Show that $\operatorname {trace}(X\rightarrow R(X,Y)Z)=\sum_{i=1}^n\langle R(e_i,Y)Z,e_i\rangle$

Question

$\langle\,,\rangle$ is Riemannian inner,and R(X,Y)Z is curvature tensor.

How to show that $\operatorname{trace}(X\rightarrow R(X,Y)Z)=\sum_{i=1}^n\langle R(e_i,Y)Z,e_i\rangle$ ?

Answer 1

If $A:V\to W$ is an operator and $f_i$'s is an orthonormal basis for $V$ then $$ trace A=\sum_k\langle Af_k,f_k\rangle $$ so for $A(x)= R(X,Y)Z$ $trace(X\rightarrow R(X,Y)Z)=trace A=\sum_{i=1}^n<R(e_i,Y)Z,e_i>$