Check if the bilinear form given in the standard basis by $$\begin{bmatrix} 1 &4 &0\\ 4 &4 &0\\ 0 &0 &2 \end{bmatrix}$$ gives some dot product to $\mathbb R^3$.
I need confirmation that my solution is correct:
One of the conditions for the dot product is non-negative definiteness. Thus, for a given bilinear form to give some dot product, the given matrix must be non-negative definite, which is equivalent to all eigenvalues of the matrix being non-negative. Let's check: $$\left| \begin{matrix} 1-\lambda &4 &0\\ 4 &4-\lambda &0\\ 0 &0 &2-\lambda \end{matrix} \right|=(1-\lambda)(4-\lambda)(2-\lambda)-16(2-\lambda)$$ $$=(2-\lambda)(4-\lambda -4\lambda +\lambda^2 -16)=(2-\lambda)(\lambda ^2 -5\lambda -12)$$ $$\lambda_1=2, \lambda_2=\frac{\sqrt{73}+5}{2}, \lambda_3\frac{-\sqrt{73}+5}{2}$$ $\lambda_3<0$, so the matrix is not non-negative definite, i.e. the form cannot specify an dot product, and there is no need to check the rest of the product conditions.