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I can represent a tensor by a matrix. Suppose we are talking about a 2nd order tensor, and the matrix is therefore 3x3. If I find one eigenvector of that matrix; that vector represents normal vector on a plane on which there is no shear stress (the stress state was described by the 2nd order tensor). My question is, how is this happening (looking for the proof) that eigenvector points to a plane without shear stress?

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If you have 3 eigenvectors of the stress tensor/matrix $T$, you can choose them as your new basis and it will be diagonal there - and no off-diagonal elements mean no shear stress, since the shear stress on the plane in the $ij$-direction ($i \neq j$) is given by $T_{ij}$, which, for $i \neq j$, will be zero in this basis.

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Eigen vector:It is a vector that undergo only scaling and no rotation https://www.youtube.com/watch?v=-tOcSuRkIzs (for graphical intuition)

Since principle stress acts on normal direction(extends normally)without shear due to rotation eigen vectors are principle stress.

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