Consider a rank-1 matrix $\mathbf{A}=\lambda\mathbf{u}\cdot\mathbf{v}^\mathsf{T}$ where $\lambda,\mathbf{u},\mathbf{v}$ are known. Let $\mathbf{A}$ be of size $N\times N$.
(Edit: A direct consequence is that $\lambda$ is the only non-zero eigenvalue, with right and left eigenvectors $\mathbf{v},\mathbf{u}$ with $\mathbf{v}\cdot\mathbf{u}=1$.)
Is it possible to construct two sets of $N-1$ vectors that, combined with $\mathbf{u},$ and $\mathbf{v}$ form a biorthogonal basis? That is, let $\mathbf{x}_i,\mathbf{y}_i\in\mathbb{C}^N$ such that $\mathbf{x}_i\cdot\mathbf{u} = 0,\mathbf{y}_i\cdot\mathbf{v} = 0,\mathbf{x}_i\cdot\mathbf{y}_j=\delta_{ij}$.
If possible, how could I construct such a basis?
This is possible if $u,v$ are not orthogonal to each other.
We start with $u$, take any basis $x_1,\ldots, x_{n-1}$ in the orthogonal complement of $u$ in $\mathbb C^n$, then $x_1,\ldots, x_{n-1}\perp u$.
Denote $X$ as the matrix whose columns vectors are $x_1,\ldots, x_{n-1}$, and perform an SVD on $X$ to get $X=U\Sigma V$. Take its pseudo-inverse $X^{\dagger}=V^{-1}\Sigma^{\dagger}U^{-1}$ so that $X^{\dagger}X=I_{n-1}$ (the identity matrix of size $n-1$).
Denote columns in $(X^{\dagger})^*$ (Adjoint of $X^{\dagger}$) as $z_1,\ldots,z_{n-1}$, then $z_1,\ldots,z_{n-1}$ satisfy the bi-orthonormal relation $x_j\cdot z_k=\delta_{jk}$.
Now for each $k$, set $$y_k=z_k-\frac{v\cdot z_k}{v\cdot u}u,$$
then $y_1,\ldots,y_{n-1}$ still satisfy the bi-orthonormal relation (since $x_1,\ldots, x_{n-1}\perp u$), and $y_1,\ldots,y_{n-1}\perp v$ as desired.
If $u\perp v$ then such system need not exist, e.g., take $n=3$ and $u=(1,0,0)$, $v=(0,0,1)$.