A primitive matrix is an irreducible matrix such that it has a unique dominant eigenvalue( positive and real). Another definition which does not use irreducibility is: $ A \text{ primitive} \iff \exists k :A^k >>0 $, where $>>$ means it has all elements strictly positive.
Let A be a primitive matrix of order n. Let $V_A$ be the normalized positive eigenvector associated with the dominant eigenvalue $\lambda_A$, which satisfies $\lambda_A =\lambda_1> |\lambda_2| \ge... \ge |\lambda_n|$. Then if $\lambda_2 \ne0$, as $k \rightarrow \infty$, we have: $$ A^k=(\lambda_A)^k V_A(V_A)^T +O(k^{m_2}|\lambda_2|^k) $$ where $m_2$ is the algebraic multiplicity of $\lambda_2$.
I know that, if A is diagonalizable $A^k =UD^kU^{-1}$. So if we call $r_i $ the rows of $U^{-1}$, then: $$ A^k=\sum_{i=1}^n \lambda_i^kV_ir_i$$ where the $V_i $s are the columns of $U$ that is the eigenvectors of A. Now the term O(..) should appear in the general form, with generalized eigenvectors. Still I don't understand why there is $(V_A)^T$. It looks like the row correspondent to $V_A$ in $U^{-1}$ is $(V_A)^T$. This would be true if for instance A was orthogonally diagonalizable(e.g. A symmetric) but why is it true in this case? From some consequence of Perron_Frobenius I know that the eigenvectors of A different form $V_A$ are orthogonal to the dominant eigenvector of $A^T$, that is $V_{A^T}$( notice not $(V_A)^T$), I don't know if it may depend on this. Can anyone provide a complete proof of this result?
The dominant term is false except -as you said- if $A$ is normal.
The correct first term is ${\lambda_A}^kV_A{W_A}^T$, where $V_A$ (resp. $W_A$) is an eigenvector of $A$ (resp. $A^T$) associated to $\lambda_A$ and satisfying ${W_A}^TV_A=1$.