you do realize that T(a1,a2) = (a1+1,a2) is not a linear transformation, don't you?
clearly T,U must be singular. an obvious singular transformation is T(x,y) = (x,x).
can you find a U now so that UT = 0, but TU ≠ 0?
Attached is the problem statement and my attempt. I'm pretty sure my answers for the compositions are correct, but then again, I'm not sure why I can't find the correct matrices A and B from them (it looks like AB=BA=0 matrix, which contradicts the condition given in the problem statement).
Can someone clarify this for me? ( Deveno? )
Thanks!
you do realize that T(a1,a2) = (a1+1,a2) is not a linear transformation, don't you?
clearly T,U must be singular. an obvious singular transformation is T(x,y) = (x,x).
can you find a U now so that UT = 0, but TU ≠ 0?
I do nowyou do realize that T(a1,a2) = (a1+1,a2) is not a linear transformation, don't you?
I know singular means it doesn't have an inverse, but is there an easy way to tell right away if a transformation is singular or not?clearly T,U must be singular. an obvious singular transformation is T(x,y) = (x,x).
Well, it seems like U has to be something that outputs only zeros...so I gave U(x,y)=(0,0) a shot, but that messes up TU, because then TU=(0,0), which is what we don't want.can you find a U now so that UT = 0, but TU ≠ 0?
Sooo...I switched things around and got T=(a2,0) and U=(0,a2)
This seemed to work out nicely, although I'm curious if there was a way to do it with the transform you mentioned (ie, T(x,y)=(x,x)), because I couldn't get it to work with that one. What do you think?
ps. thanks for the help!
ways to tell if a transformation is singular:
it's matrix (w.r.t. some bases) is not invertible (not so easy to tell).
it's matrix (w.r.t. some bases) has a row of 0's at the bottom when row-reduced (this is a good general method).
it's matrix (w.r.t. some bases) has 0 determinant (good for 2x2 and 3x3 matrices....bigger ones are harder).
it's matrix (w.r.t. some bases) has more rows than columns.
if it is possible to find U ≠ 0 with TU = 0.
as for my example....there may have been other that would have worked, but i was thinking of U(x,y) = (x-y,0).
then UT(x,y) = U(x,x) = (x-x,0) = 0, but:
TU(x,y) = T(x-y,0) = (x-y,x-y) which is non-zero if x ≠ y.