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We know that normal matrices are diagonalizable, but the converse is not true. For example, see here. Since a diagonalizable matrix represents a scaling operation under certain basis, so I wonder what additional geometrical meanings a normal matrix processes to be distinguished from other diagonalizable matrices. In other words, how to geometrically interpret matrices that can be diagonalized by unitary and non-unitary matrices? Thanks!

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    $\begingroup$ Diagonalizing Matrix for Normal matrices is a orthogonal matrix. Orthogonal matrices represent rotation. Thus Normal matrices first rotates your vector to their co-ordinate system (given by the columns), then scales the components of rotated vector, and then rotates it back. Note that this co-ordinate system is also orthogonal. But for other matrices, the co-ordinate system in which scaling happens is not a orthogonal co-ordinate system. $\endgroup$ – dineshdileep Jan 10 '13 at 7:16
  • $\begingroup$ @dineshdileep, I wonder if "orthogonal matrix/matrices" in your description should be relaxed to "unitary matrix/matrices". $\endgroup$ – Computist Jan 15 '15 at 1:36
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So, eigenvectors are the axes where there is a magnification/contraction.

For normal matrices, these axes are orthogonal to each other, thus, a matrix is normal if and only if it represents a linear transform that scales the coordinate axes, (in a suitable chosen orthogonal basis).

A diagonalizable matrix in general is a similar linear transform, (scaling of the coordinate axes), but the basis need not to be orthogonal.

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    $\begingroup$ So is a non-diagonalizable matrix that has, for example, ones in the first super-diagonal in its Jordan form, represented by a linear transform and that added affine component? $\endgroup$ – bright-star May 30 '13 at 11:24
  • $\begingroup$ That sounds correct. $\endgroup$ – Per Alexandersson Jun 5 '13 at 18:31

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