Jordan Canonical form of the matrix Using Jordan Canonical theorem, prove that the matrix sequence $\lbrace A^k\rbrace\rightarrow 0$ (i.e matrix sequence tending to zero matrix) if and only if $|\lambda_i|<1$ i.e the absolute value of each eigenvalue (of $A$) is less than $1$.
Further prove that $\lbrace A^k\rbrace\rightarrow 0$ if $||A||<1$, where $||\cdot||$ is a subordinate matrix norm.
 A: Your condition $|λ_i| < 1$ is equivalence to $\rho(A) < 1$, where $\rho(A)$ is the spectral radius of $A$.
Proof:
($\Leftarrow$) Suppose that ${A}^k \rightarrow {O}$. Let $\lambda\in\sigma({A})$, where $\sigma(A)$ = spectrum of $A$, then there exists ${x}_{\lambda}\in\mathbb{R}^n$, ${x}_{\lambda} \neq {0}$, such that ${A}{x}_{\lambda} = \lambda{x}_{\lambda}$.
Note that ${A}^k{x}_{\lambda} = \lambda^k{x}_{\lambda}$, $\lambda^k\in\sigma({A}^k)$.
Later, by $\rho(A) \leq \|A\|$ (this is easy to prove, and you need use it in the last part), then
$$\left|\lambda^k\right|\ \leq\ \|{A}^k\|.$$
And you have that ${A}^k \rightarrow {O}$, then $\left|\lambda^k\right| \rightarrow 0$, where you have that $|\lambda| < 1$. Finally, as $\lambda$ is arbitrary, then $\rho({A}) < 1$.
($\Rightarrow$) By Jordan canonical from, ${A} =
{S}{J}{S}^{-1}$. Then, for $k \geq 1$, you have that ${A}^k
= {S}{J}^k{S}^{-1}$. So, is sufficient to prove that ${J}^k\rightarrow O$. Note that each Jordan block  ${J}_i$ is of the form
$${J}_i\ =\ \lambda_i{I} + {N}_i$$
where ${N}_i$ is a nilpotent matrix of degree $k_i$, i.e ${N}_i^{k_i} = {O}$.
So, for the binomial expansion, you can deduce that for each $k \geq k_i$,
$${J}_i^k\ =\ \left(\lambda_i{I} + {N}_i\right)^k\ =\ \sum_{j=0}^{k_i-1}\frac{k!}{j!(k-j)!}\lambda_i^{k-j}{N}_i^j,$$
therefore
$$\|{J}_i^k\|\ \leq\ \sum_{j=0}^{k_i-1}\frac{k!}{j!(k-j)!}|\lambda_i|^{k-j}\|{N}_i^j\|.$$
The convergence $\|{J}_i^k\| \rightarrow 0$ is from the fact, that for each $j \in \mathbb{N}$ fixed, the sequence $\{x^j_k\}_{k\in\mathbb{N}}$, with $k$th term,
$$x^j_k\ =\ \frac{k!}{j!(k-j)!}|\lambda_i|^{k-j},$$
converge to zero, because $\rho({A}) < 1$.
