From right-eigenvectors to left-eigenvectors? Given an irreducible non-symmetric, non-negative real matrix $A$ with eigenvalues $\lambda_1 \geq \lambda_2 \geq \lambda_3 \geq \cdots$ and $k$ of its largest right eigenvectors:
\begin{eqnarray}
%
A v_1 &= \lambda_1 v_1 \\
A v_2 &= \lambda_2 v_2 \\
\vdots & \vdots \\
A v_k &= \lambda_k v_k \\
%
\end{eqnarray}
What (if anything) can we say about it's $k$ largest left eigenvectors?
\begin{eqnarray}
%
w_1 A &= \lambda_1 w_1 \\
w_2 A &= \lambda_2 w_2 \\
\vdots & \vdots \\
w_k A &= \lambda_k w_k \\
%
\end{eqnarray}
 A: If $wA = \lambda w$, we have, using transposition, $A^T w^T = \lambda w^T$. 
Assuming $A$ is diagonalizable, we have $A = T^{-1} D T$ for an invertible matrix $T$ and a diagonal matrix $D$, so $A^T = T^T D^T (T^{-1})^T = T^T D (T^T)^{-1}$.
Therefore, $\lambda w^T = A^T w^T = T^T D (T^T)^{-1} w^T$, or equivalently, $\lambda (T^T)^{-1} w^T =  (T^T)^{-1}A^T w^T = D (T^T)^{-1} w^T$. But $\lambda v = D v$ iff $T^{-1} v$ is an eigenvector of $A$ (simple calculation), so $T^{-1} (T^T)^{-1} w^T = (T^T T)^{-1} w^T$ is an eigenvector of $A$.
Thus, $(T^T T)^{-1} w^T = v$ for some right eigenvector $v$, or equivalently, $w = ((T^T T) v)^T = v^T T^T T$. This should generalize to the non-diagonalizable case using arguments of the Jordan form.
A: Your left eigenvectors are characterized by the null space of $V$, as explaned in the following.
The matrix form of your known equations is $$AV = VD$$ where $D = diag(\lambda_i)$ for $1 \le i \le k$ and the columns of $V$ are the known eigenvectors.

Writing in terms also of the column matrix of unknown vectors, call it $W$, we have
$$A\pmatrix{V & W} = \pmatrix{V & W}\pmatrix{D & \mathbf{0} \\ \mathbf{0} & D_w}$$

Since $\pmatrix{V & W}$ is invertible (it is a basis) it diagonalizes A, and we may write
$$\pmatrix{V & W}^{-1}A\pmatrix{V & W} = \pmatrix{D & \mathbf{0} \\ \mathbf{0} & D_w}$$

Denote $\pmatrix{V & W}^{-1}=\pmatrix{\tilde{V}^T \\ \tilde{W}^T}$ to obtain (using $n$ as the dimension of $A$)
$$\pmatrix{\tilde{V}^T \\ \tilde{W}^T}\pmatrix{V & W} = \pmatrix{\mathbf{I}_k & \mathbf{0} \\ \mathbf{0} & \mathbf{I}_{n-k}} =\pmatrix{\tilde{V}^T V & \tilde{V}^TW \\ \tilde{W}^TV & \tilde{W}^T W}\tag{1}$$
In terms of the $k$ known right eigenvectors $V$:
$$\pmatrix{\tilde{V}^T \\ \tilde{W}^T} V = \pmatrix{\mathbf{I}_k \\ \mathbf{0}}$$
From this we see that $n-k$ of the left eigenvectors $\tilde{W}^T$ are contained in the perpendicular space to $V$ since $\tilde{W}^TV = \mathbf{0}$.
