Let $H$ be the quadratic form in vector space $\mathbb{R}^3$ s.t $$H=x^2+2y^2+3z^2+2xy+2xz+2yz.$$
Let $\varphi$ be the symmetric bilinear form on $\mathbb{R}^3$ and $\varphi:\mathbb{R}^3\times \mathbb{R}^3\to\mathbb{R}$ s.t $$H(v)=\varphi(\overrightarrow{v},\overrightarrow{v})\quad\forall \overrightarrow{v}\in\mathbb{R}^3.$$
Given $\psi$ is the symmetric bilinear form on $\mathbb{R}^3.$
Does there exit $t_0\in\mathbb{R}$ such that the bilinear form $\psi +t_0\varphi$ is a dot product on $\mathbb{R}^3$ and why?
P/s: I already found $\varphi((x_1,y_1,z_1),(x_2,y_2,z_2))=x_1x_2+2y_1y_2+3z_1z_2+x_1y_2+x_2y_1+y_1z_2+y_2z_1+z_1x_2+z_2x_1.$
Since $\varphi$ and $\psi$ are both symmetric bilinear form, $\psi +t_0\varphi$ is also a symmetric bilinear form. So we need to check whether there exits $t_0\in\mathbb{R}$ such that $\psi +t_0\varphi$ is positive definite.
Can anyone continue to solve this problem?