I have to show that if $\displaystyle A$ is a $\displaystyle (1,1)$ tensor then $\displaystyle (tr A)^2$ is a linear function of $\displaystyle A \otimes A$.

My approach:

If I define a function $\displaystyle f$ by $\displaystyle f(A \otimes A) = (A \otimes A)^{ij}_{ij} $ (einstein summation is used, and $\displaystyle (A \otimes A)^{ij}_{ij} $) is the coefficient respect to a basis, then I have:

$\displaystyle f((A \otimes A)) = (A \otimes A)^{ij}_{ij} = A^i_i A^j_j = (tr A)^2$

and

$\displaystyle f(\alpha A \otimes A + B \otimes B) = (\alpha A \otimes A + B \otimes B)^{ij}_{ij} =^* \alpha (A \otimes A)^{ij}_{ij} + (B \otimes B)^{ij}_{ij} = $

$\displaystyle \alpha f(A \otimes A) + f(B \otimes B)$

I'm not quite sure about the identity I have used at the $\displaystyle ^*$, but maybe it follows from:

Let $\displaystyle e_j and \varepsilon^i$ be a basis such that

$\displaystyle A = A^i_j e_i \otimes \varepsilon^j$

and

$\displaystyle B = B^i_j e_i \otimes \varepsilon^j $

then

$\displaystyle \alpha A + B = \alpha A^i_j e_i \otimes \varepsilon^j + B^i_j e_i \otimes \varepsilon^j = (\alpha A^i_j + B^i_j) e_i \otimes \varepsilon^j $

by uniqueness of the coefficients i get:

$\displaystyle (\alpha A + B)^i_j = \alpha A^i_j + B^i_j$

I think I could proove this in the more general case wich I have used where I use a theorem that states that for any tensor (r,s) there are unique coefficients $\displaystyle A^{i_1 ... i_r}_{j_1 ... j_s}$, but there are so much indices so I don't want to wright it here.