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.