How to prove the inequality $\det\left(\sum_{i=1}^k{A_i^TA_i}\right) \geq 0$ for matrices $A_1, \dots, A_k\in M_n(\mathbb{R})$?

Question

I am exploring a mathematical problem involving two natural numbers, $n$ and $k$ ($n, k \geq 1$), and a collection of matrices $A_1, \dots, A_k \in M_n(\mathbb{R})$. The goal is to establish the validity of the following inequality:

$$\det\left(\sum_{i=1}^k{A_i^TA_i}\right) \geq 0.$$

Despite my efforts, I have been unable to devise a proof for this inequality. I have examined properties of matrices in the form $A_i^TA_i$ and attempted to leverage determinant formulas and properties without success. I would greatly appreciate any assistance or hints that could guide me toward a solution for this problem. Your insights are immensely valuable to my understanding of this mathematical challenge.

Answer 1

Let $$ B = \sum_{i=1}^k A_i^T A_i. $$

We immediately see that $B$ is symmetric as it is the sum of symmetric matrices.

We also have that for any $x \in \mathbb{R}^n$ $$ x^T B x = x^T \left( \sum_{i=1}^k A_i^T A_i \right) x = \sum_{i=1}^k x^T A_i^T A_i x = \sum_{i=1}^k \|A_ix\|^2 \geq 0, $$ so $B$ is positive semidefinite.

$B$ being symmetric positive semidefinite means all its eigenvalues $\lambda_1, \dots, \lambda_n$ are nonnegative.

We can then conclude that $$ \det B = \lambda_1 \cdot \lambda_2 \cdot \ldots \cdot \lambda_n \geq 0$$