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If I multiply a point by a translation matrix the point is translated. If I multiply a point by a rotation matrix the point is rotated. These two transformations are intuitive. But, when I multiply a point by a scaling matrix it is not scaled (it cannot be), rather it is like "transformed"! This is counter-intuitive! Can anyone please explain this to me? And how is scaling done on a complex 3D model with texture applied on it in computer graphics?

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  • $\begingroup$ Well aren't scaling matrix defined with respect to a certain "fixed" point? (usually (0,0,0) ) $\endgroup$ – AnalysisStudent0414 Aug 31 '16 at 13:31
  • $\begingroup$ Yes, but since I'm new to linear algebra, can you explain it more? That doesn't answer my question. $\endgroup$ – Shuvo Sarker Aug 31 '16 at 13:33
  • $\begingroup$ I posted it as a comment to "hint" you at the correct explanation, which AOrtiz excellently gave $\endgroup$ – AnalysisStudent0414 Aug 31 '16 at 15:12
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A scaling transformation $T$, as you call it has the property that for $v\in\Bbb R^3$, $Tv = \alpha v$, for some $\alpha \in\Bbb R.$ To see how this looks in matrix form, we use the fact that a linear transformation is determined on a basis, say $e_1, e_2, e_3$, the standard basis. $Te_1 = (\alpha, 0, 0)$, $Te_2 = (0,\alpha, 0)$, $Te_3 = (0,0,\alpha)$. Hence, the matrix $A$ of $T$ is: $$ A=\begin{pmatrix} \alpha & 0 & 0 \\ 0 & \alpha & 0 \\ 0 & 0 & \alpha \end{pmatrix}. $$ So, if you take any point $v\in\Bbb R^3$ and compute $Av$, you will find that the result is equivalent to applying the transformation $\alpha\,\cdot\,\colon\Bbb R^3\to\Bbb R^3$ that "scales" a vector $v$ via standard scalar multiplication. In other words, $Av = \alpha \cdot v$.

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  • $\begingroup$ That explains a lot. But, suppose that I have a line consisting of 1000 points and I want to scale it to make it appear larger. If I do scaling for 1000 points one by one I don't get more than 1000 points. Then how is the scaling happening?! $\endgroup$ – Shuvo Sarker Aug 31 '16 at 14:03
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    $\begingroup$ @ShuvoSarker, if the scaling is still of this type, then the scaling is happening in the exact same way. That is, for each of the points $v_1, v_2,\ldots, v_{1000}$, we compute $Av_1,Av_2,\ldots,Av_{1000}.$ This is the same "line," with each point on the line scaled by the appropriate transformation. $\endgroup$ – Alex Ortiz Aug 31 '16 at 14:12
  • $\begingroup$ @AOritz... But how!? The number of points is the same in both cases. The second case makes the line appear larger with the same number of points just in different locations! I cannot picture it in my mind! $\endgroup$ – Shuvo Sarker Aug 31 '16 at 14:22
  • $\begingroup$ The interval $[0,1]$ contains as many points as the whole $\mathbb{R}$. Weird things can happen when infinity is involved. Another example: natural numbers are just as many as natural even numbers. Just consider the bijection $n \to 2n$ $\endgroup$ – AnalysisStudent0414 Aug 31 '16 at 15:13
  • $\begingroup$ @ShuvoSarker, If you are talking about line segments, then you are right, the length of the new line is $\alpha$ times longer, and they are not the same segment. However, if you are referring to a line defined by $1000$ points, then the transformed line is exactly the same as the untransformed line. $\endgroup$ – Alex Ortiz Aug 31 '16 at 15:31

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