Find the standard matrix of the linear transformation

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  • #1
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Homework Statement


20180424_120108_Film4.jpg


Homework Equations


None.

The Attempt at a Solution


I know that the standard matrix of a counterclockwise rotation by 45 degrees is:
[cos 45 -sin 45]
[sin 45 cos 45]
=[sqrt(2)/2 -sqrt(2)/2]
[sqrt(2)/2 sqrt(2)/2]
But the problem says "followed by a projection onto the line y=-2x", how do I find that?
 

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  • #2
Ray Vickson
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Homework Statement


View attachment 224780

Homework Equations


None.

The Attempt at a Solution


I know that the standard matrix of a counterclockwise rotation by 45 degrees is:
[cos 45 -sin 45]
[sin 45 cos 45]
=[sqrt(2)/2 -sqrt(2)/2]
[sqrt(2)/2 sqrt(2)/2]
But the problem says "followed by a projection onto the line y=-2x", how do I find that?

Where does the point ##(x,y) = (a,b)## end up after the transformation?
 
  • #3
mathwonk
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do you know about orthonormal bases?, i.e. a basis of length one vectors and mutually perpendicular? for such a basis, it is easy to expand a vector as a linear combination of the basis vectors, using dot products. And the point is that the component in the direction of say the first vector, will be the orthogonal projection onto the line spanned by that vector. to be precise, if u is any unit length vector, and v is any other vector, then |u.v| is the length of the projection of v onto the line spanned by u.
 
  • #4
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So how do I find the point end up after the transformation?
 
  • #5
Ray Vickson
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So how do I find the point end up after the transformation?

You work it out!

Where does ##(x,y) = (a,b)## go after the first step (the rotation)? So, if the rotation takes ##(a,b)## over to ##(a', b')##, where does ##(a', b')## go after the second step (the projection)?

You really are required to do the work; we are not allowed to do it for you.
 

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