MHB Please Critique My Solution Involving Linear Independence, Linear Dependence and Span

[bwpbruce]

Problem:

True or False? If $x$ and $y$ are linearly independent, and if $\{\textbf{x}, \textbf{y}, \textbf{z}\}$ is linearly dependent, then $\textbf{z}$ is in Span $\{\textbf{x},\textbf{y}\}$

Solution:
$\textbf{True}$. If $a\textbf{x} + b\textbf{y} = \textbf{0}$ is true and if $a\textbf{x} + b\textbf{y} + c\textbf{z} = \textbf{0}$ is true, then $-a\textbf{x} -b\textbf{y} = \textbf{z}$ must also be true because when $c = 1$: \begin{align*}a\textbf{x} + b\textbf{y} + (1)\textbf{z} = \textbf{0}\end{align*}, we have the non-trivial solution. Furthermore,
\begin{align*}a\textbf{x} + b\textbf{y} + \textbf{z} = \textbf{0}\end{align*} becomes
\begin{align*}a\textbf{x} + b\textbf{y} = -\textbf{z}\end{align*} which simplifies to
\begin{align*}-a\textbf{x} + -b\textbf{y} = \textbf{z}\end{align*}

 

[Evgeny.Makarov]

I'll use regular (not bold) letters from the end of the alphabet for vectors and letters from the beginning of the alphabet for numbers.

You may have shown that if $ax+by=0$ and $ax+by+cz=0$, then $z\in\operatorname{Span}\{x,y\}$. But the assumption in the problem was simply $ax+by+cz=0$ for some $a,b,c$; you were not authorized to assume that $ax+by=0$, especially with the same coefficients $a,b$. But you don't seem to actually use the assumption $ax+by=0$ later in the proof.

Further, what right do you have to assume that $c=1$? More precisely, of course, you can assume anything, but for the proof to be complete you also need to consider the case $c\ne 1$. Instead of assuming $c=1$ try dividing the equation by $c$. But first you need to prove that $c\ne0$.

A minor note is that to make a proof easy to read, it should start with assumptions, use them to deduce intermediate statements, use those to deduce other statements and progress in that way until you arrive at the conclusion. You, in contrast, started with assumptions and then announced a goal that you seem to have proved only at the end. In other words, if you are using words like "therefore", "so" and "thus", you are moving forward; but if you are using "because", then you reversed your direction and are now justifying facts announced ahead of time. It's a matter of style and readability, so this is not necessarily bad, but if you are learning to write proofs, the forward-moving style is preferable.

 

[bwpbruce]

Can you please provide me with an example of how to write a proof in the manner that you suggest. I'm awful with it and my proofs often ends up getting ripped to shreds in the same manner you just did.

 

[Evgeny.Makarov]

Assumptions:

(A1) $x,y$ are linearly independent, i.e., $ax+by=0$ implies $a=b=0$ for all $a,b$.

(A2) $x,y,z$ are linearly dependent, i.e., there exists $a,b,c$ such that $ax+by+cz=0$ and it is not the case that $a=b=c=0$.

Take some $a,b,c$ guaranteed to exist by (A2). Assume that $c=0$; then (A2) implies that $ax+by=0$, but it is not the case that $a=b=0$ (otherwise we would have $a=b=c=0$). But this contradicts (A1); therefore, the assumption $c=0$ was false and $c\ne0$. Dividing both sides of $ax+by+cz=0$ by $c$ we get $\frac{a}{c}x+\frac{b}{c}y+z=0$, or
\[
z=-\frac{a}{c}x-\frac{b}{c}y.
\]
Therefore, $z\in\operatorname{Span}\{x,y\}$.