Find the necessary and sufficient condition for $A^m\to0$ Let $A$ be $n\times n$ matrix on $\mathbb{C}$. Find a necessary and sufficient condition for $A^m\to0$ as $m\to\infty$. 
My thought: I think it should be that eigenvalues of $A$ are less than $1$. But I am unable to prove it. 
 A: Certainly when $\| A\| <1$, since in that case 
$$
\lim_{n\to \infty } \| A^n \| \leq \lim_{n\to \infty } \| A \|^n = 0
$$
Therefore $A^n \to 0$ as $n \to \infty$
Conversely, if $A^n \to 0$ as $n \to \infty$, continuity of the norm gives that $\|A^n\| \to 0$ as $n \to \infty$
A: We know from the theorem on Jordan decompositions there exist a diagonalizable matrix $D$ and a nilpotent matrix $N$ such that $A=D+N$ and $DN=ND$, and then $A^k=\sum_{i=0}^k\binom{k}{i}D^{k-i}N^{i}$. Since $N^n=0$, this means that in fact $$A^k=\sum_{i=0}^{n-1}\binom{k}{i}D^{k-i}N^{i}.$$
Use what you know about diagonalizable matrices to conclude that the condition that the eigenvalues of $A$ —which are, as you know, the same as those of $D$— be of absolute value less than $1$ is a sufficient condition; notice this last expression has a fixed number of terms, independent of $k$.
If there is an eigenvalue $\lambda$ with $|\lambda|\geq1$ and $v\in\mathbb C^n$ is a non-zero vector such that $Av=\lambda v$, then we have $A^kv=\lambda^kv$, which has norm not going to zero if $k$ goes to infinity. This implies that $A^k$ does not go to zero as $k\to\infty$.
A: We use $l_2$ norm for matrix as

$$\|A\|=\sqrt{\sum\limits_{i,j=1}^{n}|a_{ij}|^2}\tag{1}$$

First we define inequality for matrix as

Let $A$ be $n\times n$ matrix on $\mathbb{C}$ and $B$ be $n\times n$ non-negative matrix.
  $$A\leqslant B \hspace{5 mm} \text{iff } \hspace{5 mm} |a_{ij}|\leqslant b_{ij}, \hspace{2 mm} i,j=1,\cdots,n \tag{2}$$

We can prove following lemmas.

Lemma:
$1. \hspace{2 mm} A\to0 \iff \|A\|\to0$
$2.\hspace{2 mm} A\leqslant B\implies \|A\|\leqslant\|B\|$
$3.\hspace{2 mm} A\leqslant B\implies A^m\leqslant B^m$
$4.\hspace{2 mm}\|AB\|\leqslant\|A\|\|B\|$
$5.\hspace{2 mm} A\leqslant B\land B\to0\implies A\to0$

Prove of $1$ and $2$ are trivial.
Prove of $3$:
We only prove $m=2$ and rest follows from induction.
$$|(A^2)_{ij}|=\left|\sum\limits_{k=1}^{n}a_{ik}a_{kj}\right|\leqslant\sum\limits_{k=1}^{n}|a_{ik}||a_{kj}|\leqslant\sum\limits_{k=1}^{n}b_{ik}b_{kj}=(B^2)_{ij}$$
Prove of $4$:
\begin{align}
\|AB\|^2&=\sum\limits_{i,j=1}^{n}\left|\sum\limits_{k=1}^na_{ik}b_{kj}\right|^2
\\
&\leqslant\sum\limits_{i,j=1}^{n}\left(\sum\limits_{k=1}^n|a_{ik}|^2\sum\limits_{k=1}^n|b_{kj}|^2\right)\tag{Cauchy-Schwarz}
\\
&=\sum\limits_{i,j=1}^{n}\left(\sum\limits_{k,l=1}^n|a_{ik}|^2|b_{lj}|^2\right)
\\
&=\sum\limits_{i,k=1}^{n}|a_{ik}|^2\sum\limits_{l,j=1}^n|b_{lj}|^2
\\
&=\|A\|^2\|B\|^2
\\
\end{align}
Prove of $5$:
By $1$ and $2$, $\|A\|\to0$. And by $1$ again, $A\to0$.
The following theorem is the key to prove.

Theorem:
  $A^m \to 0$ as $m\to\infty$ iff any eigenvalue of $A$, $|\lambda_i|<1, i=1,\cdots,n$.

Prove:
Let $J_A$ be Jordan Canonical Form and there is invertible $P$ such that 
$$
P^{-1}AP=J_A=\pmatrix{J_{\lambda_1}\\&\ddots \\&& J_{\lambda_l}}
$$
where $\lambda_i,1\leqslant i\leqslant l$ are distinct eigenvalues of $A$ and $J_{\lambda_i}$ are Jordan Blocks. 
If all $|\lambda_i|<1$, then for any $|\lambda_i|$, there is a $\mu_i$ such that $|\lambda_i|<\mu_i<1$. So
$$
J_A=\pmatrix{J_{\lambda_1}\\&\ddots \\&& J_{\lambda_l}}\leqslant \pmatrix{\mu_1 \hspace{5 mm}1\\\ddots \\& \mu_{n-1} \hspace{5 mm}1 \\& \hspace{10 mm}\mu_n }=S
$$
Note that $\mu_i$ are set to be all distinct (even for same $\lambda_i$) so that $S$ is diagonalizable. Let
$$
Q^{-1}SQ=\text{diag}(\mu_1\cdots\mu_n)
$$
So
$$
S^m=Q\text{diag}(\mu_1^m\cdots\mu_n^m)Q^{-1}\to0
$$
By lemma $3$
$${J_A}^m\leqslant S^m$$
And by lemma $5$
$${J_A}^m\to0$$
Thus
$$A^m=P{J_A}^mP^{-1}\to0$$ 
Conversely if ${A}^m\to0$, then 
$${J_A}^m=P^{-1}A^mP\to0$$
So $|\lambda_i|<1, i=1,\cdots,n$.
