Let $g_0$ be a Riemannian metric on the unit sphere and $Ric(g_0)$ be its Ricci curvature tensor. How may I show that $Ric(g_0) = (n-1)g_0$?
In particular, I ignore how to compute the curvature of any submanifold of $\mathbb{R}^n$ and I was therefore looking for some example.
Fix $x\in S^n$ Since ${\rm Ric}$ is symmetric so there exists $\{ e_i\}_{i=1}^{n}$ on $T_xS^n$ In further $$ R_{ij}=0,\ i\neq j $$ $$ R_{ii}=\sum_j K_{ij}$$ where $K_{ij}$ is sectional curvature
Since $K=K_{ij} $ for all $i\neq j$ (note that a round sphere has a symmetry), then $$ R_{ii}= (n-1) K $$
That is $$ {\rm Ric}=(n-1)K g_0 $$
Remaining thing is to determine a sectional curvature : A $2$-dimensional round sphere $S:=S^2\subset S^n$ is totally geodesic since a geodesic between any two points in $S$ is in $S$ Hence we suffice to find a sectional curvature for $S$ : $$ \int_S K\ d{\rm vol}=2\pi \chi (S) $$
Hence $K=1$