The topic of real number sequence has been discussed rather thoroughly in Analysis. We here extends the topic to the matrices field.
The Frobenius norm is defined as:
$$\norm{ A }_F = \sqrt[ ]{ \sum_{ i=1 }^{ m } \sum_{ j = 1 }^{ m } \abs{ a_{ij}}^2 } $$
We say that a sequence $\set{ \b{ A }_k }$ of $m \times n $ matrices converges to the $m \times n $ matrix $\b{ A } $ if
$$\limu{ k }{ \infty } \norm{ \b{ A } - \b{ A }_k } = 0$$
A Jordan normal form of a linear operator on a finite-dimensional vector space is an upper triangular matrix of a particular form called a Jordan matrix, representing the operator with respect to some basis.
Consider the following matrix:
$$A=\left[{\begin{array}{*{20}{r}}5&4&2&1\br0&1&-1&-1\br-1&-1&3&0\br1&1&-1&2\end{array}}\right]$$
Including multiplicity, the eigenvalues of $\b{ A } $ are $λ = 1, 2, 4, 4$. The dimension of the eigenspace corresponding to the eigenvalue $4$ is $1$(and not $2$), so $\b{ A }$ is not diagonalizable. However, there is an invertible matrix $\b{ P } $ such that $\b{ A } = \b{ PJP }^{-1}$, where
$$ J = \begin{bmatrix} 1 & 0 & 0 & 0 \br[2pt] 0 & 2 & 0 & 0 \br[2pt] 0 & 0 & 4 & 1 \br[2pt] 0 & 0 & 0 & 4 \end{bmatrix}. $$
The matrix $\b{ J } $ is almost diagonal. This is the Jordan normal form of $\b{ A }$.
Let $\b{ A} \in \R ^{n \times n}.$
$$\limu{ k }{ \infty } \b{ A }_k = \b{ O } \iff\abs{ \lambda_i(\b{ A })} < 1, i = 1, …, n$$
The series of $n \times n $ matrices
$$\b{ I }_n + \b{ A } + \b{ A }^2 + … + \b{ A }^k + … $$
converges if and only if $\limu{ k }{ \infty } \b{ A }^k = \b{ O } $.
In this case, the sum of the series equals $\inv{(\b{ I }_n - \b{ A })} $.
Let $\b{ A } : \R^r \to \R ^{ n \times n }$ be an $ n \times n $ matrix-valued function that is continuous at $\xi_0 $. If $\inv{ \b{ A }(\xi)} $ exists, then $\inv{ \b{ A }(\xi)}$ exists for $\xi $ sufficiently close to $\xi_0 $ and $\inv{ \b{ A }(\cdot)} $ is continuous at $\xi_0 $.