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...)