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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...)

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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

1 Answer 1

up vote 2 down vote accepted

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.

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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
@mercurial Yes, that's right. –  David Mitra May 9 '12 at 16:11

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