Find the rank of the tensor I need to find the rank of the tensor $t = a \otimes a \otimes b + a \otimes b \otimes a + b \otimes a \otimes a$. For simplicity let $a = (1, 0)^T$ and $b = (0, 1)^T$.
I know the answer, rank equals 3. And moreover as I know rank equals 3 for every pair of linear independent vectors $a$ and $b$. But I do not know how to prove that rank equals 3 even in this particular case.
I tried to write an assumption that $t = u_1 \otimes v_1 \otimes w_1 + u_2 \otimes v_2 \otimes w_2$ and write elementwise equations. But I got eight equations, so it looks difficult to solve:
\begin{cases}
t_{111} = u_{11}v_{11}w_{11} + u_{21}v_{21}w_{21} = 0\\
t_{112} = u_{11}v_{11}w_{12} + u_{21}v_{21}w_{22} = 1\\
t_{121} = u_{11}v_{12}w_{11} + u_{21}v_{22}w_{21} = 1\\
t_{122} = u_{11}v_{12}w_{12} + u_{21}v_{22}w_{22} = 0\\
t_{211} = u_{12}v_{11}w_{11} + u_{22}v_{21}w_{21} = 1\\
t_{212} = u_{12}v_{11}w_{12} + u_{22}v_{21}w_{22} = 0\\
t_{221} = u_{12}v_{12}w_{11} + u_{22}v_{22}w_{21} = 0\\
t_{222} = u_{12}v_{12}w_{12} + u_{22}v_{22}w_{22} = 0\\
\end{cases}
Thanks for the help, do not judge strictly it is my first experience with tensors.
 A: $8=2^2+2^2$
Let $\mathbf V=\left[\begin{matrix}v_{11}&v_{12}\\v_{21}&v_{22}\end{matrix}\right]$ and 
$\mathbf W=\left[\begin{matrix}w_{11}&w_{12}\\w_{21}&w_{22}\end{matrix}\right]$.
Then the system above can be rewritten as
\begin{equation*}
\begin{split}
\mathbf V^T diag(u_{11},u_{21})\mathbf W = 
\left[\begin{matrix}t_{111}&t_{112}\\t_{121}&t_{122}\end{matrix}\right]=
\left[\begin{matrix}0&1\\1&0\end{matrix}\right],\\
\mathbf V^T diag(u_{12},u_{22})\mathbf W = 
\left[\begin{matrix}t_{211}&t_{212}\\t_{221}&t_{222}\end{matrix}\right]=
\left[\begin{matrix}1&0\\0&0\end{matrix}\right].
\end{split}
\end{equation*}
From the first equation it follows that $\det \mathbf V\cdot u_{11}\cdot u_{21}\cdot\det\mathbf W=-1\ne 0$.Hence, 
$$
\mathbf V^T =  
\left[\begin{matrix}0&1\\1&0\end{matrix}\right] \mathbf W^{-1}
diag(u_{11}^{-1},u_{21}^{-1}),
$$
and from the second equation,
$$
\left[\begin{matrix}0&1\\1&0\end{matrix}\right] \mathbf W^{-1}
diag(u_{11}^{-1},u_{21}^{-1})diag(u_{12},u_{22})\mathbf W = \left[\begin{matrix}1&0\\0&0\end{matrix}\right].
$$
Finally,
$$
\mathbf W^{-1}
diag(u_{11}^{-1}u_{12},u_{21}^{-1}u_{22})\mathbf W = \left[\begin{matrix}0&0\\1&0\end{matrix}\right],
$$
a contradiction (we have obtained that a diagonal matrix is similar to a jordan block).
