Convergence of Matrices
Definition1
Let $\left\{ A_{n} \right\}$ denote a sequence of real (or complex) matrices. The statement that $\left\{ A_{n} \right\}$ converges to the matrix $A$ means that each sequence of entries $\left\{ [A_{n}]_{ij} \right\}$ of $A_{n}$ converges to the corresponding entry $[A]_{ij}$ of $A$ (see convergence).
$$ \lim\limits_{n \to \infty} A_{n} = A, \quad \text{if } \lim\limits_{n \to \infty} [A_{n}]_{ij} = [A]_{ij} \quad \forall i,j $$
$$ \left\{ A_{n} \right\} \to A \text{ as } n \to \infty, \quad \text{if } \left\{ [A_{n}]_{ij} \right\} \to [A]_{ij} \text{ as } n \to \infty \quad \forall i,j $$
Remark
In short, componentwise convergence is the convergence of matrices. Because convergence is defined componentwise, the properties of limits of real sequences (see properties of limits of real sequences) carry over unchanged.
Properties
Below, assume $A$ and $B$ are matrices for which addition and multiplication are well-defined in the given context. For $\lim\limits_{n \to \infty} A_{n} = A$ and $\lim\limits_{n \to \infty} B_{n} = B$ the following hold.
(a) $\lim\limits_{n \to \infty} A_{n} = A \iff \| A_{n} - A \| \to 0$
(b) $\lim\limits_{n \to \infty} (A_{n} + B_{n})$
(c) $\lim\limits_{n \to \infty} (A_{n} B_{n}) = \left( \lim\limits_{n \to \infty} A_{n} \right) \cdot \left( \lim\limits_{n \to \infty} B_{n} \right) = AB$
(d) If $\lim\limits_{n \to \infty} A_{n} = A$, then $\lim\limits_{n \to \infty} (A_{n})^{\ast} = A^{\ast}$
Proof
(b)
$$ \begin{align*} \lim\limits_{n \to \infty} [A_{n}+B_{n}]_{ij} &= \lim\limits_{n \to \infty} \left( [A_{n}]_{ij} + [B_{n}]_{ij} \right) \\ &= \lim\limits_{n \to \infty} [A_{n}]_{ij} + \lim\limits_{n \to \infty} [B_{n}]_{ij} \\ &= [A]_{ij} + [B]_{ij} = [A+B]_{ij} \end{align*} $$
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(c)
$$ \begin{align*} \lim\limits_{n \to \infty} [A_{n}B_{n}]_{ij} &= \lim\limits_{n \to \infty} \sum_{k} [A_{n}]_{ik} [B_{n}]_{kj} \\ &= \sum_{k} [A]_{ik} [B]_{kj} \\ &= [AB]_{ij} \end{align*} $$
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(d)
The conjugate transpose is continuous, and by the definition of continuity the following holds.
$$ \lim\limits_{n \to \infty} (A_{n})^{\ast} = \left( \lim\limits_{n \to \infty} A_{n} \right)^{\ast} = A^{\ast} $$
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Brian C. Hall. Lie Groups, Lie Algebras, and Representations (2nd), p ↩︎
