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Geometrically, what is the difference between a diagonizable matrix and an orthogonally diagonizable matrix?

I understand the difference algebraically, as explained here and many other places. But I'm struggling to see the difference geometrically: both will not rotate (assuming the field is the reals), but will stretch by different amounts along different axes (Wikipedia calls this a inhomogenous dilation). Graphically, I can't see the difference between, say, a circle or ellipse dilated inhomogenously along orthogonal axes versus non-orthogonal axes: in either case, it will remain an ellipse.

SRobertJames
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  • @AnneBauval As far as I can tell, even a inhomogeneous dilation along orthogonal axes won't preserve rectangles, if the axes are not the principal ones of the rectangle. – SRobertJames May 17 '23 at 03:49

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In fact: There is no geometric difference! Every plane transform of a diagonizable matrix is geometrically congurent to an orthogonally diagonizable matrix, via the polar decomposition. See Conjecture: Every shear transformation of the plane is congruent to a dilation .

(I'm using congruence in its geometric meaning: $F$ is congruent to $F'$ if there is an isometry sending $F \mapsto F'$.)

SRobertJames
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