Eigenvectors of a tensor in $\mathbb{C}^2 \otimes \mathbb{C}^2 \otimes \mathbb{C}^2$ I want to find the critical point of tensor $f=a_0b_0c_0 + a_1b_1c_1$ in $\mathbb{C}^2 \otimes \mathbb{C}^2 \otimes \mathbb{C}^2$, and I followed this construction:
First, I take the following partial derivative:
With respect to $a$,  $\cfrac{\partial f }{\partial a_0}=b_0c_0, \cfrac{\partial f }{\partial a_1}=b_1c_1,$ and with respect to $b$,  $\cfrac{\partial f }{\partial b_0}=a_0c_0, \cfrac{\partial f }{\partial b_1}=a_1c_1,$ and with respect to $c$,  $\cfrac{\partial f }{\partial c_0}=a_0b_0, \ \ \cfrac{\partial f }{\partial c_1}=a_1b_1.$
Then I need to find $2 \times 2$-minors of the following matrices:
$$
\begin{bmatrix}
    b_0c_0 & b_1c_1  \\
    a_0 & a_1  \\
\end{bmatrix},
\begin{bmatrix}
    a_0c_0 & a_1c_1  \\
    b_0 & b_1  \\
\end{bmatrix},
\begin{bmatrix}
    a_0b_0 & a_1b_1  \\
    c_0 & c_1  \\
\end{bmatrix},
$$
Then I have the following system of equation to solve:
\begin{equation*}
    \left\{
    \begin{alignedat}{3}
        % R & L   &  R & L   &  R & L 
        a_1b_0c_0 & - a_0b_1c_1   = 0 \\
        b_1a_0c_0 & - b_0a_1c_1   = 0 \\
        c_1a_0b_0 & - c_0a_1b_1   = 0
    \end{alignedat} \ ,
    \right.
    \end{equation*}
There are 6 variables and 3 equations, and by solving this system we get 23 group of solutions, then we have many infinite solutions, I am wondering if there is another way. Solution has the form:
$$(\alpha_0 a_0 + \alpha_1 a_1) \otimes (\beta_0 b_0 + \beta_1 b_1) \otimes (\gamma_0 c_0 + \gamma_1 c_1).$$
 A: Recall. Consider a $d$-dimensional rectangular tensor $T$ in $\mathbb{K}^{n_1 \times \dots \times n_d}$. It corresponds to a multilinear form:
$$T=\sum_{i_1=1}^{n_1} \sum_{i_2=1}^{n_2} \dots \sum_{i_n=1}^{n_d} t_{i_1 \dots i_d}x_{i_1}x_{i_2} \dots x_{i_d}.$$
The singular vector tuples (critical points) of $T$ are the fixed points of the gradient
map $$\nabla T: \mathbb{P}^{{n_1}-1} \times \dots \mathbb{P}^{{n_d}-1} \dashrightarrow \mathbb{P}^{{n_1}-1} \times \dots \mathbb{P}^{{n_d}-1}  .$$
$f$ is considered as a $2 \times 2 \times 2$-tensor. The gradient $\nabla f$ of this trilinear form is the rational map:
$$\nabla T: \mathbb{P}^{1} \times \mathbb{P}^{1} \times \mathbb{P}^{1}  \dashrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \  \  \mathbb{P}^{1} \times \mathbb{P}^{1} \times \mathbb{P}^{1} $$
$$((a_0:a_1),(b_0:b_1),(c_0:c_1))  \mapsto  \ \ ((b_0c_0:b_1c_1),(a_0c_0:a_1c_1),(a_0b_0:a_1b_1)),$$
then we need to find fixed points of the map $\nabla T$. As we see in the question this leads us to the written system of equations, by solving it we get exactly $6$ solutions which lie on $\mathbb{P}^{1} \times \mathbb{P}^{1} \times \mathbb{P}^{1}$(Segre variety), it means $(a_0,a_1),(b_0,b_1),(c_0,c_1)\neq(0,0)$. The solutions are:
$$((1:0),(1:0),(1:0)), \  ((0:1),(0:1),(0:1)), \  ((1:1),(1:1),(1:1)), \  ((1:1),(1:−1),(1:−1)), \ ((1:−1),(1:1),(1:−1)), \ ((1:−1),(1:−1),(1:1)).$$
The expected number of singular vector triples is predicted by the following
theorem.
Theorem.[Friedland and Ottaviani] For a general $n_1 \times \dots \times n_d$-tensor $T$ over an
algebraically closed field $\mathbb{K}$, the number of singular vector tuples is the coefficient of the
monomial $z_{1}^{n_1-1} \dots z_{d}^{d_1-1}$ in the polynomial
$$\prod_{i=1}^{n} \cfrac{\hat{z_i}^{n_i} - {z_i}^{n_i} }{\hat{z_i}-z_i},$$
where $\hat{z_i}=z_1+ \dots +z_{i-1}+z_{i+1}+ \dots + z_d.$
Here our case is $2 \times 2 \times 2$-tensors, so we will have 6 singular triples.
