Undergrad Why Is Reflection in a Hyperplane a Linear Function?

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Reflection in a hyperplane is considered a linear function because it preserves the linearity of transformations, as illustrated by the mirror analogy. This means that if an object is stretched or rotated, its reflection behaves in the same linear manner. However, translations do not maintain this property, as they alter the position of the object without preserving the origin. It is crucial to specify that the reflection occurs in a hyperplane that includes the zero vector to ensure it qualifies as a linear transformation rather than an affine one. Understanding these distinctions clarifies why reflections are categorized as linear functions in mathematical terms.
Aleoa
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Is it possible to understand intuitively (without using a formal proof ) why a reflection is a linear function ?
 
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Well, you look in the mirror. If you behave 'linearly' (whatever this means), then your reflection does, too.

I'm not responsible for any misunderstandings stemming from intuitive logic :olduhh:
 
Aleoa said:
Is it possible to understand intuitively (without using a formal proof ) why a reflection is a linear function ?
Yes, @nuuskur's mirror is a good analogy: whether you stretch something or add lines with whatever angle in between, the mirror image does the same. The same can be said about rotations and stretches by their own. E.g. translations are not linear: if you stretch a line from one given and fixed point in a certain direction, a translated line segment will become something else than the not translated line segment. The point here is the fixed point.
 
technically be a little careful. a linear transformation is one that sends parallelograms to parallelograms, but it also sends the zero vector to itself. so you probably should say "reflection in a hyperplane that passes through the zero vector", otherwise it is not linear but only affine linear.
 
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Thread 'How to define a vector field?'

Thread 'How to define a vector field?'

Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...
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