Answer
True
Work Step by Step
Assume $A$ is $n \times n$ matrix with diagonal elements $a_1,a_2,...a_n$
Then, $\det (A-\lambda I)=0\\
(a_1-\lambda)(a_2-\lambda)...(a_n-\lambda)=0$
The eigenvalues are $\lambda=a_1,a_2,...a_n$.
Hence, the statement is true.
