Vector Spaces :State True or False? 
*

*A and B are subsets of vector spaces, then A≠B → L(A) ≠L(B).

*A set containing a linearly independent set of vectors is itself linearly independent.


I think these two statements are obvious as it seem but the answer to them is false. Please Explain.
 A: In order to prove that a statement is false, providing a counter example is enough.
For 1. I suggest you look, for $V$ some vector space of non-null dimension of your choice, $A=V$ and $B=V\setminus \{0\}$ (if your sets are required to be finite, which is not specified here, look two a basis and one additional non-null vector for each)
For 2. I suggest you look at some vector space $V$ (if your set is requires to be finite, which is not specified here, look at a basis and an additional non-null vector)
A: 
(1) Consider vector space $\mathbb R^3$.

Now,consider $P$={$ \left[
    \begin{array}{cc|c}
      1\\
      0\\
0
    \end{array}
\right] $,$ \left[
    \begin{array}{cc|c}
      0\\
      1\\
0
    \end{array}
\right] $,$ \left[
    \begin{array}{cc|c}
      0\\
      0\\
1
    \end{array}
\right] $} & $Q$={$ \left[
    \begin{array}{cc|c}
      1\\
      0\\
0
    \end{array}
\right] $,$ \left[
    \begin{array}{cc|c}
      0\\
      1\\
0
    \end{array}
\right] $,$ \left[
    \begin{array}{cc|c}
      0\\
      0\\
2
    \end{array}
\right] $}
Then  $P \neq Q $,but the space spanned by vectors of both these sets is $\mathbb R^3$.Hence,statement (1) is NOT TRUE in  general.

(2) Consider vector space $\mathbb R^2$.

Now,consider the set $T$={$A= \left[
    \begin{array}{cc|c}
      1\\
      0
    \end{array}
\right] $,B=$ \left[
    \begin{array}{cc|c}
      0\\
      1
    \end{array}
\right] $,C=$ \left[
    \begin{array}{cc|c}
      2\\
      3
    \end{array}
\right] $},
$ A$ & $B$ are linearly independent,but $C=2A+3B$ which means the set $T$ is linearly dependent.Hence,statement (2) is NOT TRUE in general.
A: I.
Is $L$ a linear transformation?  If so then think about what must be true for it to do so.  Does a linear transformation have to be invective? (hint: recall that all linear transformations have matrix representations)
Does $\bar{u} =\begin{pmatrix} 1 \\ 0 \\ \end{pmatrix} = \begin{pmatrix} 3 \\ 0 \\ \end{pmatrix} = \bar{v}?$
Yet, consider the linear transformation represented by the matrix: $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix} $
Note that $A\bar{u} = A\bar{v} = \bar{0}$
II.
As to the second the question is being tricky, reread it carefully and think about how set containment works.  If $A \subset B$ does it follow that $A = B$ ?
