Is taking the inverse of a matrix a convex operation?

Question

Let $\mathbf{X,Y}$ be two positive definite matrices. Can we obtain the following Jensen-like inequality $$(1-\lambda)\mathbf{X}^{-1}+\lambda\mathbf{Y}^{-1} \succeq((1-\lambda)\mathbf{X}+\lambda\mathbf{Y})^{-1}$$ for any $\lambda \in (0,1)$, where the notation $\mathbf{A}\succeq \mathbf{B}$ represents matrix $(\mathbf{A-B})$ is positive semi-definite?

Answer 1

Assume $X, Y$ are positive definite and $\lambda \in [0,1]$. We know $Z(\lambda) = (1-\lambda) X + \lambda Y$ is positive definite and so does $Z^{-1}(\lambda)$. Let us use $(\ldots)'$ to represent derivative with respect to $\lambda$, we have:

$$ Z Z^{-1} = I_n \implies Z'Z^{-1} + Z ( Z^{-1} )' = 0_n \implies (Z^{-1})' = - Z^{-1} Z' Z^{-1} $$ Differentiate one more time and notice $Z'' = 0_n$, we get: $$(Z^{-1})'' = - (Z^{-1})' Z' Z^{-1} - Z^{-1}Z' (Z^{-1})' = 2 Z^{-1}Z' Z^{-1} Z' Z^{-1}\tag{*1}$$

Pick any random non-zero vector $u$ and consider following pair of vector/matrix valued functions:

$$v(\lambda) = Z'(\lambda) Z^{-1}(\lambda) u\quad\quad\text{ and }\quad\quad \varphi(\lambda) = u^T Z^{-1}(\lambda) u$$

$(*1)$ tell us

$$\varphi''(\lambda) = u^T (Z^{-1})''(\lambda) u = 2 v^T(\lambda) Z^{-1}(\lambda) v(\lambda) \ge 0\tag{*2}$$

because $Z^{-1}(\lambda)$ is positive definite. From this we can conclude $\varphi(\lambda)$ is a convex function for $\lambda$ over $[0,1]$. As a result, for any $\lambda \in (0,1)$, we have:

$$\begin{align}&(1-\lambda)\varphi(0) + \lambda\varphi(1) - \varphi(\lambda) \ge 0\\ \iff& u^T \left[ (1-\lambda) X^{-1} + \lambda Y^{-1} - ((1-\lambda) X + \lambda Y)^{-1}\right] u \ge 0\tag{*3} \end{align}$$ Since $u$ is arbitrary, this implies the matrix within the square bracket in $(*3)$ is positive semi-definite and hence:

$$(1-\lambda) X^{-1} + \lambda Y^{-1} \succeq ((1-\lambda) X + \lambda Y)^{-1}$$

Please note that when $Z' = Y - X$ is invertible, $v(\lambda)$ is non-zero for non-zero $u$. The inequalities in $(*2)$ and $(*3)$ become strict and the matrix within the square bracket in $(*3)$ is positive definite instead of positive semi-definite.

Answer 2

Let $\mathbf{P}=\mathbf{X}^{-1/2}\mathbf{Y}\mathbf{X}^{-1/2}$. By left- and right- multiplying both sides of the equation by $\mathbf{X}^{1/2}$, the inequality is equivalent to $(1-\lambda)I+\lambda\mathbf{P}^{-1} \succeq((1-\lambda)I+\lambda\mathbf{P})^{-1}$. Since $\mathbf P$ is positive definite, it can be unitarily diagonalised and hence we may assume that it is a diagonal matrix. So, the inequality reduces down to the scalar case $(1-\lambda)+\lambda p_{ii}^{-1} \ge ((1-\lambda)+\lambda p_{ii})^{-1}$, which is true because the function $f(t)=\frac1t$ is convex for $t>0$.

Answer 3

Let $ S^{n}_{++} $ be the set of $ n \times n $ positive definite matrices, and $ S^{n}_{+} $ be the set of $ n \times n $ positive semi-definite matrices.

$ \forall X, Y \in S^{n}_{++}, \theta \in [0, 1], $

$$ \left(\begin{array}{2} X^{-1} & I \\ I & X \end{array}\right), \left(\begin{array}{2} Y^{-1} & I \\ I & Y \end{array}\right) \in S^{2n}_{+}, $$

Therefore, $$ M = \left(\begin{array}{2} \theta X^{-1} + (1 - \theta)Y^{-1} & I \\ I & \theta X + (1 - \theta)Y \end{array}\right) \in S^{2n}_{+}. $$

Let $ S = (\theta X + (1 - \theta)Y) - (\theta X^{-1} + (1 - \theta)Y^{-1})^{-1} $, which is the schur complement of $ \theta X^{-1} + (1 - \theta)Y^{-1} $ in $ M $.

Since $$ \theta X^{-1} + (1 - \theta)Y^{-1} \in S^{n}_{++}, $$ $$ M \in S^{2n}_{+}, $$ we have $ S \in S^{n}_{+}, $ or $$ (\theta X + (1 - \theta)Y) - (\theta X^{-1} + (1 - \theta)Y^{-1})^{-1} \succeq O. $$

This step follows https://math.stackexchange.com/a/2835328/1459075.

Then, $$ \theta X^{-1} + (1 - \theta)Y^{-1} \succeq (\theta X + (1 - \theta)Y)^{-1} $$

Inspired by POSITIVE DEFINITE MATRICES (Rajendra Bhatia)

Answer 4

Edit: The proof is wrong as pointed out by @mewmew. I am tryin to fix it.

Let $\lambda\in [0,1]$. Let, $$A=(1-\lambda){X}^{-1}+\lambda {Y}^{-1}\\ B=\left((1-\lambda){X}+\lambda Y\right)$$

Then, we note that $A$ and $B$ are positive definite since $X,\ Y$ are positive definite. Hence $X,Y,A,B$ are invertible.

Then \begin{align} AB-I=&\left((1-\lambda){X}^{-1}+\lambda {Y}^{-1}\right)\left((1-\lambda){X}+\lambda Y\right)-I\\ \ =& \left(1-\lambda\right)^2I+\lambda(1-\lambda)\left(X^{-1}Y+Y^{-1}X\right)+\lambda^2I-I\\ \ =& 2\lambda(\lambda-1)I+\lambda(1-\lambda)\left(X^{-1}Y+Y^{-1}X\right)\\ \ =& \lambda(1-\lambda)\left(-X^{-1}X-Y^{-1}Y+X^{-1}Y+Y^{-1}X\right)\\ \ =& \lambda(1-\lambda)(Y^{-1}-X^{-1})(X-Y)\quad \\ \end{align} So, for any vector $u\ne 0$ \begin{align} u^T(B^TAB-B^T)u=& u^TB^T(AB-I)u=\lambda(1-\lambda)u^TB^T(Y^{-1}-X^{-1})(X-Y)u\\ \ =&\lambda(1-\lambda)u^T((1-\lambda)X^T+\lambda Y^T)(Y^{-1}-X^{-1})(X-Y)u \end{align} Now, for any vector $v\ne 0\ \exists$ a vector $u\ne 0$ such that $$Bu=v$$. Hence, for any vector $v\ne 0$\begin{align} v^T(A-B^{-1})v=& (Bu)^T(A-B^{-1})Bu\\ \ =& u^T(B^TAB-B^T)u\ge 0 \end{align}

Hence $$A\ge B^{-1}\\ \Rightarrow (1-\lambda)X^{-1}+\lambda Y^{-1}\ge \left((1-\lambda){X}+\lambda Y\right)^{-1}\quad \forall \lambda\in [0,1]\hspace{0.6cm}\Box$$