Basis for the set of all covariant $k$-tensors on V Here's a proposition from Lee's Smooth Manifolds:

Let $V$ be a real vector space of dimension $n$, let $(E^i)$ be any basis for $V$, and let $(\epsilon^i)$ be the dual basis. The set of all $k$-tensors of the form $\epsilon^{i_1}\otimes...\otimes\epsilon^{i_k}$ for $1 \leq i_1,...,i_k \leq n$ is a basis for the set of all covariant $k$-tensors on $V$, denoted by $T^k(V)$, which therefore has dimension $n^k$.

I want to construct an example based on this proposition. Let $V=\mathbb{R}^3$, and let $e_1, e_2, e_3$ denote the usual basis for $\mathbb{R}^3$. Then the dual basis $(\epsilon^i)$ is $dx, dy$ and $dz$. Then $dx\otimes dy, dx\otimes dz, dy\otimes dz, dy\otimes dx, dz\otimes dy, dz\otimes dx, dx\otimes dx, dy\otimes dy$ and $dz\otimes dz$ form a basis for $T^2(\mathbb{R}^3)$.
Is the example correct until this point? If so, which of them are the symmetric 2-tensors?
 A: Just to elaborate on the comment of Santiago Canez:
We say that a covariant $k$-tensor $T$ is symmetric if it is unchanged under permutation of arguments, i.e:
$$T(X_1, \ldots, X_k) = T(X_{\sigma(1)}, \ldots, X_{\sigma(k)})$$
where $\sigma$ is any permuation of $k$ elements.
So it is immediate that $dx\otimes dx$, $dy\otimes dy$ and $dz\otimes dz$ are symmetric. Furthermore, there is a natural projection from $T^k(V)$ to the set of symmetric tensors called symmetrization. Let $S_k$ be the group of permutations of $\{1, \ldots, k\}$ and for any $\sigma\in S_k$ define
$$T^\sigma(X_1, \ldots, X_k) = T(X_{\sigma(1)}, \ldots, X_{\sigma(k)}).$$
Then we define $\text{Sym}: T^k(V) \to\text{symmetric tensors}$, by
$$\text{Sym}\, T = \frac{1}{k!}\displaystyle\sum_{\sigma\in S_k}T^\sigma.$$
Then $\text{Sym}\, T$ is a symmetric tensor. 
A: Yes, you are correct. As you have quoted, $T^k(V)$ has dimension $n^k$. Now $n=3$ and $k=2$. Therefore $3^2=9$ is the number of bases and there are $9$ elements in 
$$\{dx\otimes dy, dx\otimes dz, dy\otimes dz, dy\otimes dx, dz\otimes dy, dz\otimes dx, dx\otimes dx, dy\otimes dy, dz\otimes dz\}.$$
None of them is sysmmetric 2-tensor. By definition, symmemetric tensor is a tensor $T$ that is invariant under a permutation of its vector arguments. Therefore, permutate $x$ and $y$, we have
$$dx\otimes dy\leftrightarrow dy\otimes dx,\hspace{2mm}dx\otimes dz\leftrightarrow dy\otimes dz, \hspace{2mm}
 dz\otimes dy\leftrightarrow dz\otimes dx,\hspace{2mm} dx\otimes dx\leftrightarrow dy\otimes dy.$$
On the other hand, permutate $x$ and $z$, we have
$$dx\otimes dx\leftrightarrow dz\otimes dz.$$
Therefore, none of them is symmetric. However, you can find symmetric 2-tensor by forming the linear combination of them (since they form a basis), e.g.
$$dx\otimes dy+dy\otimes dx+dx\otimes dz+dz\otimes dx+dy\otimes dz+dz\otimes dy$$
or 
$$dx\otimes dx+dy\otimes dy+dz\otimes dz$$
are symmetric 2-tensors.
