Stretching space or blender scale inherritance Let's say we have an object a plane. Then I stretch the plane along its Y axis until the object doubles in size along this axis.
I can say I have applied a multiplication vector of (1,2) to my object.
Now we consider our object has it's own reference frame. If it is aligned with the plane frame, then the scaling vector is still (1,2). But if I turn the object 90° (clockwise or counter-clockwise, doesn't matter), plane Y axis is my object X axis, and in the frame of the object the scaling is (2,1)
If my object was rotated 45°, what scaling does it have now ? Given an angle wrt the plane frame, what scaling is applied ?
A bit of context : what I am trying to describe here is the scale of a child object in blender when it's parent change in size. I tried to illustrate the problem here
 A: You have two coordinate frames, and for simplicity we can assume that the origins of both coincide.  The child (the object) frame is related to the parent as follows:
If $r_0$ is the coordinate of a point with respect to the parent frame, and $r_1$ is the coordinate of the same point with respect to the child frame, then
$r_0 = R r_1 $
For some rotation matrix $R$.
The above equation implies $r_1 = R^T r_0 $
Now in your questions, you want to scale $r_0$ by a certain matrix $ A $ so that its image is
$r_0' = A r_0  $
The coordinate of $r_0'$ in the child reference frame is
$r_1' = R^T r_0' = R^T A r_0 = R^T A R r_1 $
Thus the scaling that has occurred in the child frame is given by $B=R^T A R$
To verify this, using the example mentioned in the question, we have
$A = \begin{bmatrix} 1&& 0 \\ 0 && 2 \end{bmatrix} $
If the child frame is not rotated with respect to the parent frame then $R = I$ the identity matrix, and it follows that $B = A $
If the child frame is rotated by $\pm 90^\circ$ degrees then,
$R = \begin{bmatrix} 0 && \mp 1 \\ \pm 1 && 0 \end{bmatrix} $
Thus
$ B = \begin{bmatrix} 0 && \pm 1 \\ \mp 1 && 0 \end{bmatrix} \begin{bmatrix} 1 && 0 \\ 0 && 2 \end{bmatrix} \begin{bmatrix} 0 && \mp 1 \\ \pm 1 && 0 \end{bmatrix} $
Multiplying,
$ B = \begin{bmatrix} 0 && \pm 1 \\ \mp 1 && 0 \end{bmatrix}  \begin{bmatrix} 0 && \mp 1 \\ \pm 2 && 0 \end{bmatrix} $
And finally,
$ B = \begin{bmatrix} 2 && 0 \\ 0 && 1 \end{bmatrix} $
As mentioned in the question.
