How can $\int_{t}^{t+T}xx^Td\tau \geq \alpha_1 I$, where $x \in \mathbb{R}^n$, especially when the determinant of outer product or $xx^T$ is 0?

Question

I am quite perplexed, I wish to prove one of the fundamental lemmas in adaptive control, i.e., \begin{equation} \alpha_1 I \leq \int_t^{t+T}xx^Td\tau \leq \alpha_2 I \end{equation} where, $\alpha_1,\alpha_2,T$ are positive constants and $x \in \mathbb{R}^n$. Now, how is this possible when we know that $xx^T$ is the outer product which has determinant 0? Please shed some light on this.

Answer 1

It is true that at every time instance the matrix $x\,x^\top$ is singular (it can even at most only have rank one). However, an integral can be seen as a summation and summing multiple singular matrices can become a nonsingular matrix. For example the two by two identity matrix can be written as

$$ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix}^\top + \begin{bmatrix} 0 \\ 1 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix}^\top. $$