Rank 1 tensors, how to describe them? (specific case)

Question

I want to undestand a specific case. I consider two $\mathbb{C}$-vectorial spaces, $\mathbb{C}^2$ both. Then, I want to work with $\mathbb{C}^2\otimes \mathbb{C}^2$. Now, I consider basis for each vectorial space $\langle a,b\rangle=\mathbb{C}^2$ and $ \langle u,v\rangle=\mathbb{C}^2$. In this point, I consider a tensor in $\mathbb{C}^2\otimes \mathbb{C}^2$ expressed in terms of these basis as follows: $$x (a\otimes u)+y(a\otimes v)+w( b\otimes u)+ z( b\otimes v)$$ and I'd like to see when it is a rank 1 tensor, so it has to be like this: $$\lambda_1a+\lambda_2b\otimes \gamma_1u+\gamma_2v$$
From these and making some algebra, I got $$\frac{x}{\gamma_1}=\frac{y}{\gamma_2};\frac{w}{\gamma_1}=\frac{z}{\gamma_2}$$
$$\frac{x}{\lambda_1}=\frac{w}{\lambda_2};\frac{y}{\lambda_1}=\frac{z}{\lambda_2}$$
And it says me if consider this tensor as a matrix, then this matrix has rank 1. But I'd like to get some equations in terms just of $x,y,w$ and $z$, and if it's possible get some characterization of rank 1 tensors in terms of these equations

Answer 1

If you make the matrix $$ \begin{bmatrix} x & w \\ y & z \end{bmatrix} $$ the condition for your tensor to be rank-1 is that this matrix must be rank-1, as you said. So the same equations that guarantee you have a rank-1 matrix are necessary and sufficient. That means that one of the eigenvalues is $0$, hence the determinant must equal $0$. But the matrix is not the $0$-matrix. So you could write $xz-wy=0$ because that is the formula for the determinant. But also $|x|^2+|y|^2+|z|^2+|w|^2\neq 0$ because that is equivalent to saying that the matrix is not $0$.