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.


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 $D=\operatorname{diag}(x,1)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.