Exercise 5.A.19

Proof. Suppose $\lambda$ is an eigenvalue of $T$. Then

That implies $x_k=\frac{1}{\lambda }(x^1+\cdots +x_n)$ for each $k=1,\dots ,n$. Hence all the $x$’s are equal. Therefore $\lambda =n$ and the corresponding eigenvectors are those which all entries equal to each other. Since every eigenvector of $T$ must satisfy such property and eigenvectors corresponding to distinct eigenvalues are linearly independent, it follows that $T$ has no other eigenvalues. □