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Let $P= P^{\top} \in \mathbb{R}^{n \times n}$ be positive definite.

Prove that there exists a diagonal, invertible matrix $D \in \mathbb{R}^{n \times n}$ such that the matrix

$$ D^{\top} \, P \, D $$

has (positive) eigenvalues such that the maximum eigenvalue is less than $2$ times the minimum eigenvalue.

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It's not true. Suppose $\pmatrix{a&b\\ b&d}$ is positive definite. Then its maximum eigenvalue is less than the double of its minimum eigenvalue if and only if $$ a+d + \sqrt{(a-d)^2 + 4b^2} < 2\left(a+d - \sqrt{(a-d)^2 + 4b^2}\right),\tag{1} $$ meaning that $9\left[(a-d)^2+4b^2\right]<(a+d)^2$, or $8(a-d)^2+36b^2<(a+d)^2-(a-d)^2$, or $$ 2(a-d)^2 + 9b^2 \le ad.\tag{2} $$ Now consider $P=\pmatrix{2&1\\ 1&1}$. By scaling $D$, we may assume that $D=\operatorname{diag}(1,x)$ or $\operatorname{diag}(x,1)$ for some $x\ne0$. Therefore $$ D^\top PD = \pmatrix{2&x\\ x&x^2} \text{ or } \pmatrix{2x^2&x\\ x&1}. $$ In both cases, we would have $9b^2 > ad$ in $(2)$, regardless of $x$. Hence $(2)$ can never be satisfied. According to WolframAlpha, the minimum ratio is $3+2\sqrt{2}$ when $D=\operatorname{diag}(1,x)$ or $D=\operatorname{diag}(x,1)$.

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