# Diagonal elements of the inverted correlation matrix

Is it true that the diagonal elements of the inverted correlation matrix will always be larger than 1? Why?

Yes, it is true: the diagonal elements can never be less than unity.

By permuting the order of the variables, any diagonal element can be made to appear in the upper left corner, so it suffices to study that element. The statement is trivially true for $n=1$. For $n\gt 1$, any $n$ by $n$ correlation matrix can be written in block form as

$$\mathbb C = \pmatrix{ 1 & \mathbf {\vec e} \\ \mathbf e &\mathbb D}$$

where $\mathbb D$ is the correlation matrix of variables $2, 3, \ldots, n$ and $\mathbf {\vec e}$ is the transpose of the column vector $\mathbf e$ containing the correlations between the first variable and the remaining variables.

Assume for the moment that $\mathbb C$ is invertible. By Cramer's Rule, the upper left corner of its inverse is

$$\left(\mathbb C^{-1}\right)_{11} = \det \mathbb D / \det \mathbb C.$$

If we can prove that this ratio cannot be less than $1$, we are done in the general case (even for singular $\mathbb C$), because the entries in the inverse are continuous functions of $\mathbb C$ and the non-invertible ones form a lower-dimensional submanifold of the space of all such $\mathbb C$.

The problem is reduced, then, to showing that determinants of invertible correlation matrices cannot increase as the number of variables is increased. Invertibility of $\mathbb C$ implies $\mathbb D$ is invertible, enabling allows row-reduction to simplify the top row. This will help us relate the determinant of $\mathbb C$ to that of $\mathbb D$. Reduction of that row amounts to left-multiplication by the inverse of

$$\mathbb P = \pmatrix{ 1 & \mathbf {\vec e}\,\mathbb D ^{-1} \\ \mathbf 0 &\mathbf 1_{n-1,n-1}},$$

showing that

$$\mathbb C = \mathbb P\pmatrix{ 1 - \mathbf {\vec e}\,\mathbb D ^{-1}\, \mathbf e &\mathbf {\vec 0} \\ \mathbf e &\mathbb D}.$$

Taking determinants yields

$$\det \mathbb C = \det \mathbb P \det \pmatrix{ 1 - \mathbf {\vec e}\,\mathbb D ^{-1} \,\mathbf e &\mathbf {\vec 0} \\ \mathbf e &\mathbb D} = \left( 1 - \mathbf {\vec e}\,\mathbb D ^{-1}\, \mathbf e \right)\det \mathbb D$$

because $\det \mathbb P = \det(1)\det \mathbf{1}_{n-1,n-1}=1$.

Since $\mathbb D$, being an invertible correlation matrix in its own right, is positive-definite, and $\mathbb C$ is positive definite, we immediately deduce that

1. $\mathbf {\vec e}\,\mathbb D ^{-1}\, \mathbf e \ge 0$.

2. $\det \mathbb D \gt 0$.

3. $\det \mathbb C \gt 0$.

Therefore $1 \ge 1 - \mathbf {\vec e}\,\mathbb D ^{-1}\, \mathbf e \gt 0$, whence $\det \mathbb C \le \det \mathbb D$, QED.