In Dajczer's Submanifolds and Isometric Immersions, a couple paragraphs before presenting the Fundamental Theorem for Submanifolds, the author presents Ricci's equation: $$(\widetilde{R}(X,Y)\xi)^\perp=R^\perp(X,Y)\xi+\alpha(A_\xi X,Y)-\alpha(X,A_\xi Y)\text{, }\forall X,Y\in T_xM, \xi\in T_xM^\perp$$
(where $\alpha$ is second fundamental form and $A$ is the shape operator)
He points out that if $\widetilde{M}$ has constant sectional curvature, Ricci's formula becomes: $$R^\perp(X,Y)\xi=\alpha(X,A_\xi Y)-\alpha(A_\xi X, Y)$$
Then he says: "notice that $R^\perp_x=0$ if and only if there exists an orthogonal basis for $T_xM$ that diagonalizes simultaneously all $A_\xi, \xi\in T_xM^\perp$"
If I understand correctly, he means there is a basis $\{v_1,...,v_n\}$ of $T_xM$ with $\langle v_i,v_j\rangle=\delta_{ij}$ and such that $A_\xi v_i=\lambda_iv_i$ for all $i$ and for all $\xi\in T_xM^\perp$.
The way he mentions it makes it look like something obvious, but I really can't see how this relates to $R_x^\perp=0$. Am I missing something?
Take the inner product of the Ricci formula with another normal vector $\eta\in T_x M^\perp$. You get $$ \begin{align*} \langle R^\perp(X,Y)\xi,\eta\rangle &= \langle \alpha(X,A_\xi Y),\eta \rangle - \langle \alpha(A_\xi X,Y),\eta \rangle \\ &= \langle A_\eta X, A_\xi Y \rangle - \langle A_\eta A_\xi X, Y \rangle \\ &= \langle A_\xi A_\eta X, Y \rangle - \langle A_\eta A_\xi X, Y \rangle \\ &= \langle [A_\xi, A_\eta]X, Y\rangle. \end{align*} $$ Now you see that $R^\perp_x$ vanishes iff the commutators $[A_\xi, A_\eta]$ of the shape operators vanish for all $\xi, \eta \in T_x^\perp M$. Recall that the shape operators are diagonalisable, since they are symmetric. It is a well-known fact that diagonalisable linear maps can be diagonalised simultaneously if and only if they commute. This shows the statement.