The Student Room Group

Linear Transformations

"Show the following transformation is a linear transformation:
T :Map(R,R)→Map(R,R),
T(f(x))=f(x+1)"

I have tried but seem to be struggling:
T(f(X1)+f(X2))=T(f(X1+X2))=f(X1+X2+1)?
T(f(aX1))=f(aX1+1)?

Is this correct? I haven't done seen a proof of a linear transformation before so I'm unsure when doing the addition whether it is
T(f(X1+X2)) or T(f(X1)+f(X2)) or whether these are equal or when doing scalar whether it is
T(aF(X)) or T(F(aX1)) or whether these are also equal?
Reply 1
Original post by mollyjordansmith
"Show the following transformation is a linear transformation:
T :Map(R,R)→Map(R,R),
T(f(x))=f(x+1)"

I have tried but seem to be struggling:
T(f(X1)+f(X2))=T(f(X1+X2))=f(X1+X2+1)?
T(f(aX1))=f(aX1+1)?

Is this correct? I haven't done seen a proof of a linear transformation before so I'm unsure when doing the addition whether it is
T(f(X1+X2)) or T(f(X1)+f(X2)) or whether these are equal or when doing scalar whether it is
T(aF(X)) or T(F(aX1)) or whether these are also equal?


When showing that something is a linear transformation, you want T(λf(x)+μg(x))=λT(f(x))+μT(g(x))T(\lambda f(x) + \mu g(x)) = \lambda T(f(x)) + \mu T(g(x)).

In particular here, you want to show that T(λf(x)+μg(x))=λf(x+1)+μg(x+1)T(\lambda f(x) + \mu g(x)) = \lambda f(x+1) + \mu g(x+1)
Reply 2
I think it may be easiest if you let h(x)=λf(x)+μg(x) h(x)=\lambda f(x) + \mu g(x) .
Then of course T(h(x))=h(x+1)=... T(h(x))=h(x+1)=... .
Original post by Zacken
When showing that something is a linear transformation, you want T(λf(x)+μg(x))=λT(f(x))+μT(g(x))T(\lambda f(x) + \mu g(x)) = \lambda T(f(x)) + \mu T(g(x)).

In particular here, you want to show that T(λf(x)+μg(x))=λf(x+1)+μg(x+1)T(\lambda f(x) + \mu g(x)) = \lambda f(x+1) + \mu g(x+1)


So does this show it:
IMG_0384.jpg
Original post by B_9710
I think it may be easiest if you let h(x)=λf(x)+μg(x) h(x)=\lambda f(x) + \mu g(x) .
Then of course T(h(x))=h(x+1)=... T(h(x))=h(x+1)=... .


So does this show it:
IMG_0384.jpg
Reply 5
Original post by mollyjordansmith
So does this show it:


Yes.

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