What does it mean for $A(t)$ to be continuous, where $A(t)$ is a matrix?

Question

A theorem I encountered in my differential equations class (during the section on systems of DEs, where linear algebra is used a lot) is:

Let $A(t)$ be a continuous ($n\times n$)-matrix on an open interval $I$. If $\vec x_1(t),\dotsc,\vec x_n(t)$ are linearly independent solutions to the homogenous system $\vec x'(t) = A(t)\vec x(t)$ on $I$, then every solution has the form $\vec x(t) = c_1 \vec x_1(t) + \dotsb + c_n \vec x_n(t)$

What does it mean for a matrix to be continuous? This doesn't seem to be a common term because I couldn't find a thing about this online...

Answer 1

It means that $t \mapsto A(t)$ as a mapping $I \to \mathbb{C}^{n \times n}$ is continuous. Here $\mathbb{C}^{n \times n}$ can be endowed with the standard euclidean norm on $\mathbb{C}^{n^2}$, though it does not matter which norm you choose since all norms are equivalent in this case.

Answer 2

It also means that the maps $t \mapsto a_{ij}(t)$ are all continuous, where the $a_{ij}(t)$ are the entries of $A(t)$.