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:
$$(\alpha\_0 a\_0 + \alpha\_1 a\_1) \otimes (\beta\_0 b\_0 + \beta\_1 b\_1) \otimes (\gamma\_0 c\_0 + \gamma\_1 c\_1).$$



Eigenvectors of a tensor in $\mathbb{C}^2 \otimes \mathbb{C}^2 \otimes \mathbb{C}^2$
**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.


