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FP1 matrix

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Can someone explain why (yx) -> (y+12x) is not a linear transformation and why (yx) -> (3y-x) is linear transformation?thank you

Original post by alesha98

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Can someone explain why (yx) -> (y+12x) is not a linear transformation and why (yx) -> (3y-x) is linear transformation?thank you



When is something a linear transformation?
Reply 2
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Zacken
22
Original post by alesha98

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Can someone explain why (yx) -> (y+12x) is not a linear transformation and why (yx) -> (3y-x) is linear transformation?thank you



Does the first one satisfy the property that T(ax)=aT(x)\mathbf{T}(ax) = a\mathbf{T}(x)?
Reply 3
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alesha98
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Original post by Zacken
Does the first one satisfy the property that T(ax)=aT(x)\mathbf{T}(ax) = a\mathbf{T}(x)?


Am i on the right line?
For x:
T(kx) = Tk(x)
1(1x) = 1(2)(x)
(x) = 2(x)
Reply 4
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Zacken
22
Original post by alesha98
Am i on the right line?
For x:
T(kx) = Tk(x)
1(1x) = 1(2)(x)
(x) = 2(x)


Yes, so the 'x' component works.

What about the "y+1" bit? Does T(ay) = aT(y)?
Reply 5
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alesha98
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Original post by Zacken
Yes, so the 'x' component works.

What about the "y+1" bit? Does T(ay) = aT(y)?

T(ky) = Tk(y)
(y) = (y+1)
In this case, i cant move out +1 from the bracket , so it doesnt satisfy?
Reply 6
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Zacken
22
Original post by alesha98
T(ky) = Tk(y)
(y) = (y+1)
In this case, i cant move out +1 from the bracket , so it doesnt satisfy?


pretty much
Reply 7
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alesha98
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Original post by Zacken
pretty much


so when i am doing this kind of questions, i need to solve x and y separately?
Can u show me a working example?
Reply 8
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Zacken
22
Original post by alesha98
so when i am doing this kind of questions, i need to solve x and y separately?
Can u show me a working example?


Uh, you should just be able to spot whether they're a linear transform or not? :confused: not sure what you want
Reply 9
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alesha98
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4
Original post by Zacken
Uh, you should just be able to spot whether they're a linear transform or not? :confused: not sure what you want


ah ok, so i dont need to prove it.
for linear transformation, it cant be
1. ie (x,y) -> (xy,y). So no xy element in the transformation
2. ie (x,y)->(x+1,y). So no constant in the transformation?

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