Proving $x_1y_1 +x_2y_2 + kx_1y_2 + kx_2y_1$ defines a scalar product in $\mathbb{R}^2$ if $|k|$ <1

Question

Prove $x_1y_1 +x_2y_2 + kx_1y_2 + kx_2y_1$ defines a scalar product in $\mathbb{R}^2$ if $\left|k\right|<1$

Using definition that scalar product is a bilinear function which satisfies linearity, symmetry and positive definiteness

Answer 1

To prove that $$\langle x,y\rangle_{k}=x_{1}y_{1}+x_{2}y_{2}+kx_{1}y_{2}+kx_{2}y_{1}$$ is an inner product we have to show three things. For all $x,y,z\in \mathbb{R}^{2}$ and $\lambda\in\mathbb{R}$ we have:

1) $\langle x,y\rangle_{k}=\langle y,x\rangle_{k}$.

2) $\langle \lambda x,y\rangle_{k}=\lambda\langle x,y\rangle_{k}$ and $\langle x+y,z\rangle_{k}=\langle x,z\rangle_{k}+\langle y,z\rangle_{k}$.

3) $\langle x,x\rangle_{k}\geq0$ and $\langle x,x\rangle_{k}=0$ if and only if $x=0$.

1) is obvious from the definition. For 2) note that $$\langle \lambda x,y\rangle_{k}=\lambda x_{1}y_{1}+\lambda x_{2}y_{2}+\lambda kx_{1}y_{2}+\lambda kx_{2}y_{1}=\lambda\langle x,y\rangle_{k}$$ and $$\langle x+y,z\rangle_{k}=(x_{1}+y_{1})z_{1}+(x_{2}+y_{2})z_{2}+k(x_{1}+y_{1})z_{2}+k(x_{2}+y_{2})z_{1}$$ $$=x_{1}z_{1}+x_{2}z_{2}+kx_{1}z_{2}+kx_{2}z_{1}+y_{1}z_{1}+y_{2}z_{2}+ky_{1}z_{2}+ky_{2}z_{1}=\langle x,z\rangle_{k}+\langle y,z\rangle_{k}$$ and finally by Cauchy-Schwarz we have that $|\langle x,(x_{2},x_{1})\rangle|=|x_{1}x_{2}+x_{2}x_{1}|\leq x_{1}^{2}+x_{2}^{2}$, hence $$\langle x,x\rangle_{k}=x_{1}^{2}+x_{2}^{2}+kx_{1}x_{2}+kx_{2}x_{1}\geq x_{1}^{2}+x_{2}^{2}-|k||x_{1}x_{2}+x_{2}x_{1}|$$ $$\geq x_{1}^{2}+x_{2}^{2}-|k|(x_{1}^{2}+x_{2}^{2})=(1-|k|)(x_{1}^{2}+x_{2}^{2})\geq0$$ with equality if and only if $x_{1}=x_{2}=0$. Note that it is in the last step where we use that $|k|<1$.