Determine the kernel and range. HELP???

The transformation takes the vector #((x),(y))# to the vector #((x^'),(y^'))# where

#((x^'),(y^')) = ((6,4),(3,2))((x),(y)) implies #

#x^' = 6x+4y#
#y^' = 3x+2y#

The range
It is easy to see that #x^'=2y^'#, so that vectors in the range of the map must be of the form #((2a),(a)), a in RR#. Conversely, if a vector is of the form #((2a),(a))# with #a in RR#, then any of the infinite vectors #((x),(y)) # with #3x+2y=a# will map into it. Thus

#color(red)("range"(T) = {((2a),(a))| a in RR})#

It is easy to see that all the elements of #"range"(T)# are multiples of the single vector #((2),(1))#. The single element set #{((2),(1))}# is thus a spanning set for #"range"(T)#. Since it is obviously linearly independent, this set provides a basis for #"range"(T)#. Hence we have

#color(red)("dim"\ "range"(T) = 1)#

The Kernel

The kernel of the transformation is defined by

#"ker"(T) equiv {v in RR^2|Tv=0}#

So, if #v=((x),(y)) in "ker"(T)#, we have

# 6x+4y = 0#
# 3x+2y = 0#

Which will be satisfied by #x = 2t,\ y=-3t# for any #t in RR#. So

#color(red)("ker"(T) = {((2t),(-3t))| t in RR})#

It is obvious that #{((2),(-3))}# is a basis for #"ker"(T)#. Hence

#color(red)("dim"\ "ket"(T) = 1)#

Rank-nullity theorem

Since #"domain"(T)=RR^2#, we have #dim "domain"(T) = 2#

Thus the rank-nullity theorem

#dim"range"(T)+dim"ker"(T) = dim"domain"(T)#

is verified by #1+1 = 2#