# Proving dependent columns when the rows are dependent

Tags:
1. Sep 3, 2015

### kostoglotov

I feel like I almost understand the solution I've come up with, but a step in the logic is missing. I'll post the question and my solution in LaTeX form.

Paraphrasing of text question below in LaTeX. Text question can be seen in its entirety via this imgur link: http://i.imgur.com/41fvDRN.jpg

$$\ if \begin{pmatrix} a\\ b \end{pmatrix} \ is \ a \ multiple \ of \begin{pmatrix} c\\ d \end{pmatrix} \ with \ abcd \neq 0, show \ that \begin{pmatrix} a\\ c \end{pmatrix} \ is \ a \ multiple \ of \begin{pmatrix} b\\ d \end{pmatrix}$$

My solution so far

$$assume \ \lambda \begin{pmatrix} c\\ d \end{pmatrix} = \begin{pmatrix} a\\ b \end{pmatrix} \rightarrow \begin{matrix} a = \lambda c\\ b = \lambda d \end{matrix}$$

$$now \ assume \ \gamma \begin{pmatrix} b\\ d \end{pmatrix} = \begin{pmatrix} a\\ c \end{pmatrix} \rightarrow \begin{matrix} a = \gamma b\\ c = \gamma d \end{matrix}$$

So I'm making two assumptions

Let's take the assumptions and put them into a system of four equations

$$1: \ a = \lambda c \ \ \ \ 2: \ b = \lambda d \\ 3: \ a = \gamma b \ \ \ \ 4: \ c = \gamma d$$

Now if we sub 3 into 1 to get A, and sub 2 into A to get B and then sub 4 into B to get C

$$C \rightarrow \lambda \gamma d = \lambda \gamma d$$

Similarly if we, sub 2 into 3 to get A, and 1 into A to get B, and 4 into B to get C

$$C \rightarrow \lambda \gamma d = \lambda \gamma d$$

I want to stop trying all the possible ways to get C now, because I want to look for a generalized way to show that they will all end up at the same point.

But more than this...what is the step of logic that connects the final equation C to proving the first two assumptions. I feel like this should prove the assumptions, but I don't know how exactly, or how exactly to express it.

Thanks :)

2. Sep 3, 2015

### mathman

$\gamma = \frac {c}{d}$
Then $\frac {a}{b} =\frac {\lambda c}{\lambda d} = \frac{c}{d}$.

3. Sep 3, 2015

### Staff: Mentor

You have to prove those assumptions.
And it is hard to prove things that are wrong...
Actually, there are just two special cases where this assumption holds.

If you think you need "assumptions", look for counterexamples first. They are easy to find here, and they save a lot of work.

4. Sep 3, 2015

### jbunniii

With this second assumption, you are assuming the fact that you are supposed to prove! That won't get you anywhere. Instead, work with the first assumption, which is given: $(a,b)$ is a multiple of $(c,d)$. So that means there is some constant $\lambda$ such that $a = \lambda c$ and $b = \lambda d$.

By the way, note that the condition $abcd \neq 0$ means that all four of $a,b,c,d$ are nonzero, and therefore $\lambda$ is also nonzero.

Now we can rewrite the matrix as
$$\begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} = \begin{pmatrix} \lambda c & \lambda d \\ c & d \\ \end{pmatrix}$$
From this, you can easily see that first column is a constant multiple of the second column. (What is the constant?)

5. Sep 3, 2015

### kostoglotov

Nice one! Very clear explanation, thanks :)