For our discussion, we'll assume that we're working with $\mathbb{R}^m$ only, but much or all of the following discussion should be carried over immediately to any finite dimensional inner product space.

For completeness, we'll state a few definitions, that'll be used in the questions that follow.

**1) Tensors:** We'll call **$T$ a tensor of order $k$ (or just a $k$-tensor) on $\mathbb{R}^m$** if $T:(\mathbb{R}^m)^k\to \mathbb{R}$ is a $k$-linear map.

**2) Tensor product of two $1$-linear forms on $\mathbb{R}^m:$**

Let $A, B: \mathbb{R}^m\to \mathbb{R}$ be two $1$-linear forms. Then we define their tensor product $A\otimes B$ to be the bilinear form: $A\otimes B: (v,w)\mapsto A(v)B(w).$

**3) Constructing a $k$-tensor from just one vector:** Identifying $\mathbb{R}^m$ with its dual thanks to the canonical inner product on $\mathbb{R}^m,$ we can think of any vector $v$ as a $1$-linear form, and $v\otimes v\otimes \dots \otimes v =:v^{\otimes k}$ as a $k$-linear form (i.e. a $k$-tensor).

**4) Symmetry:** We define $T$ to be **symmetric** if $T(\sigma(v_1),\dots \sigma(v_n))=T(v_1\dots v_n)$ for every permutation $\sigma.$

**5) Diagonalizability and orthogonal diagonalizability :** Let $\{v_1\dots v_m\}$ be a basis for $\mathbb{R}^m.$ We say that $T$ is **diagonalizable** if there exist $\lambda_i\in \mathbb{R}, 1\le i \le m,$ so that $T=\sum_{i=1}^{k}\lambda_iv_i^{\otimes k}.$ Furthermore, we define $T$ to be **orthogonally diagonalizable** (in short, odeco, following this paper) if $v_i's$ form an orthonormal basis for $\mathbb{R}^m.$ Note that $T$ is odeco implies that there exists $R\in O(n), R:=[v_1\dots v_m],$ so that $T(x\dots x)=\sum_{i=1}^{m}\lambda_i(Rx)_i^k.$

P.S. The paper linked above, as well this paper do *not* define **diagonalizability** of a tensor, although they do **orthogonally diagonalizable**. I made up the former definition just by dropping the assumption that the basis need *not* be orthonormal. Also they used the word "orthogonally decomposable" instead of "orthogonally diagonalizable."

**Here are my questions:**

**By the definition above, all diagonalizable tensors are symmetric**(but this is not the case for matrices, so I'm having trouble here -**did I define diagonalizability of a tensor correctly?**)- When $k=2,$ a tensor becomes a bilinear form, that correspond to a matrix after choosing a basis. We know, thanks to the spectral theorem, that
**symmetry of a matrix is**. However, the papers linked above claim that there that this is not so for higher order tensors.*equivalent*to orthogonal diagonalizability**What's an example of a symmetric tensor that is not orthogonally diagonalizable (with proof), please?** **Density(?):**What's an appropriate topology (ideally a metrizable one) on the space of $k$-tensors, and in this topology, can we approximate a symmetric tensor by a sequence of orthogonally diagonalizable tensors? If no, then how about a sequence of diagonalizable tensors?

Relevant references very highly appreciated! I'll take the liberty to add a few, as it appears that tensors have been of interest in the signal processing community more than the math community.