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If you know the eigenvectors of a specific matrix, is that enough information to get the original matrix back? So for example, for some 2x2 matrix, it is known that its two distinct eigenvectors are [1, u] and [1, v] after normalization, can the original matrix be recovered?

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up vote 8 down vote accepted

No, it's not enough information. For instance, both $$\left(\begin{array}{cc} 1 & 0\\ 0 & 2 \end{array}\right)\quad\text{and}\quad\left(\begin{array}{cc} 7 & 0\\ 0 & -1 \end{array}\right)$$ have $[1,0]$ and $[0,1]$ as eigenvectors, but they are not even similar matrices (let alone equal).

However, if you have enough eigenvectors ($n$ linearly independent eigenvectors in the case of an $n\times n$ matrix) and you know their corresponding eigenvalues, then you can recover the original matrix.

If $\mathbf{v}_1,\ldots,\mathbf{v}_n$ are linearly independent eigenvectors with corresponding eigenvalues $\lambda_1,\ldots,\lambda_n$, then letting $P$ be the matrix that has $\mathbf{v}_1,\ldots,\mathbf{v}_n$ as its columns, in that order, then the "original" matrix will be $$A = P\left(\begin{array}{cccc} \lambda_1 & 0 & \cdots & 0\\ 0 & \lambda_2 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0\cdots & \lambda_n \end{array}\right)P^{-1}.$$

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It is not enough, as Arturo points out. Lets stick with $2-$D for a geometric explanation. The geometric reason for this is as follows. The eigenvectors give the direction of maximum and minimum scaling. However, without the eigenvalues we do not know what the scaling is going to be. So we could have two different matrices whose directions of maximum and minimum scaling are the same but the scale factors along these directions are different. So one matrix could deform a circle into an ellipse of high eccentricity while another matrix could deform into an ellipse of low eccentricity though the axes for both the ellipse are the same.

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