How do we get $A^k = \lambda_1 u_1\cdot v_1^T + \mathcal{O}(k^{m_2-1}|\lambda_2|^k)$?

I just have a quick question to the following approximation.

Let $A \in \mathbb{R}^{n \times n}$ be a primitive Matrix with Eigenvalues $\lambda_1,\dots,\lambda_n$ ordered this way, $\lambda_1 > | \lambda_2 | \geq \cdots \geq | \lambda_n |$.

$A^k = \lambda_1 u_1\cdot v_1^T + \mathcal{O}(k^{m_2-1}|\lambda_2|^k)$,
where $u_1, v_1$ are Right- and Lefteigenvector with Eigenvalue $\lambda_1$. If $|\lambda_2|=\cdots = |\lambda_j|$ for some $j \geq 3$ then $m_2 \geq m_j$ where $m_j$ is the algebraic multiplicity of the Eigenvalue $\lambda_j$ .
I am not sure if we mean by $m_2$ the algebraic multiplicity of $\lambda_2$ and why we say that $m_2 \geq m_j$. Additonally I don't know how we are getting $\mathcal{O}(k^{m_2-1}|\lambda_2|^k)$. If $m_2$ is the algebraic multiplicity of $\lambda_2$ then I understand where the Landau comes from, as this is related to the convergence of a Jordanblock with length $m_2$.