Find an inner product on \(\RR^2\) such that \((\epsilon_1, \epsilon_2) = 2\).
Solution
Define \[ ((x_1,x_2)\mid(y_1,y_2)) =x_1y_1+2x_1y_2+2x_2y_1+5x_2y_2. \] Equivalently, its Gram matrix in the standard basis is \[ G=\begin{pmatrix}1&2\\2&5\end{pmatrix}. \] This form is symmetric and bilinear. Moreover, \[ (x\mid x) =x_1^2+4x_1x_2+5x_2^2 =(x_1+2x_2)^2+x_2^2, \] which is strictly positive for every nonzero \(x\). Hence it is an inner product [@HK71]. Finally, \[ (\epsilon_1\mid\epsilon_2)=G_{12}=2. \] Thus this inner product has the required property.