Linear Algebra – Eigenvalues of a Weighted Mean with Positive Definite Matrices

Question

Suppose I have some positive definite matrices $A_1, A_2, \dots A_k \in \mathbb{R}^{n \times n}$ and some values $s_1 \leq s_2\dots \leq s_k \in \mathbb{R}$. Consider the matrix $A = (\sum_{k=1}A_k s_k)(\sum_{k=1} A_k)^{-1}$. For $n=1$ (the scalar case), $A$ becomes a weighted mean of the $s_i$'s, and thus necessarily satisfies $s_1 \leq A \leq s_k$. For higher $n$, can we say that the eigenvalues of $A$ must lie between $s_1$ and $s_k$? I'm particularly interested if this can be shown (or disproven) in the special case where each $A_k$ is of the form $A_k = x_k x_k^\top$.

Answer 1

This is true. We first observe the following: for two matrices $A$ and $B$, the spectra of $AB$ and of $BA$ coincide. Indeed, $\sigma(AB)\setminus\{0\} = \sigma(BA)\setminus\{0\}$ is true in all unital algebras. For matrices, $AB$ is invertible iff $A$ and $B$ are both invertible iff $BA$ is invertible, so we also have $\sigma(AB)$ contains $0$ iff $\sigma(BA)$ contains $0$.

Applying to the current case, we have the spectrum of $A = (\sum_{i=1}^k A_i s_i)(\sum_{i=1}^k A_i)^{-1} = (\sum_{i=1}^k A_i s_i)(\sum_{i=1}^k A_i)^{-1/2}(\sum_{i=1}^k A_i)^{-1/2}$ coincides with the spectrum of $A’ = (\sum_{i=1}^k A_i)^{-1/2}(\sum_{i=1}^k A_i s_i)(\sum_{i=1}^k A_i)^{-1/2}$. Observe that $A’$ is self-adjoint. Furthermore, because $A_i$ are positive and $s_1 \leq s_i \leq s_k$ for all $i$, we have,

$$\begin{split} (\sum_{i=1}^k A_i)^{-1/2}(\sum_{i=1}^k A_i s_i)(\sum_{i=1}^k A_i)^{-1/2} &\leq (\sum_{i=1}^k A_i)^{-1/2}(\sum_{i=1}^k A_i s_k)(\sum_{i=1}^k A_i)^{-1/2}\\ &= s_k (\sum_{i=1}^k A_i)^{-1/2}(\sum_{i=1}^k A_i)(\sum_{i=1}^k A_i)^{-1/2}\\ &= s_k I \end{split}$$

And similarly $(\sum_{i=1}^k A_i)^{-1/2}(\sum_{i=1}^k A_i s_i)(\sum_{i=1}^k A_i)^{-1/2} \geq s_1 I$, i.e., $s_1 I \leq A’ \leq s_k I$. Since $A’$ is self-adjoint, $\sigma(A’) \subset [s_1, s_k]$, so the same holds for $\sigma(A)$ as well, i.e., eigenvalues of $A$ lie within $[s_1, s_k]$.