##### Practice Midterm #4
nabihahshuaib
How do you prove that the transformations are linear?
abbyto1607: May 4, 2015, 5:38 p.m.
There are two conditions for a transformation to be linear. First, $T(v+w) = T(v)+ T(w)$ Second, $kT(v)=T(kv)$ Prove that the transformation holds true for both of these conditions to prove that it is linear. A simple example would be $T((a,b))=(2a+b)$ To determine if this is linear or not, test if the two conditions hold true. $T((a_{1},b_{1}))= (2a_{1} +b_{1})$ $T((a_{2},b_{2}))= (2a_{2} +b_{2})$ $T((a_{1},b_{1}))+T((a_{2},b_{2}))$ $=(2a_{1} +b_{1})+(2a_{2} +b_{2})$ $=(2a_{1}+2a_{2}) +(b_{1}+b_{2}))$ $=(2(a_{1}+a_{2}) +(b_{1}+b_{2}))$ Now, see that, $T((a_{1},b_{1}) + (a_{2},b_{2}))= T((a_{1}+a_{2},b_{1}+b_{2}))$ $=(2(a_{1}+a_{2}) +(b_{1}+b_{2}))$ Thus, $T((a_{1},b_{1}) + (a_{2},b_{2}))= T((a_{1},b_{1}))+T((a_{2},b_{2}))$ A similar process can be applied for the condition of multiplication by a scalar, as well as for other transformations.
nabihahshuaib: May 4, 2015, 7:16 p.m.
Thank you abbto1607!
GarimaLunawat: May 4, 2015, 8:41 p.m.
Does the zero vector also need to be included for linear transformations or does that only apply to proving sets?
ericpan64: May 4, 2015, 9:49 p.m.
^I think the zero vector should be included because it can be considered a "scalar multiple" if you pick the constant 0 Also quick question: Would it be right to say that 4d is a linear function because it a linear combination of $a+bx+cx^2$? Also does anyone know how to find the matrix of the transformation for 4d? I'm not sure how to find a matrix without assuming a basis of $(1, x, x^2)$ Thanks!
jared_brock: May 4, 2015, 10 p.m.
@ericpan64, if you prove both conditions stated by abbyto1607, the system proves to be linear, I don't think anything has to be done to a+bx+c\$x^{2}.
jared_brock: May 4, 2015, 10:01 p.m.
didn't come out right... but hopefully my meaning is clear!
danielkim: May 4, 2015, 10:42 p.m.
To see if something is a linear equation or not, you can also just see if anything is a polynomial or has a constant attached to it. Ie T(a,b) = (a^2 , a+b+2) is not linear.
aleksanderjanczewski: May 5, 2015, 12:43 a.m.
@GarimaLunawat: I guess your think about the importance of zero vector when you consider subspaces of vector spaces, then each subspace must cross the origin - in other words must include zero vector.
aleksanderjanczewski: May 5, 2015, 12:45 a.m.
Did you guys do part b) of problem 4, I am not sure but I get that it is nor linear transformation, can someone confirm that? Thanks
DonnaBranchevsky: May 5, 2015, 7:46 p.m.
I also got that part b is not a linear transformation because it fails T(v+w)=T(v)+T(w)