1. linear Algebra (1.14)

Let V and W be vector spaces and T:V --> W be linear.

a) Prove that T is one-to-one if and only if T carries linearly independent subsets of V onto linearly independent subsets of W.

b)Suppose that T is one-to-one and that S is a subset of V. Prove that S is linearly independent if and only if T(S) is linearly independent.

c)suppose B={v1,v2,....,vn} is a basis for V and T is one-one and onto. Prove that T(B)= {T(v1),T(V2),....,T(Vn)}.

2. Originally Posted by JCIR
Let V and W be vector spaces and T:V --> W be linear.

a) Prove that T is one-to-one if and only if T carries linearly independent subsets of V onto linearly independent subsets of W.
Here is one direction. Let $\displaystyle S = \{\bold{v}_1,...,\bold{v}_k\}$ be linearly independent. Then $\displaystyle T(S) = \{ T(\bold{v}_1),...,T(\bold{v}_k)\}$. Write $\displaystyle \Sigma_j k_jT(\bold{v}_j) = \bold{0}$ and we will show $\displaystyle k_j$ are all zero to prove independence. This sum is the same as $\displaystyle \Sigma_j T(k_j\bold{v}_j) = \bold{0}$. But $\displaystyle T$ is one-to-one so the kernel is trivial, thus, $\displaystyle \Sigma_j k_j\bold{v}_j = \bold{0}$. However, $\displaystyle S$ is linearly independent so $\displaystyle k_j =0$.

You should post what you tried for the other problems.

3. Originally Posted by JCIR
Let V and W be vector spaces and T:V --> W be linear.

a) Prove that T is one-to-one if and only if T carries linearly independent subsets of V onto linearly independent subsets of W.

b)Suppose that T is one-to-one and that S is a subset of V. Prove that S is linearly independent if and only if T(S) is linearly independent.

c)suppose B={v1,v2,....,vn} is a basis for V and T is one-one and onto. Prove that T(B)= {T(v1),T(V2),....,T(Vn)}.
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.

4. 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.
Your first part is okay but for your second part you should be choosing y in S, not y in the nullspace of T. You want to show that S is linear independent, not the nullspace of T.

In fact, for the first part of (b), you could have just made use of the result of (a), no need to do any extra work.

ThePerfectHacker did the first part of (a). For the second part of part (a), suppose T preserves linear independence. Let $\displaystyle v_1,v_2\in V$ and suppose $\displaystyle v_1\ne v_2$. Then $\displaystyle v_1-v_2\ne 0_V$ and so $\displaystyle \{v_1-v_2\}$ is a linearly independent set in V. Thus $\displaystyle \{T(v_1-v_2)\}$ is linearly independent in W. Therefore $\displaystyle T(v_1-v_2)\ne0_W$, i.e. $\displaystyle T(v_1)\ne T(v_2)$; in other words, T is one-to-one.

For part (c), to show that T(B) is a basis for W, just show that T(B) spans W. By (b) you already know it's linearly independent. To show that it spans W, use the fact that T is onto.