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.
 A: 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 $k>2$.
A: A symmetric tensor $T:V^n\to k$ descends to a linear functional $\tilde{T}:\mathrm{Sym}^nV\to k$ on the symmetric algebra. Denote the diagonal $V^{[n]}=\{v^n:v\in V\}$. As illustrated in the other answer, when $n=2$,
$$xy=\frac{(x+y)^2-(x^2+ y^2)}{2},$$
demonstrating that every element in $\mathrm{Sym}^2(V)$ can be written as a combination of elements from our set of squares $V^{[2]}$. Similarly, for arbitrary $n$, it suffices for our purposes to show that $\langle V^{[n]}\rangle$ is always $\mathrm{Sym}^nV$, because then every element is a linear combination of the $n$th powers, and so $\tilde{T}$ is uniquely determined by its values on $V^{[n]}$.
Equivalently, we seek to prove that in the polynomial ring $k[x_1,\cdots,x_n]$, the elementary symmetric polynomial $e_n$ can be written as a sum of $n$th powers of degree $1$ homogeneous (but not necessarily symmetric!) polynomials. If we can do this, we can write $x_1\cdots x_n$ as a linear combination of terms from $V^{[n]}$ by mimicking the formula for the polynomial ring.
This is an incomplete answer. I'm not sure if the above thoughts make this question (in the general case) easier, or harder, but it seems like a natural route to follow. I will ask a follow-up question...
A: Suppose there is another symmetric tensor $(\text{Sym}~\alpha)'$ such that $$(\text{Sym}~\alpha)'(v,\dots,v) = \alpha (v,\dots,v),~ \forall v\in V$$
Let's define a tensor $\beta\in T^k(V^*)$ by: $\beta = (\text{Sym}\alpha) - (\text{Sym}\alpha)' $. We observe that $$\beta(v,\dots,v)=0, ~\forall v \in V$$
Moreover, we clearly see that $\beta$ is actually symmetric, since $\Sigma^k(V^*)$ is a vector space. Define $\gamma_v:(-\epsilon,\epsilon)\ni t\mapsto v+t w_1 \in V$, where $v,w_1\in V, ~\epsilon >0$. Then, we have:
$$ \beta(\gamma_v(t),\dots,\gamma_v(t)) = 0, \quad \forall t\in (-\epsilon,\epsilon) \implies \frac{d}{dt}(\beta(\gamma_v(t),\dots,\gamma_v(t))) = 0, \forall t\in (-\epsilon,\epsilon)$$
Now, we compute:
$$\frac{d}{dt}(\beta(\gamma_v(t),\dots,\gamma_v(t)))\big|_{t=0} = \beta(w_1,v,\dots,v) + \beta(v,w_1,v,\dots,v) + \cdots + \beta(v,\dots,v,w_1) = 0 $$
Since $\beta$ is symmetric, we conclude that $\beta(w_1,v,\dots,v)=0, \forall w_1,v\in V$. We repeat this argument $k-1$ times, each time replacing $v$ at some slot with an arbitrary $w_i\in V$ until we conclude that $\beta(w_1,\dots,w_k)=0, ~\forall w_i\in V$. This means that $\beta\equiv 0$, i.e. $(\text{Sym}\alpha) = (\text{Sym}\alpha)'$
