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If we are given a set of vectors $v_1$, $v_2$....$v_n$ how many matrices can we construct with the given set of vectors as the eigenvectors of the matrix?

If a set of eigenvalues and eigenvectors were given then we could have found a unique matrix. In this case no information on eigenvalues is given and hence they can be chosen arbitrarily. So, I think there are an infinite number of matrices which satisfies the required condition. Is my conclusion correct?

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    $\begingroup$ Since $Iv=v$ then your eigenvector will be an eigenvector of the identity matrix with eigenvalue 1. Now take $2Iv=2v$ therefore it is an eigenvector of $2I$ with eigenvalue 2. And so on. $\endgroup$ – Artem Nov 27 '16 at 15:38
  • $\begingroup$ @Artem If we choose n numbers $x_1,x_2,...x_n$ as the eigenvalues then we can write down a diagonal matrix with the eigenvalues as the diagonal elements. Then, we can use a matrix S with its columns as the eigenvectors and use it to find the matrix in the standard basis (reverse process of diagonalization). But since, $x_1,x_2,...x_n$ are arbitrary we have an infinite number of such matrices. Is my argument correct? $\endgroup$ – Rajath Krishna R Nov 27 '16 at 15:44
  • $\begingroup$ "If a set of eigenvalues and eigenvectors were given then we could have found a unique matrix." Be careful, unless you add additional constraints you can also have infinitely many matrices for a specified set of eigenvalues and eigenvectors. $\endgroup$ – Christiaan Hattingh Nov 27 '16 at 17:55

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