Linear Mapping/Matrices Proof At first look a rather logical question which has till date stumped many of us attempting to solve it. Hmm, hope you guys could offer some brain power here :)
$A$ is a matrix from $\mathbb{R}^{2,2}$, ${v_{1}}$ and ${v_{2}}$ are vectors from $\mathbb{R}^{2,2}$. Proof or disprove the following statements:
a) ${v_{1}}$ and ${v_{2}}$ are linear depedent, then $A{v_{1}}$ and $A{v_{2}}$ are linear dependent.
b) $A{v_{1}}$ and $A{v_{2}}$ linear dependent $\Rightarrow$ ${v_{1}}$, ${v_{2}}$ linear dependent
c) ${v_{1}}$ and ${v_{1}}$ linear independent $\Rightarrow$ $A{v_{1}}$ and $A{v_{2}}$ linear independent
d) $A{v_{1}}$ and $A{v_{2}}$ linear independent $\Rightarrow$ ${v_{1}}$ and ${v_{2}}$ linear independent.
My attempt
only tried it out for a), cause using my method, it appears all statements are proved. A little too fishy to be true.
$
\alpha{v_{1}} + \beta{v_{2}} = 0$, where
$$\begin{array}{rcl}
{v_{1}} &=& \left[ \begin{matrix} x_{1}\\ y_{1} \end{matrix} \right]\\
{v_{2}} &=& \left[ \begin{matrix} x_{2}\\ y_{2} \end{matrix} \right]\end{array}
\Rightarrow x_{1} =\left ( \frac{-\beta}{\alpha} \right )x_{2}\tag{1}$$ 
$$y_{1}=\left ( \frac{-\beta}{\alpha}\right )y_{2}\tag{2}$$ 
Assuming statement a) is true,
$\alpha A{v_{1}} + \beta A{v_{2}} = 0$, 
$A = \left[ \begin{matrix} a_{1}&a_{2}\\ b_{1}&b_{2} \end{matrix} \right]\Rightarrow\left[ \begin{matrix} a_{1}x_{1} + a_{2}y_{1}\\ b_{1}x_{1} + b_{2}y_{1}\end{matrix} \right] = \left(\frac{-\beta}{\alpha}\right)\left[ \begin{matrix} a_{1}x_{2} + a_{2}y_{2}\\ b_{1}x_{2} + b_{2}y_{2} \end{matrix} \right]$
Observing only the first row, we have
$$a_{1}x_{1} + a_{2}y_{1} = \left(\frac{-\beta}{\alpha}\right)a_{1}x_{2} + a_{2}y_{2}\tag{3}$$ 
Substituting (1) and (2) into (3), we have
$a_{1}\left ( \frac{-\beta}{\alpha} \right )x_{2} + a_{2}\left ( \frac{-\beta}{\alpha}\right )y_{2} = \left(\frac{-\beta}{\alpha}\right)a_{1}x_{2} + a_{2}y_{2}$ -- LHS = RHS, therefore statement proved (or is it?)
(Thanks for your patience...)
 A: Your statement tagged (2) is not true if $\alpha=0$.  
Immediately afterwards, you assume that a) is true. This, of course, is cheating. You need to prove that a) is true; and to do that you assume its hypotheses holds.
$\def\b#1{{\bf #1}}$
More generally, it's hard to see what you're trying to do in your argument. I think it's ok (aside from the division by $\alpha$), but you leave many things unstated... 
If I surmise correctly  what you've shown, with some corrections, is that if $\alpha \b v_1+\beta \b v_2=\b0$, then
$\alpha A\b v_1=-\beta A\b v_2$.
From this it will follow that a) is true; but you need to explain why...  

In my opinion, at this level you should be writing out exactly what's going on in each step.
A straightforward argument proving a) might run as follows: start by saying
$\ \ \ \ \ $"Assume that $\b v_1=(x_1,y_1)$ and $\b v_2=(x_2,y_2)$ are dependent".  
OK, now is the time to write out that the equation
$\alpha \b v_1+\beta \b v_2=\b0$ has a nontrivial solution  and (with a bit of hindsight) what this implies. You'd say 
$\ \ \ \ \ $"Then we may, and do, select $a$ and $b$ not both zero, so that $a\b v_1+b\b v_2=\b0$. Note then that both  $\ \ \ \ \ \ \ ax_1+bx_2$ and $ay_1+by_2$ are  zero"  
Now you have to show that the equation $\alpha A\b v_1+\beta A\b v_2=\b0$ has a non-trivial solution. Towards this end, present the computation that shows $\alpha=a$ and $\beta=b$ works.
$\ \ \ \ \ $"Let $A=\Bigl[\matrix{a_1&a_2\cr b_1&b_2 }\Bigr]$.  We will show that  $a A\b v_1+b A\b v_2=\b0$. Indeed:
$$
a A\b v_1\!+\!b A\b v_2 \! =\!a\Bigl[\matrix{a_1x_1+a_2y_1\cr b_1x_1+b_2y_1 }  \Bigr]
\!+\!b\Bigl[\matrix{a_1x_2+a_2y_2\cr b_1x_2+b_2y_2 }  \Bigr]
\!=\!\Bigl[\matrix{ a_1(ax_1+bx_2)\! + \! a_2(ay_1+by_2)  \cr  b_1(ax_1\!+\!bx_2) +  b_2(ay_1\!+\!by_2)}  \Bigr]=\Bigl[\matrix{0\cr0}\Bigr],
$$
$\ \ \ \ \ \ $since $ax_1+bx_2$ and $ay_1+by_2$ are both zero.
Finally state that you've accomplished your task: 
$\ \ \ \ \ \ \ $"Then since $a$ and $b$ are not both zero, $A\b v_1$ and $A\b v_2$ are dependent, as desired."

As Andre points out in the comments, b) is false. 
You are actually (almost) done at this point, if you observe  that c) and b) are contrapositives of each other and that  d) and a) are contrapositives of each other.
