Inequality for a positive matrix

Question

I consider an $n\times n$ real positive-definite matrix $A$, which I rewrite in its diagonalized form: $A=O^TDO$, where $O$ is an orthogonal matrix and $D$ is diagonal. Since $A$ is positive-definite, I can define $B=O^TD^{1/2}O$ as the “square root” of $A$. I am trying to understand whether or not the following inequality is true:$$\sum_{i=1}^n(A_{ii}(B^{-1})_{ii}-B_{ii})\ge0.$$ I have tested the inequality with random matrices numerically and I can't find a counterexample, but I don't manage to prove that the inequality holds either. Using Cauchy-Schwarz inequality is very tempting but can't make it work unfortunately. Anyone has an idea on how to tackle this problem?

Answer 1

Since $O$ is orthogonal, $OO^T=I$, so $$B^2=O^TD^{1/2}OO^TD^{1/2}O=O^TDO=A.$$ Note also that, since $D$ has all positive diagonal entries, $D^{1/2}$ does too, so $B$ is positive definite. It thus suffices to prove that $$\sum_{i=1}^n (B^2)_{ii}(B^{-1})_{ii}\geq \sum_{i=1}^n B_{ii}$$ for any positive definite matrix $B$. Let $\{v_1,\dots,v_n\}$ be an orthonormal eigenbasis of $B$ with eigenvalues $\lambda_1,\dots,\lambda_n$, and let $e_i=a_{i1}v_1+\cdots+a_{in}v_n$ (all this information is essentially that contained in $O$ and $D$, just repackaged slightly). Then for any $m\in\mathbb Z$ $$(B^m)_{ii}=e_i^TB^me_i=\sum_{j=1}^n\sum_{k=1}^n a_{ij}a_{ik}v_j^TB^mv_k=\sum_{j=1}^n a_{ij}^2\lambda_j^m.$$ So, for $m_1,m_2\in\mathbb Z$, $$\sum_{i=1}^n (B^{m_1})_{ii}(B^{m_2})_{ii}=\sum_{i=1}^n\sum_{j=1}^n\sum_{k=1}^na_{ij}^2a_{ik}^2\lambda_j^{m_1}\lambda_k^{m_2}.$$ Now, note that for any $j$ and $k$ $$\lambda_j^2\lambda_k^{-1}+\lambda_k^2\lambda_j^{-1}\geq\lambda_j+\lambda_k,$$ since $$\left(\lambda_j^2\lambda_k^{-1}+\lambda_k^2\lambda_j^{-1}\right)-\left(\lambda_j-\lambda_k\right)=\frac{(\lambda_j+\lambda_k)(\lambda_j-\lambda_k)^2}{\lambda_j\lambda_k}\geq 0.$$ So, $$\sum_{i=1}^n (B^2)_{ii}(B^{-1})_{ii}\geq \sum_{i=1}^n\sum_{j=1}^n\sum_{k=1}^n a_{ij}^2a_{ik}^2\lambda_j^2\lambda_k^{-1}\geq \sum_{i=1}^n\sum_{j=1}^n\sum_{k=1}^n a_{ij}^2a_{ik}^2\lambda_j=\sum_{i=1}^n B_{ii}I_{ii}=\sum_{i=1}^n B_{ii}.$$