To expand on vela's response- presumably you know that if A has n independent eigenvectors (where A is an n by n matrix) then, with P the matrix having those eigenvectors as columns, [itex]PAP^{-1}= D[/itex] where D is the diagonal matrix with the eigenvalues of A on the diagonal. From that, [itex]A= P^{-1}DP[/itex]. If you are given the eigenvalues and eigenvectors of A, you can form both P and D from that information and so find A.
If A is not diagonalizable (does not have n independent eigenvectors), then it is a little harder but the same idea- D will be the Jordan Normal Form matrix with eigenvalues along the diagonal and possibly "1"s above the diagonal. There will be fewer than n eigenvectors so you will have to supplement them with "generalized eigenvectors" to form the matrix P. Fortunately, the generalized eigenvectors of A are the same as those of D so that can be done.