# Diagonal entries of inverse larger than 1 over those for the original matrix

A friend asked me if, given $A \in \mathcal{S}^n_{++}$ (positive definite), is it true $\forall$ j=1,...,n that $$(A^{-1})_{jj} \geq \frac{1}{A_{jj}}$$ Not sure if it's true but we haven't found counterexamples yet.

Attempt (assuming true): Let $A_{-ij} \in \mathbb{R}^{n-1 \times n-1}$ be A w/ the ith row and jth column removed. It's equivalent to show $A_{jj}(A^{-1})_{jj} \geq 1$. Using this , it's equivalent to show $$\frac{A_{jj} det(A_{-jj})}{\sum_{i=1}^n (-1)^{i+j}A_{ij}det(A_{-ij})} \geq 1$$ Note the numerator and denominator both have terms $A_{jj} det(A_{-jj})$ so it's equivalent to show $\sum_{i \neq j}(-1)^{i+j}A_{ij}det(A_{-ij}) \leq 0$, and here is where I'm stuck.
I'm not sure if the attempt is the best approach, so alternate proposals are also much appreciated.

Write $A=P^TDP$, so $A^{-1}=P^TD^{-1}P$. Now $A_{ii} = P_i^T D P_i$ (where $i$ indexes the columns), while $(A^{-1})_{ii} = P_i^T D^{-1} P_i$

Your question boils down to whether $P_i^T D^{-1} P_i \geq \frac{1}{P_i^T D P_i}$, or, equivalently, to whether $\sum_{j} \frac{P_{ij}P_{ij}}{D_{jj}} \geq \frac{1}{\sum_{j} P_{ij}P_{ij}D_{jj}}$. This is the celebrated mean vs. harmonic mean inequality.

Since $A$ is symmetric positive definite, then there exists $Q$ orthogonal matrix such that \begin{align} A = Q^TDQ \end{align} where $D$ is a diagonal matrix with positive entries. Then it follows \begin{align} A^{-1} = Q^TD^{-1}Q. \end{align} In particular, we see that \begin{align} (A^{-1})_{ij} = \sum_{k, \ell} (Q^T)_{i\ell}(D^{-1})_{\ell k}(Q)_{kj} = \sum_{k, \ell} (Q^T)_{i\ell}(Q)_{kj}\frac{1}{\lambda_k}\delta_{\ell, k} = \sum_{k} (Q)_{ki}(Q)_{kj}\frac{1}{\lambda_k} \end{align} which means \begin{align} (A^{-1})_{ii}= \sum_{k}\frac{1}{\lambda_k}(Q)_{ki}(Q)_{ki}=\sum_{k}\frac{1}{\lambda_k}(Q)_{ki}^2. \end{align} Likewise, we could show \begin{align} (A)_{ii} = \sum_{k}\lambda_k(Q)_{ki}^2. \end{align} Next, observe \begin{align} (A^{-1})_{ii}(A)_{ii} = \sum_k\lambda_k (Q)_{ki}^2\sum_\ell \frac{1}{\lambda_\ell}(Q)_{\ell i}^2 \geq \sum_k (Q)_{ki}^2 \end{align} where the last inequality is a Cauchy-Schwarz. Using the unit property of the columns and rows of $Q$, we have our desired result \begin{align} (A^{-1})_{ii}(A)_{ii} \geq 1. \end{align}

Answer assuming $A$ is symmetric:

If $A$ is symmetric and positive definite, then there exists a symmetric positive definite $B$ such that $B^2 = A$. Note that $A_{jj} = e_j^TAe_j = e_j^TB^TBe_j = \|Be_j\|^2$ and similarly $(A^{-1})_{jj} = \|B^{-1}e_j\|^2$. By Cauchy-Schwarz, $$\langle Be_j, B^{-1}e_j\rangle^2 \le \|Be_j\|^2\|B^{-1}e_j\|^2 = A_{jj}(A^{-1})_{jj}.$$ But $\langle Be_j, B^{-1}e_j\rangle = e_j^T(B^{-1})^TBe_j = e_j^TB^{-1}Be_j = e_j^Te_j = 1$. Thus $A_{jj}(A^{-1})_{jj}\ge 1$, as desired.

• Same proof with nicer notation. =) – Jacky Chong Sep 28 '16 at 21:51

For a statistical approach to this question, note that $A$ can be interpreted as the covariance matrix of an $n$-vector of Gaussian random variables $X \sim \mathcal N(0, A)$. Now, some basic facts about such random vectors:

• $A_{ii}$ is the marginal variance of $X_i$.
• $1/(A^{-1})_{ii}$ is the conditional variance of $X_i$ given $X_{-i} = (X_1, \ldots, X_{i-1}, X_{i+1}, \ldots, X_n)$.
• The conditional variance of $X_i$ does not depend on the value of $X_{-i}$. Hence, from the conditional variance formula $\mbox{Var}(X_i) = \mbox{Var}(X_i | X_{-i}) + \mbox{Var}(E(X_i \mid X_{-i}))$.

Combining these facts we immediately get $A_{ii} \ge 1/(A^{-1})_{ii}$, or equivalently $$(A^{-1})_{ii} \ge \frac{1}{A_{ii}}$$ as desired.