Matrices that are not diagonal or triangular, whose eigenvalues are the diagonal elements

I am not aware of any relevant research. Yet, for any $n\ge3$, there always exists a matrix that is non-triangular but whose eigenvalues are any $n$ given scalars $\lambda_1,\lambda_2,\ldots,\lambda_n$. The construction is recursive. First, we begin with a triangular matrix $$A_2=\pmatrix{\lambda_1&1\\ 0&\lambda_2}.$$ Now, if $n\ge3$ is odd, we define $$A_n=\pmatrix{A_{n-1}&0\\ \mathbf1^T&\lambda_n},$$ where $\mathbf 1$ is a vector of ones of appropriate length. If $n\ge3$ is even, define $$A_n=\pmatrix{A_{n-1}&\mathbf1\\ 0&\lambda_n}.$$ To illustrate, we have $$ A_4=\pmatrix{\lambda_1&1&0&1\\ 0&\lambda_2&0&1\\ 1&1&\lambda_3&1\\ 0&0&0&\lambda_4}. $$ Clearly, $A_n$ is not triangular (although it is block triangular) when $n\ge3$, because it has both subdiagonal and superdiagonal nonzero elements. Furthermore, as $A_n$ is block triangular, its eigenvalues are $\lambda_n$ and those eigenvalues of $A_{n-1}$. In turn, $\lambda_1,\lambda_2,\ldots,\lambda_n$ are eigenvalues of $A_n$.

matrix example of this type : Let $A$ and $B$ are two Jordan matrices form not diagonalizable, and $M$ the matrix obtained by concatenating $A$ and $^tB$, then $M$ is a matrix with two block in its diagonal and this block one are upper triangular matrix and the other are lower triangular matrix. Example $M =\left( \begin{array}{cccc} \alpha & 1 &0&0\\ 0 & \alpha&0&0\\ 0 & 0&\beta&0\\ 0&0&1&\beta \end{array} \right) $