Originally Posted by

**squarerootof2** i will do part (b),

=>

first suppose that S is a linearly independent set.

then for s1,..,sn in S and scalars c1,...,cn, we know c1s1+...+cnsn=0 implies that c1,...,cn=0.

now consider T(c1s1+...cnsn)=0. this means that c1s1+...+cnsn is in the nullspace of T.

since T is one-to-one, we know that the nullspace is empty, so c1s1+...cnsn=0.

we know that S is linearly independent, so c1,...,cn=0

<=

conversely, suppose that T(S) is linearly independent and choose some y in the nullspace of T.

this means that there exists unique scalars c1,...,cn such that y=c1s1+...+cnsn

also, 0=T(y)=T(c1s1+...cnsn)=c1T(s1)+...cnT(sn) (last equality by linearity and first equalities by the definition of nullspace)

this means that c1T(s1)+...cnT(sn)=0, and since T(S) is a linearly independent set, we know that c1,...,cn=0, which is what we wanted to show.