Linear Algebra - Linear transformations

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  • #1
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Homework Statement



which of the following are linear transformations.

a) L(x,y,z) = (0,0)
b) L(x,y,z) = (1 ,2, -1)
c) L(x,y,z) = (x^2 + y, y - z)

The Attempt at a Solution



I know that L is a linear transformation if L(u + v) = L(u) + L(v) and L(ku) = kL(u).

I am not sure how to apply that to those equations.

Please help.
 

Answers and Replies

  • #2
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Homework Statement



which of the following are linear transformations.

a) L(x,y,z) = (0,0)
b) L(x,y,z) = (1 ,2, -1)
c) L(x,y,z) = (x^2 + y, y - z)

The Attempt at a Solution



I know that L is a linear transformation if L(u + v) = L(u) + L(v) and L(ku) = kL(u).

I am not sure how to apply that to those equations.

Please help.
Start with a couple of vectors, say u = <x1, y1, z1> and v = <x2, y2, z2>.

For a), see if L(u + v) = L(u) + L(v) and L(ku) = kL(u). Do the same for b) and c).
 
  • #3
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Start with a couple of vectors, say u = <x1, y1, z1> and v = <x2, y2, z2>.

For a), see if L(u + v) = L(u) + L(v) and L(ku) = kL(u). Do the same for b) and c).

I guess my problem is, I don't know how to apply that to L(x,y,z) = (0,0). If it was L(v) = Av I would understand. I just don't understand the transformations themselves and how to apply them.
 
  • #4
Office_Shredder
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Well if it was L(v)=Av it wouldn't be interesting to ask if the transformation was linear!

For example for the first one if (x,y,z)=(1,2,4)

L(1,2,4)=(0,0).

And if (x,y,z)=(1,1,1)

L(1,1,1)=(0,0)

So L(1,2,4)+L(1,1,1)=(0,0)+(0,0)=(0,0).

Is this equal to L(2,3,5)? ((1,2,4)+(1,1,1))
 
  • #5
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Well if it was L(v)=Av it wouldn't be interesting to ask if the transformation was linear!

For example for the first one if (x,y,z)=(1,2,4)

L(1,2,4)=(0,0).

And if (x,y,z)=(1,1,1)

L(1,1,1)=(0,0)

So L(1,2,4)+L(1,1,1)=(0,0)+(0,0)=(0,0).

Is this equal to L(2,3,5)? ((1,2,4)+(1,1,1))

Ok, I think I understand. so that is linear because L(2,3,5) = (0,0). And L(ku) = kL(u) = (0,0).

But for b), the second part doesn't hold up because L(ku) = (1,2,-1) but kL(u) = k(1,2,-1) and they are not equal for all k.

and c probably shows something similar. I will have to write it out in a bit in Maple.

Thanks!
 

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