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?)

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For (b), for example, let $A$ be the zero matrix. Take any two vectors $v_1$ and $v_2$ in $\mathbb{R}^2$. It is clear that $Av_1$ and $Av_2$ are linearly dependent. There are less extreme cases, such as $A$ the matrix that projects onto the $x$-axis. This should also help with (c). – André Nicolas May 8 '12 at 21:11
You might note that c) and b) are contrapositives of each other. The same for d) and a). – David Mitra May 8 '12 at 21:23
What does *$A$ mean? – Dylan Moreland May 8 '12 at 22:39
@Dylan: OP had *A* instead of putting A in math mode, then attempted to use a mixture of text, italic text and math mode, which completely messed up the formatting. – Arturo Magidin May 9 '12 at 0:04
@GerryMyerson yep just did so for the previous questions! Thanks for pointing it out anyway – mercurial May 9 '12 at 5:51

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.

$\ \ \ \ \ \ \$"Then since $a$ and $b$ are not both zero, $A\b v_1$ and $A\b v_2$ are dependent, as desired."
Ah i get it now.. So the main rectification was to ensure that in the case that $\beta$ = 0 and $\alpha$ $\neq$ 0 or vice versa, (or simply put, a and b cannot be simultaneously 0) the proof still holds, right? Thanks! – mercurial May 9 '12 at 15:54