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Suppose we have a rotation matrix $P$ and a diagonal matrix $D$. How do you explain the effect of $D$ as a linear transformation on the plane $\mathbb{R}^2$ geometrically? What about the effect of $P^{-1}DP$?

My approach: Well whenever we apply a diagonal matrix \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix} on a vector, we just rescale the vector thus we just stretch the $\mathbb{R}^2$ plane.

Then for $P^{-1}DP$ we first rotate the vector, stretch and then bring it all back. So in this case what is the answer?

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    $\begingroup$ It is a stretch along axes, but not the vertical/horizontal axes. $\endgroup$ – vadim123 Mar 24 '18 at 15:09
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The effect of $D$ is a rescaling with dilation/contraction of the $2$ axes.

The effect of $P^{-1}DP$ is a rescaling with dilation/contraction in the direction $-\theta$ and $-\theta + 90°$.

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  • $\begingroup$ Why in that direction? $\endgroup$ – mandella Mar 24 '18 at 15:55
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    $\begingroup$ @mandella Since with $P$ we rotate by $\theta$ than we apply $D $the we rotate of $-\theta$ by $P^{-1}$ $\endgroup$ – user Mar 24 '18 at 15:56
  • $\begingroup$ Makes sense. What about an explanation with eigenvectors/orthobasis ? Do you have any idea about that $\endgroup$ – mandella Mar 24 '18 at 16:03
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    $\begingroup$ @mandella finally did you get it? $\endgroup$ – user Mar 27 '18 at 20:44
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    $\begingroup$ @mandella Well done! I'm very happy for that! Bye $\endgroup$ – user Mar 27 '18 at 20:46

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