Isometry for polynomials of degree at most $n$

isometrylinear algebravector-spaces

Let $P_n$ be the set of all polynomials of at most $n$ degrees such that $p(x)=a_n x^n +\cdots + a_1x+a_0$ and $q(x)=b_n x^n + \cdots + b_1x + b_0.$ Let $\langle p, q\rangle = a_n b_n + \cdots + a_1 b_1 + a_0 b_0$. For every $q \in P_2$ define the mapping $F_q : P_3 \to P_5$ such that $F_q(p) = qp.$ For what polynomials $q$ the mapping $F_q$ is an isometry from $(P_3, \langle \cdot, \cdot\rangle)$ to $(P_5, \langle \cdot, \cdot\rangle)$?

Every $q$ has the following form $q(x)= b_2x^2+b_1x+b_0$ so I have that $F_q(p) =(b_2x^2+b_1x+b_0)(a_3x^3+a_2x^2+a_1x+a_0)$ which indeed results in a polynomial of degree $5$, but I'm not sure how to find what polynomials makes this map isometric. I would need to have that $$\langle F_q(p), F_q(p')\rangle_{P_{5}} = \langle p, p'\rangle_{P_3}$$ where $p,p' \in P_3$? Any hints would be appreciated.

Best Answer

This is my approach

The mapping $F_q$ is an isometry from $(P_3, \langle \cdot, \cdot\rangle)$ to $(P_5, \langle \cdot, \cdot\rangle)$ mean $\forall p,p' \in P_3$ $$\langle F_q(p), F_q(p')\rangle_{P_{5}} = \langle p, p'\rangle_{P_3}$$

So, I think, we can make it easy by choose particular values of $p,p' \in P_3$ and make equations of $b_0,b_1,b_2$

  • With $p=1,p'=1 \rightarrow 1=b_0^{2}+b_1^{2}+b_2^{2}$
  • With $p=1,p'=x+1 \rightarrow 1=b_0^{2}+b_1^{2}+b_2^{2}+2b_1b_2$
  • With $p=1,p'=x^{2}+1 \rightarrow 1=b_0^{2}+b_1^{2}+b_2^{2}+b_0b_2$
  • With $p=x,p'=x+1 \rightarrow 1=b_0^{2}+b_1^{2}+b_2^{2}+b_0b_1+b_1b_2$

Therefore, $b_0b_1=b_1b_2=b_0b_2=0$, $1=b_0^{2}+b_1^{2}+b_2^{2}$ $\rightarrow (b_0,b_1,b_2)$ must have 2 values = 0, and the other equal 1 or -1

So, q can be $1,-1,x,-x,-x^{2},x^{2}$ ( Test back and it's true )

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