I have to show that transformation T defined by T(x1,x2)=(2x1-3x2,x1+4,5x2) is not linear?. I know by definition that T(u+v)= T(u)+T(v) and that T(cv)=cT(v). But how do I use this to prove the question?

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- Sep 15th 2012, 07:11 PMbonfire09linear transformations
I have to show that transformation T defined by T(x1,x2)=(2x1-3x2,x1+4,5x2) is not linear?. I know by definition that T(

**u**+**v**)= T(**u**)+T(**v**) and that T(c**v**)=cT(**v**). But how do I use this to prove the question? - Sep 15th 2012, 07:51 PMTheEmptySetRe: linear transformations
You just need to show that either of the two equations does not hold for ALL vectors.

Hint: Try the values

$\displaystyle \mathbf{u}=\mathbf{v}=\vec{0}$

and show that

$\displaystyle T(\mathbf{u}+\mathbf{v}) \ne T(\mathbf{u})+T(\mathbf{v})$

Remeber to disprove a universal statement (for all) you need to only produce one counter example. - Sep 15th 2012, 08:13 PMDevenoRe: linear transformations
one simple property of ANY linear transformation T is that: T(0) = 0 (where these are 0-vectors, not 0-scalars).

here is the proof:

T(0) = T(0+0) (since 0 = 0+0)

= T(0) + T(0) (since T is linear).

subtracting T(0) from both sides, we get:

0 = T(0) - T(0) = [T(0) + T(0)] - T(0) = T(0) + [T(0) - T(0)] = T(0) + 0 = T(0).

now let's look at YOUR function:

T(x,y) = (2x-3y,x+4,5y).

the 0-vector of R^{2}is (0,0), and T(0,0) = (2*0-3*0,0+4,5*0) = (0,4,0) ≠ (0,0,0), so.... - Sep 15th 2012, 08:13 PMbonfire09Re: linear transformations
I tried it with arbitrary values actually. I let

**u**=(x1,x2) and**v**=(y1,y2). So T(**u**+**v**)=T(x1+y1,x2+y2). So it then becomes T=(2x1+2y1-3x2-3y2,x1+y1+4,5x2+5y2 which is the same as T(**u**)+T(**v**)=(2x1-3x2,x1,5x2)+(2y1-3y2,y1+4,5y2) but i dont know where to take it from here? As you can see T(**u**) /= T(**v**), - Sep 16th 2012, 12:52 AMemakarovRe: linear transformations
- Sep 16th 2012, 08:44 PMbonfire09Re: linear transformations
My bad I think I figured it out. Since T(

**u+v)**=T(**u**)+T(**v**), im supposed to evaluate each vector separately on the RHS of the equation. So vector u=[x1,x2] and v=[y1,y2]. T(**u**)=[2x1-3x2,x1+4,5x2] and T(**v**)=[2y1-3y2,y1+4,5y2] so T(**u**)+T(**v**)=[2x1-3x2+2y1-3y2,x1+y1+8.,5x2+5y2]. And that does not equal the LHS of the equation making this not a linear transformation. I know when writing vectors im not supposed to use comma's to separate the entries but I

don't know latex yet. (Headbang)