Unique symmetric covariant $k$-tensor satisfying $(\operatorname{Sym} T)(A,...,A)=T(A,...,A)$ for all $A \in V$ Let $T$ be a covariant $k$-tensor on a finite dimensional vector space $V$. I want to prove that the symmetrization of $T$ is the unique symmetric $k$-tensor satisfying the following condition:
$(\operatorname{Sym} T)(A,\ldots,A)=T(A,\ldots,A)$ for all $A \in V$. 

Definition. Symmetrization of $T$ is defined as $$(\operatorname{Sym}T)(A_1,\ldots,A_k)=\frac{1}{k!} \sum_{\sigma \in S_k} T(A_{\sigma(1)},\ldots,A_{\sigma(k)})$$ where $S_k$ is the symmetric group on $k$ letters.

I assumed that there exists another symmetric $k$-tensor $\tilde{T}$ which satisfies the condition. Since $\tilde{T}$ is symmetric, it is equal to its symmetrization $\operatorname{Sym} \tilde{T}$. Then I tried to show that $(\operatorname{Sym}T)(A_1,\ldots,A_k)=\tilde{T}(A_1,\ldots,A_k)$, or equivalently, $(\operatorname{Sym}T)(A_1,\ldots,A_k)=(\operatorname{Sym}\tilde{T})(A_1,\ldots,A_k)$ but I couldn't. 
Thanks in advance.

Here's an analysis of the case $k=2$. Suppose $\bar T$ is another symmetric tensor satisfying $\bar T(A,A)= T(A,A)$ for all $A$.
Now $$T(x+y,x+y)=\bar T(x+y,x+y)= \bar T(x,x)+\bar T(x,y)+\bar T(y,x)+\bar T(y,y).$$ This equals $T(x,x)+2\bar T(x,y)+T(y,y)$. Thus
$$\bar T(x,y)=\frac{1}{2}(T(x+y,x+y)-T(x,x)-T(y,y)).$$
So $\bar T$ is uniquely determined by $T$. Since $Sym(T)$ is symmetric and also satisfies the formula $Sym(T)(A,A)=T(A,A)$, $\bar T$ and $Sym(T)$ must be equal.
A similar, but more complicated, argument works in the case of $k>2$.
