner product on $V$ .

Answer 1

An inner product on the vector space $V$ needs to satisfy three conditions:

1. Conjugate symmetry: $\langle x,y\rangle=\overline{\langle y,x\rangle}$ for all $x,y\in V$.
2. Linearity in the first argument: $\langle kx,y\rangle=k\langle x,y\rangle$ and $\langle x+y,z\rangle=\langle x,z\rangle+\langle y,z\rangle$ for all $x,y,z\in V$ and $k\in\mathbb{C}$.
3. Positive-definite for $\langle x,x\rangle$: $\langle x,x\rangle>0$ for all $x\in V\setminus\{0\}$.

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The first condition can be shown to hold with the help of a few properties of complex conjugation. In case you are unfamiliar with them, I will denote real part with $_r$ and imaginary part with $_i$ and quickly prove them:

1. $\overline{uv}=\bar{u}\cdot\bar{v}$
   Proof:
   $\begin{align*}\overline{uv}&=(uv)_r-(uv)_i\cdot i=((u_r+u_i\cdot i)((v_r+v_i\cdot i))_r-((u_r+u_i\cdot i)((v_r+v_i\cdot i))_i\cdot i\\&=(u_rv_r+u_rv_i\cdot i+u_iv_r\cdot i-u_iv_i)_r-(u_rv_r+u_rv_i\cdot i+u_iv_r\cdot i-u_iv_i)_i\cdot i\\&=u_rv_r-u_iv_i-(u_rv_i+u_iv_r)\cdot i=u_rv_r-u_rv_i\cdot i-u_iv_r\cdot i+u_iv_i\cdot i^2\\&=u_r(v_r-v_i\cdot i)-u_i\cdot i(v_r-v_i\cdot i)=(u_r-u_i\cdot i)(v_r-v_i\cdot i)=\bar{u}\cdot\bar{v}\end{align*}$
2. $\overline{u+v}=\bar{u}+\bar{v}$
   Proof:
   $\begin{align*}\overline{u+v}&=(u+v)_r-(u+v)_i\cdot i=u_r+v_r-(u_i+v_i)\cdot i=u_r-u_i\cdot i+v_r-v_i\cdot i=\bar{u}+\bar{v}\end{align*}$
3. $\bar{\bar{z}}=z$
   Proof:
   $\begin{align*}\bar{\bar{z}}&=(\bar{z})_r-(\bar{z})_i\cdot i=(z_r-z_i\cdot i)_r-(z_r-z_i\cdot i)_i\cdot i=z_r-(-z_i)\cdot i=z_r+z_i\cdot i=z\end{align*}$

Now we can show that the first condition is fulfilled. Set $p(z)=az+b$ and $q(z)=cz+d$. Then $$\langle p(z),q(z)\rangle=p(0)\overline{q(0)}+p(2)\overline{q(2)}=b\cdot\bar{d}+(2a+b)(\overline{2c+d})$$ $$\langle q(z),p(z)\rangle=q(0)\overline{p(0)}+q(2)\overline{p(2)}=d\cdot\bar{b}+(2c+d)(\overline{2a+b})$$ Now, $$\overline{\langle q(z),p(z)\rangle}=\overline{d\cdot\bar{b}+(2c+d)(\overline{2a+b})}$$ Using the first property, we get: $$=\overline{d\cdot\bar{b}}+\overline{(2c+d)(\overline{2a+b})}$$ The second property then gives: $$=\bar{d}\cdot\bar{\bar{b}}+(\overline{2c+d})(\overline{\overline{2a+b}})$$ Finally, the third property allows us to obtain: $$=\bar{d}\cdot b+(\overline{2c+d})(2a+b)=\langle p(z),q(z)\rangle$$ Thus proving that $\langle p(z),q(z)\rangle=\overline{\langle q(z),p(z)\rangle}$.

The remaining two conditions can be proven using similar properties to these ones.