$Alt(T)=0$ if $T$ is a symmetric tensor Question is to prove that $Alt(T)=0$ if $T$ is a symmetric tensor.
We have $$Alt(T)=\sum_{\sigma}sgn(\sigma)T^{\sigma}$$
As $T$ is symmetric we have $T^{\sigma}=T$ for all $\sigma$.
So, we have $$Alt(T)=\sum_{\sigma}sgn(\sigma)T=T\sum_{\sigma}sgn(\sigma)$$
As no of odd permutations is same as that of no of odd permutations we have
$\sum_{\sigma}sgn(\sigma)=0$ thus, $Alt(G)=0$.
Now, another question is, does the converse hold?
Suppose  $Alt(T)=0$ do we have that $T$ is a symmetric tensor? 
I see that this is true for $2$ tensor. Suppose $T$ is a $2$ tensor then,
$0=Alt(T)=sgn(1)T+sgn(1~2)T=T-T^{\sigma}$  implies that $T=T^{\sigma}$.
Thus, $T$ is a symmetric tensor.
I am sure this is true for all $k$ tensors but i am unable to prove.
Do i have to go by contradiction or what... I am not able to guess some path..
Please provide some hints..
 A: There is nice characterisation of tensors which $Alt(T)=0.$

Thm. Let $V$ be a vector space over $\mathbb{R}$ (without any additional assumptions). Consider subspace $N^n(V)$ of $\otimes^nV$ generated by elements $v_1\otimes\dots\otimes v_n$ such that $v_i=v_j$ for at least one pair $i\neq j.$ We have that
  $$\ker(Alt)=N^n(V).$$
  Equivalently. For every $T\in\otimes^nV$
  $$Alt(T)=0\iff T\in N^n(V).$$

It requires some work, but you can find it in Greub's Multilinear Algebra.

EDIT Explicit example
Consider space $V=(\mathbb{R}^2)^*.$ Then $3-$tensors in $V$ can be treated as multilinear maps
$$T:\mathbb{R}^2\times\mathbb{R}^2\times\mathbb{R}^2\rightarrow\mathbb{R}.$$
Set $T$ by formula
$$T((a,b),(c,d),(e,f))=ace+bde-acf-bdf.$$
Multilinear: easy to verify.
Symmetric: Nope
$$T((1,1),(1,1),(1,0))=2\neq 0=T((1,0),(1,1),(1,1))$$
$Alt(T)=0.$ Yes. See first that $$T((a,b),(c,d),(e,f))=T((c,d),(a,b),(e,f))$$
(this is the main idea, compare with the thm. in quotation above). In fact
$$T((a,b),(c,d),(e,f))=ace+bde-acf-bdf=cae+dbe-caf-dbf=T((c,d),(a,b),(e,f)).$$
Now $Atl(T)$ has $6$ elements. To permutation $\sigma=id$ corresponds $\gamma=(12),$ such that $$T^\sigma=T^\gamma\hspace{5pt}\text{and}\hspace{5pt}sgn(\sigma)=-sgn(\gamma).$$ Similarly to $\sigma=(13)$ you have $\gamma=(123)$ and to $\sigma=(23)$ you have $\gamma=(132).$ Hence the whole sum $Alt(T)$ cancels out.
