TITLE: Sequence of matrices, equivalent conditions
QUESTION [5 upvotes]: I am trying to prove that:

Let $B$ be a square matrix. The following conditions are equivalent:

$\lim\limits_{k\rightarrow\infty}B^k = 0$
$\lim\limits_{k\rightarrow\infty}B^kv = 0$ for every vector $v$
$\rho(B)<1$
$\|B\| < 1$ for at least one subordinate matrix norm $\|\cdot\|$
The matrix $(I-B)$ is invertible and $$(I-B)^{-1}\ =\ \lim_{k\rightarrow\infty}(I + B + \ldots + B^k).$$
The matrix $(I-B)$ is invertible and all the eigenvalues of the matrix $(I+2(B-I)^{-1})$ have negative real part.
There exists a positive definite Hermitian matrix $H$ such that the (Hermitian) matrix $(H-B^*HB)$ is positive definite.
Given any matrix norm $\|\cdot\|$, there exists an integer $l$ such that $\|B^l\| < 1$.


I already proved almost all, the only proof that I need is (whatever) $\Rightarrow (7)$,
but I really don't know how do it. Please, somebody help me. Thanks in advance.

REPLY [2 votes]: We can show that (3) implies (7). Since every complex matrix can be unitarily triangularised and
$$U^\ast(H-B^\ast HB)U=(U^\ast HU)-(U^\ast B^\ast U)(U^\ast HU)(U^\ast BU)\tag{a}$$
for any unitary matrix $U$, you may assume that $B$ is already upper triangular. Now, let $H=D^{-2}$, where $D=\operatorname{diag}(1,\varepsilon,\varepsilon^3,\ldots,\varepsilon^{n-1})$ with $\varepsilon>0$. Then
$$H-B^\ast HB=D^{-1}\left(I-(D^{-1}BD)^\ast (D^{-1}BD)\right)D^{-1}.\tag{b}$$
As $B$ is upper triangular, $\lim_{\varepsilon\to0^+}D^{-1}BD$ is the diagonal of $B$. Therefore the RHS of $(b)$ is positive definite when $\varepsilon$ is small because $\rho(B)<1$.