# Tensor symmetry: A symmetric mixed $(1,1)$ tensor is necessarily a multiple of the identity tensor

I found this observation in a book. "It is not possible to have an invariant definition of symmetry in one contravariant and one covariant index". That's all right, my problem is that to show how restrictive is to require symmetry on a mixed tensor $$A^i_j$$ the authors ask to resolve the following problem: If a tensor $$A$$ of type $$(1,1)$$ is symmetric in its indices with respect to every basis, that is, $$A^i_j=A^j_i$$, then $$A$$ is a multiple of the identity tensor, $$A^i_j=\alpha\delta^i_j$$.

I can prove that if $$i\neq j\implies A^i_j=0$$, but when $$i=j$$ I obtain different values instead of a constant value, in other words, the matrix $$A^i_j$$ is diagonal having different values along the diagonal.

Can anybody comment on this?

• Sounds like a job for Schur's lemma. Jul 18, 2022 at 20:18
• The book is "Tensor Analysis on Manifolds" by Bishop and Goldberg. It does not assume representation theory, so it is an easy exercise that should be easily resolved directly. However, there is always the possibility of being a mistake. Jul 18, 2022 at 20:42
• $\uparrow$ Which page? Jul 19, 2022 at 4:18
• Page 88, problem 2.15.1 "This Dover edition, first published in 1980, is an unabridged and corrected republication of the work originally published by The Macmillan Company in 1968" Jul 19, 2022 at 20:36

Consider the change of basis with the matrix $$\begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}$$. $$\begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0\\ 0 & a \end{pmatrix} \begin{pmatrix} 1 & -1\\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & a-1\\ 0 & a \end{pmatrix}$$ is symmetric if and only if $$a=1$$. In more than two dimensions just use this change of basis in a two-dimensional subspace and the identity in the normal space to it.
Sketched proof: The mixed tensor components $$A^i{}_j$$ transform (in matrix notation) as $$A^{\prime}=SAS^{-1}$$, where $$S\in GL(n,\mathbb{R})$$ is an arbitrary invertible matrix. Let $$S$$ be symmetric from now on. Then transposition yields $$A^{\prime}=S^{-1}AS$$ because all matrices are symmetric. Elimination of $$A^{\prime}$$ yields $$S^2A=AS^2$$. Now let $$S=\exp(ts)$$, where $$t\in\mathbb{R}$$, and $$s$$ is an arbitrary symmetric (not necessarily invertible) matrix. If follows that $$sA=As$$ commute. Let $$e_{ij}$$ denote the matrix whose $$(i,j)$$th matrix entry is $$1$$, and all other entries are $$0$$. By picking $$s=e_{ii}$$ we see that $$A$$ is diagonal. By picking $$s=e_{ij}+e_{ji}$$ we see that the diagonal elements of $$A$$ must be the same, i.e. $$A$$ is a multiple of the identity matrix. $$\Box$$
Thank you, guys. I think you both tackled special cases. However, your matrix approaches led me to a general solution I believe is also correct. From the general transformation $$A'=S^{-1}AS$$ we have $$SA'=AS$$. Assuming we already proved that $$A$$ is diagonal, let $$A=diag\{\lambda_1\ldots\lambda_n\}$$ and $$A'=diag\{\lambda'_1\ldots\lambda'_n\}$$, then we have $$(SA')_{ik}=\sum_j s_{ij}a'_{jk}=s_{ik}\lambda'_k$$ $$(AS)_{ik}=\sum_j a_{ij}s_{jk}=\lambda_i s_{ik}$$ Since the problem states the condition must hold for arbitrary bases, we can assume that for a given $$k,\quad s_{ik}\neq0$$ for al $$i$$, then we have $$\lambda_i=\lambda'_k=\alpha,\qquad i=1,2,\ldots,n$$ Since $$A^i_j=0$$ for $$i\neq j$$ we have $$A^i_j=\alpha\delta^i_j$$