# Linear Transformations

• Nov 11th 2008, 04:13 PM
Hellreaver
Linear Transformations
I have to either show that a transformation is either linear or not due to the additivity or the homogeneity.

The first question is:
T(x,y)=(x+1,y)

T(x)+T(y)= mx+my
m(x+1)+my= mx+m +my
mx+my =/= mx+m+my
therefore
T(x,y)=(x+1,y) is not linear under additivity.

How would I prove or disprove that the transformation is linear under homogeneity?
• Nov 11th 2008, 04:46 PM
whipflip15
Your proof does not make sense because T(x) is not defined.

However if $\vec a=(a_1,a_2) \mbox{ and }\vec b=(b_1,b_2)$
So
$T(\vec a + \vec b)=T(a_1+b_1,a_2+b_2)=(a_1+b_1+1,a_2+b_2)$
Whilst
$T(\vec a) + T(\vec b)=(a_1+b_1+2,a_2+b_2)$

For the second part.
$T(c\vec a)=T(ca_1,ca_2)=(ca_1+1,ca_2)$
but
$cT(\vec a)=(ca_1+c,ca_2)$
• Nov 11th 2008, 05:25 PM
Hellreaver
Quote:

Originally Posted by whipflip15
Your proof does not make sense because T(x) is not defined.

However if $\vec a=(a_1,a_2) \mbox{ and }\vec b=(b_1,b_2)$
So
$T(\vec a + \vec b)=T(a_1+b_1,a_2+b_2)=(a_1+b_1+1,a_2+b_2)$
Whilst
$T(\vec a) + T(\vec b)=(a_1+b_1+2,a_2+b_2)$

For the second part.
$T(c\vec a)=T(ca_1,ca_2)=(ca_1+1,ca_2)$
but
$cT(\vec a)=(ca_1+c,ca_2)$

So I'm guessing I sorta combined the two into an epic fail? I think I understand... Probably not.

A second question is:
T(x,y)=(y,y)

For this, I would have:
let $\vec a=(a_1,a_2) \mbox{ and }\vec b=(b_1,b_2)$
then
$T(\vec a + \vec b)=T(a_1+b_1,a_2+b_2)$
and
T(\vec a) + T(\vec b)=?
I'm not sure what needs to be done here...

For the second part, I'm not sure how to do what you did above using two different vectors, because I would be comparing x to y at some point, correct?
Sorry for my idiocy, by the way.