Note: lowercase v is a vector Solution for kernel, l(0) T for v=0,hence kernel = {0} range = {all v V} 1)L-1(0) is in T 2)L-1(au+bv) = aL-1(u) + bL-1(v) L(v) as au+bv V 3)as u L-1(T),L(u) = u\' L(cu) =cL(u) = cu\' as linear transformation =>cu\' L-1(T) => c(L-1(u)) L-1(T) =>cu T hence L-1(T) is a subspace of V.

1. Note: lowercase v is a vector
Solution
for kernel, l(0) T for v=0,hence kernel = {0}
range = {all v V}
1)L-1(0) is in T
2)L-1(au+bv) = aL-1(u) + bL-1(v) L(v)
as au+bv V
3)as u L-1(T),L(u) = u'
L(cu) =cL(u) = cu' as linear transformation
=>cu' L-1(T) => c(L-1(u)) L-1(T) =>cu T
hence L-1(T) is a subspace of V