Given any unit vector $V\in T_pM$, choose an orthonormal basis $\{Ei\}$ for $T_pM$ such that $E_1=V $. Then the Ricci curvature $Rc(V,V)$ at a vector $V$ is given by $$Rc(V,V)=R_{11}=R_{k11}^k=\sum_{k=1}^{n} Rm(E_k,E_1,E_1,E_k)=\sum_{k=2}^{n}K(E_1,E_k)$$ Where $R$ is the Ricci curvature tensor, $Rm$ is the Riemann curvature tensor, and $K$ is the scalar curvature.
I know that $R$ is a bilinear, symmetric tensor, so in principle it is completely determined by its values at $Rc(V,V)$ for norm $1$ vectors. Can I use the polarization identity to recover it from $Rc$?
For example suppose that $Rc(V,V)=(n-1)$ for any vector such that $g(V,V)=1$. Can I then say the following?$$\begin{align}Rc(X,Y)&=\frac{Rc(X,X)+Rc(Y,Y)-Rc(X-Y,X-Y)}{2}=\\&=(n-1)\frac{g(X,X)+g(Y,Y)-g(X-Y,X-Y)}{2}=(n-1)g(X,Y)\end{align}$$