Given eigenvectors $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n$ and eigenvalues $\lambda_1,\lambda_2,\dots,\lambda_n$, one can construct a matrix $\mathbf{V}:=(\mathbf{v}_1,\dots, \mathbf{v}_n)$ containing the eigenvectors and a diagonal matrix $\mathbf{D}$ containing the corresponding eigenvalues. Now a matrix $\mathbf{M}$ having these eigenvalues and vectors will comply to $\mathbf{MV}=\mathbf{VD}$. Matrix $\mathbf{M}$ can be found by solving $\mathbf{M}=\mathbf{VDV}^{-1}$.
My question is: can one construct a matrix $\mathbf{M}$ that only contains elements at specific locations $m_{i, j}$? Assuming that that the element placement permits the construction of a matrix that has $n$ independent eigenvectors. For example a tri-diagonal matrix.
I am also interested in solutions to this problem given $m$ eigenvectors and eigenvalues, $m < n$, the rest can be arbitrary.