I am trying to solve this question but am not able to understand how to approach it: What is the polar of an ellipsoid described by the equation:  {$(z_1, . . . , z_d) ∈ R^d: a_1z_1^2 + · · · + a_dz_d^{2} ≤ 1$}, where $a_1, . . . , a_d$ are positive.  I know what generally polar of a set means. Moreover, I am following Algebraic And geometric ideas in the theory of discrete optimization to understand more about the polar of a subset. This link gives a beautiful explanation of how to find polar of various set   I'm trying to approach it the way you solve for a disc but I am still unable to do it. Any help is appreciated Thank you.
The polar of a convex body $C$ is defined as:  For a convex body $C$, its polar $C^*$ is given by $$C^* = \{x\in\mathbb{R}^n: \langle x,c\rangle \le 1, \forall c\in C\}$$  If I start with an orthonormal basis $(e_i)_1^n$, positive constants $(\alpha_j)_1^n &gt; 0$ and the ellipsoid $$\varepsilon = \left\{x\in\mathbb{R}^n: \sum\limits_{j=1}^n \frac{\langle x,e_j\rangle^2}{\alpha_j^2} \leq 1  \right\}$$ how do I show that its polar ellipsoid is $$\varepsilon^* = S_1 = \left\{y\in\mathbb{R}^n: \sum_{j=1}^n \alpha_j^2 \langle y,e_j\rangle^2 \le 1\right\}$$ Using the definition, I saw that $$\varepsilon^* = S_2 = \{z\in\mathbb{R}^n: \langle z,x\rangle \le 1, \text{for all }x\in \varepsilon\}$$ Also, $\langle z,x\rangle = \sum_i\langle z,e_i\rangle\langle x,e_i\rangle$ I used Cauchy Schwartz to do one side of the proof, i.e. $S_1\subseteq S_2$: $$\left(\sum_i\langle y,e_i\rangle\langle x,e_i\rangle\right)^2 \leq \left(\sum_i\alpha_i^2 \langle y,e_i\rangle^2\right)\left(\sum_i \frac{\langle x,e_i\rangle^2}{\alpha_i^2}\right) \le 1 \implies \langle y,x\rangle \le 1$$ I need help with the other direction, i.e. $S_2\subseteq S_1$. Which other inequality might help?
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