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Any suggestion on what kind of dataset lets say $n \times d$ ($n$ rows, $d$ columns) would give me same eigenvectors?.

I believe it should be the one with same absolute values in each cell. Like alternating +1 and -1. But it seems to work otherwise.

Any pointers?

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I will restrict myself to the finite-dimensional situation.

First, if we are talking about normal eigenvalues and not generalized ones, we need a square matrix. If $M\in\mathbb{R}^{n\times n}$ (or $\mathbb{C}^{n\times n}$) is a diagonizable matrix, it has an eigendecomposition \begin{equation} M = V\Lambda V^{-1} \textrm{,} \end{equation} with $V:=\begin{bmatrix}v_{1},v_{2},\ldots,v_{n}\end{bmatrix}$ the matrix with its $n$ eigenvectors $v_{i}$ as columns and $\Lambda :=\operatorname{diag}\left(\lambda_{1},\lambda_{2},\ldots,\lambda_{n}\right)$ the diagonal matrix with the corresponding eigenvalues on its main diagonal.

As you can see from this equation if you want to get matrices with the same eigenvectors, i.e. $V$ fixed, you can only change the eigenvalues $\lambda_{i}$. In general the relationship between the eigenvalues and the entries of the original matrix $M$ is non-trivial.

The easiest situation arises if the eigenvectors $v_{i}$ form the standard basis, i.e. only $1$ at the $i$th position, $0$ otherwise. Then $M=I\Lambda I^{-1}=\Lambda $ ($I$ the identity matrix) and you can change every entry on the main diagonal without changing any eigenvector.

Edit

I understood the question differently than victor. I thought the question asks what kind of different datasets can be described by the same eigenvectors.

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The only such existing "dataset" is a 1-by-1 matrix with an arbitrary value of its sole item.

If the eigenvectors, being the columns of matrix $V$, are "the same" (precisely up to a scalar constant factor), then matrix $V$ is singular (its columns are linearly dependent; its determinant is zero), and, hence, non-invertible $\Rightarrow$ such an eigendecomposition $V \cdot \Lambda \cdot V^{-1}$ cannot exist.

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  • $\begingroup$ Could you elaborate further on your answer? I myself accept what you said, which seems reasonable to happen, but don't actually understand why it is so. $\endgroup$ – Rubens Mar 6 '15 at 9:33

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