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Math Help - verifying a linear transformation

  1. #1
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    verifying a linear transformation

    OK. I know that to verify a linear transformation you must show that L(u + v) = L(u) + L(v) and L(ku) = kL(u). for every u and v in R.

    I guess I don't quite understand this. Could someone show me how to verify L(x, y, z) = (x+y, 0, 2x-z)...

    I want to say u = (x1, y1, z1) and v=(x2, y2, z2). Right. How in the world do it make it look resemble what's given if I L( u + v) = L( x1 + x2, y1 + y2, z1 + z2) = (x1, y1, z1) + (x2, y2, z2) = L(u) + L(v).

    I was given some help, and supposedly the answer resembles:

    L( x1 + x2, y1 + y2, z1 + z2) = (x1 + x2+y1 + y2,0,2*(x1 + x2) - (z1 + z2)) = L(u) + L(v)

    You can just arbitrarily move things around to make it look like what you want??

    How would anything NOT be a transformation? Are there rules to follow?

    If this is how I verify the example I gave above...then when something NOT be a linear transformation. PLEASE HELP. I'm so confused...
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  2. #2
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    You're on the right track. You need to calculate L(u+v) and L(u)+L(v) separately. You calculated L(u+v) perfectly, now do the linear transformation of L(x_1 , y_1 , z_1)+L(x_2 , y_2 , z_2) and show that this is the same as the first transformation. You calculated the first part and then just set it equal to the second. You must justify this. That's the whole point.

    As for something not being linear, think about a trig function: T:V\rightarrow W such that T(x)=\sin(x)
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  3. #3
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    Quote Originally Posted by kpizle View Post
    You can just arbitrarily move things around to make it look like what you want??

    How would anything NOT be a transformation? Are there rules to follow?
    This transformation is taking vectors u and v, both in R to another vector space, which I'm assuming is also R. I know it can seem arbitrary at times but believe me there's nothing arbitrary about it. There are strict rules governing vector spaces which you should be pretty familiar with. Surely adding vectors together and multiplying by scalars are allowed. All that's really going on in this problem is using generic vectors and properties of a vector space to demonstrate a property about this transformation we are analyzing.
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  4. #4
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    Quote Originally Posted by Jameson View Post

    As for something not being linear, think about a trig function: T:V\rightarrow W such that T(x)=\sin(x)
    if I understand you correctly, L(blah) = blah MUST be a linear transformation as long as each component of blah is linear?

    If that is so, then I have another example which i know is not linear transformation...

    L(x, y) = (x-y, 0, 2x+3).

    why is not a linear transformation? is it because of 2x+3? that's what I'm inclined to think, but only because the left side doesn't define z. I am missing a huge objective here I guess... it's ridiculous i can't get this right now

    I didn't see your second post until now... THAT HELPS A LOT.

    Yes, I'm very familar with those rules... maybe that's the huge objective (needing to use those rules) I'm missing here. lol.
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  5. #5
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    Don't put yourself down. These concepts take time to digest. I'm taking Linear Algebra now as well and when I first went over this topic I was just as confused. Practice, study, and wait. It'll make sense in time.

    So to be a linear transformation, like you said, L(u+v)=L(u)+L(v) and L(cu)=cL(u). Sometimes this is written together as L(cu+v)=cL(u)+L(v).

    As for your example of a non-linear transformation, yes it's because of the 2x+3. When both vectors are combined and the transformation is applied to the sum, the last component has a "...+3" while when transformed separately then added the last component has a "...+6".
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  6. #6
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    Radical.

    Thank you very much Jameson.

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