Let $\langle\cdot,\cdot \rangle$ be a dot product on $\mathbb{R}^{2}$. We define a norm $\|x\|=\sqrt{\langle x,x \rangle}$. We know that:
$$ \sup_{x \in \mathbb{R}^2}{\frac{\| x\|_2}{\|x\|}}=3 ,\qquad \inf_{x \in \mathbb{R}^2}{\frac{\| x\|_2}{\|x\|}}=1 \\ \|(1,2)\| =\frac{\sqrt{5}}{3} \quad \text{and} \quad \|(-2,1)\|=\sqrt{5} .$$ Find a formula for $\|(x,y)\|$.
Could someone give me a hint to this problem?
Hint: This problem is equivalent to determining the coefficients of the quadratic form $Q(\mathbf v)=\|\mathbf v\|^2=\langle\mathbf v,\mathbf v\rangle=\mathbf v^TA\mathbf v$ for some symmetric matrix $A$. The given conditions tell you something about the eigenvalues and eigenvectors of this matrix.
Note first that the $\sup$ and $\inf$ are attained at $(1,2)$ and $(-2,1)$, respectively. Now, restricting ourselves to $\|v\|=1$, the given conditions tell us that $Q(\mathbf v)$ attains its maximum value of $1$ in the direction of $(-2,1)$ and its minimum of $\frac19$ in the direction of $(1,2)$. These values are the eigenvalues of $A$, with associated eigenvectors. With this information in hand, we can reconstruct $A$:$$ A=R\pmatrix{1 & 0 \\ 0 & \frac19}R^{-1}, $$ where $$ R=\frac1{\sqrt5}\left(\begin{array}{r r}2 & 1 \\ -1 & 2\end{array}\right), \tag{*} $$ giving $$ A=\frac1{45}\left(\begin{array}{r r}37 & -16 \\ -16 & 13\end{array}\right). $$ Thus, $$ Q(x,y)=\frac{37}{45}x^2-\frac{32}{45}xy+\frac{13}{45}y^2 $$ and finally $$ \|(x,y)\|=\left(\frac{37}{45}x^2-\frac{32}{45}xy+\frac{13}{45}y^2\right)^{\frac12}. $$
(*) We used $(2,-1)$ instead of $(-2,1)$ for the eigenvector so that $\det R=1$. This isn’t strictly necessary, but it’s convenient because it makes $R$ a rotation.