Limits and Continuity of Matrix-valued Functions
Overview
Define the limit and continuity of matrix-valued functions. The way to define the limit of a matrix-valued function is the same as for the limit of vector-valued functions. It is naturally defined by applying the limit for scalar functions to each component.
Definition
For a scalar function $a_{ij} : \mathbb{R} \to \mathbb{R}$, the following $\mathbf{A} : \mathbb{R} \to \mathbb{R}^{n \times m}$ is called a matrix-valued function.
$$ \mathbf{A}(t) = \begin{bmatrix} a_{11}(t) & \cdots & a_{1m}(t) \\ \vdots & \ddots & \vdots \\ a_{n1}(t) & \cdots & a_{nm}(t) \end{bmatrix} $$
The limit of $\mathbf{A}$ at $s$ is defined as follows.
$$ \lim\limits_{t \to s} \mathbf{A}(t) = \begin{bmatrix} \lim\limits_{t \to s} a_{11}(t) & \cdots & \lim\limits_{t \to s} a_{1m}(t) \\ \vdots & \ddots & \vdots \\ \lim\limits_{t \to s} a_{n1}(t) & \cdots & \lim\limits_{t \to s} a_{nm}(t) \end{bmatrix} $$
If each $a_{ij}$ is continuous, i.e. the following holds, then $\mathbf{A}$ is said to be continuous at $s$.
$$ \lim\limits_{t \to s} \mathbf{A}(t) = \mathbf{A}(s) = \begin{bmatrix} a_{11}(s) & \cdots & a_{1m}(s) \\ \vdots & \ddots & \vdots \\ a_{n1}(s) & \cdots & a_{nm}(s) \end{bmatrix} $$
Explanation
The limit and continuity of matrix-valued functions are essentially identical to the case of vector-valued functions; it is natural to regard a matrix as a single vector $\mathbb{R}^{n \times m} \cong \mathbb{R}^{nm}$. That is, for $\mathbf{A}(t) = \begin{bmatrix} a_{ij}(t)\end{bmatrix}$, the existence of $\lim\limits_{t \to s}\mathbf{A}(t)$ is equivalent to the existence of $$ \forall i,j, \quad \lim\limits_{t \to s} a_{ij}(t) $$ Differences from the vector-valued case concern operations defined only for matrices, such as matrix multiplication, transpose, trace, determinant, and inverse matrix.
Properties
Limits
If the limits of the matrix-valued functions $\mathbf{A}$ and $\mathbf{B}$ at $s$ exist, then the following hold.
(a) $\lim\limits_{t \to s} \left( \mathbf{A}(t) + \mathbf{B}(t) \right) = \lim\limits_{t \to s} \mathbf{A}(t) + \lim\limits_{t \to s} \mathbf{B}(t)$
(b) For a constant $c$, $\lim\limits_{t \to s} c\mathbf{A}(t) = c\lim\limits_{t \to s} \mathbf{A}(t)$
(c) $\lim\limits_{t \to s} \mathbf{A}(t)\mathbf{B}(t) = \left( \lim\limits_{t \to s} \mathbf{A}(t) \right) \cdot \left( \lim\limits_{t \to s} \mathbf{B}(t) \right)$
(d) $\lim\limits_{t \to s} \mathbf{A}(t)^{\mathsf{T}} = \left( \lim\limits_{t \to s} \mathbf{A}(t) \right)^{\mathsf{T}}$
(e) $\lim\limits_{t \to s} \tr \left( \mathbf{A}(t) \right) = \tr \left( \lim\limits_{t \to s} \mathbf{A}(t) \right)$
(f) $\lim\limits_{t \to s} \det \left( \mathbf{A}(t) \right) = \det \left( \lim\limits_{t \to s} \mathbf{A}(t) \right)$
Here, in (a) and (c) we assume the matrix sizes are such that addition and multiplication are well-defined. $\mathbf{A}^{\mathsf{T}}$ is the transpose of $\mathbf{A}$. $\tr$ denotes the trace, and $\det$ denotes the determinant.
Continuity
If $\mathbf{A}$ and $\mathbf{B}$ are continuous at $t = s$, then the following functions are also continuous.
- $\mathbf{A} + \mathbf{B}$
- $c\mathbf{A}$
- $\mathbf{A}\mathbf{B}$
- $\mathbf{A}^{\mathsf{T}}$
If $\mathbf{A}$ is continuous at $t = s$ and $\mathbf{A}(s)$ is an invertible matrix, then for $t$ sufficiently close to $s$, $\mathbf{A}(t)$ is also invertible and the following holds.
$$ \lim_{t \to s} \mathbf{A}(t)^{-1} = \left( \lim_{t \to s} \mathbf{A}(t) \right)^{-1} = \mathbf{A}(s)^{-1} $$
Proof
Matrix addition and multiplication are performed entrywise. Since the sum and product of continuous scalar functions are again continuous (sum and product of continuous scalar functions), the same properties hold for matrix-valued functions.
(d), (e), (f)
Transpose, trace, and determinant are continuous functions, and since the equivalent conditions for continuity are as follows, the assertions hold.
$$ \lim_{t \to s} f(\mathbf{A}(t)) = f\left(\lim_{t \to s} \mathbf{A}(t)\right) $$