$\{x\mid x^TAx\leq 1\} = \{x\mid x^TBx\leq 1\} \Rightarrow A = B$, where $A\succ 0, B\succ 0$

Question

I want show that $$\{x\mid x^TAx\leq 1\}_{\epsilon_A} = \{x\mid x^TBx\leq 1\}_{\epsilon_B} \Rightarrow A = B,$$ where $A\succ 0, B\succ 0$ (positive definite).

Can I prove this by arguing the following?

1. We know $\epsilon_A$, $\epsilon_B$ are ellipsoids.
2. Let eigen-decomposition of $A$ be $A = Q\Lambda Q^T$.
3. We know for the set $\epsilon_A$, the semi-axis is $a_i = \lambda_i^{-1/2}q_i$
4. Due to $\epsilon_A = \epsilon_B$, so both have the same semi-axis, i.e., $b_i = \lambda_i^{-1/2}q_i$.
5. $A$ and $B$ have the same eigenvectors and eigenvalues
6. $A = B$.

Could anyone please give me any suggestions?

Answer 1

For $x\ne 0$, the quadratic form $Q_A(x)=x^TAx$ can be evaluated by observing that $Q_A(tx)=t^2Q_A(x)$ and hence $$Q_A(x)=\inf\{t^{-2}: Q_A(tx)\le 1\} = \inf \{ t^{-2}: tx\in \epsilon_A\}.$$ Since $\epsilon_A=\epsilon_B$, the quadratic forms $Q_A(x)$ and $Q_B(x)$ are equal. Hence the corresponding bilinear forms $x^TAy$ and $x^TBy$ are equal, and finally the matrices $A$ and $B$.