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I know there are some questions about this exercise. Neverthless, I proved it by another way and I want you to check if my proof is correct, please.

Proof.

  • First part. $\left \{ u,v \right \}$ is linearly independent $\Rightarrow$ $\left \{ u+v,u-v \right \}$ is linearly independent

Let be $\lambda_1,\lambda_2 \in F$ and $u,v \in V$. We know that $\left \{ u,v \right \}$ is linearly independent, so

\begin{align*} \lambda_1 u+\lambda_2 v=0 \end{align*}

implies that $\lambda_1=\lambda_2=0=-\lambda_1=-\lambda_2$. So it satisfies this

\begin{align} \lambda_1u+\lambda_2v=\lambda_1 u+ \lambda_1 v = \lambda_1 (u+v) \ \ \ \ \ \ \ \ \ \ \ \ (1) \end{align}

By other side, we have that it satisfies

\begin{align} \lambda_1u+\lambda_2v=\lambda_2u+(-\lambda_2)v=\lambda_2(u-v) \ \ \ \ \ \ \ \ \ \ \ \ (2) \end{align}

With (1) and (2), we have

\begin{align} \lambda_1 (u+v)+\lambda_2(u-v)&=\lambda_1u+\lambda_1v+\lambda_2u-\lambda_2v\\ &=\underbrace{\lambda_1u-\lambda_2v}_{=0 \text{ since, $u,v$ are l.i.}}+\underbrace{\lambda_1v+\lambda_2u}_{=0 \text{ since $u,v$ are l.i.}}\\ &= 0+0\\ &= 0 \end{align}

Therefore,

\begin{align} \therefore \left \{ u+v,u-v \right \} \text{ is linearly independent} \end{align}

  • Second part. $\left \{ u+v,u-v \right \}$ is linearly independent $\Rightarrow$ $\left \{ u,v \right \}$ is linearly independent

Since $\left \{ u+v,u-v \right \}$ is l.i., we know that

\begin{align} \lambda_1(u+v)+\lambda_2(u-v)=0 \Longrightarrow \lambda_1=\lambda_2=0 \end{align}

This is the same: \begin{align} \lambda_1u+\lambda_1v+\lambda_2u-\lambda_2v=0 \Longrightarrow \lambda_1=\lambda_2=0\\ \end{align}

Also, this is the same: \begin{align} (\lambda_1+\lambda_2)u+(\lambda_1-\lambda_2)v=0 \Longrightarrow \lambda_1=\lambda_2=0\\ \end{align}

With $\lambda_1+\lambda_2=\gamma_1 \in F$

and $\lambda_1-\lambda_2=\gamma_2 \in F$, we have:

\begin{align} \gamma_1u+\gamma_2v=0 \Longrightarrow \lambda_1=\lambda_2=0 \Longrightarrow \gamma_1=\gamma_2=0\\ \end{align}

\begin{align} \therefore \left \{ u,v \right \} \text{ is linearly independent} \end{align}

By the first part and the second part, the proof is complete. Am I correct? I would be really very grateful if you help me to verify that it is correct.

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Neither proof looks correct to me - there's a bit of trouble here with variables sharing a name, allowing you to write a proof that looks ok, but is wrong. As far as I can tell, your first proof shows:

If $u$ and $v$ are linearly independent and $\lambda_1$ and $\lambda_2$ are such that $\lambda_1 u + \lambda_2 v = 0$ then $\lambda_1(u+v)+\lambda_2(u-v) = 0$.

But what you're supposed to show is:

If $u$ and $v$ are linearly independent and $\lambda_1$ and $\lambda_2$ are such that $\lambda_1(u+v) + \lambda_2(u-v) =0$, then $\lambda_1=0$ and $\lambda_2=0$.

These statements are really different - and I think this proof need to be entirely rewritten. Let me show the outline of how you do this; you should start your proof by just introducing the relevant variables:

Let $u$ and $v$ be linearly independent and $\lambda_1$ and $\lambda_2$ be such that $\lambda_1(u+v)+\lambda_2(u-v)=0$. We wish to show that $\lambda_1=0$ and $\lambda_2=0$.

This introduction mirrors the form of the statement you need to prove. The only thing you're allowed to do in proving this is to assume for any coefficients $\alpha_1$ and $\alpha_2$ that if $\alpha_1 u + \alpha_2 v= 0$ then $\alpha_1=\alpha_2=0$ - note that these $\alpha$'s do not have anything to do with the $\lambda$'s - they are bound variables rather than specific values.

You would probably make use of the fact that $$(\lambda_1+\lambda_2)u + (\lambda_1-\lambda_2)v = \lambda_1(u+v)=\lambda_2(u-v)=0$$ and linear independence of $u$ and $v$ to note that $\lambda_1+\lambda_2=0$ and $\lambda_1-\lambda_2=0$. Finishing from there should be easy*.

Your second proof establishes:

If $(u+v)$ and $(u-v)$ are linearly independent and $\lambda_1$ and $\lambda_2$ are such that $(\lambda_1+\lambda_2)u+(\lambda_1-\lambda_2)v=0$ then $\lambda_1+\lambda_2=0$ and $\lambda_1-\lambda_2=0$.

This is closer, but not quite right either. You need to instead show:

If $(u+v)$ and $(u-v)$ are linearly independent and $\lambda_1$ and $\lambda_2$ are such that $\lambda_1u+\lambda_2v=0$ then $\lambda_1=0$ and $\lambda_2=0$.

which is more general, since it loses the assumption that the pair $(\lambda_1, \lambda_2)$ can be written as $(x+y,x-y)$ - although, of course, it turns out that any pair can be written in that form*.

Mirroring the correct statement, your proof could begin:

Let $u$ and $v$ be vectors so that $(u+v)$ and $(u-v)$ are linearly independent. Suppose that $\lambda_1 u + \lambda_2 v = 0$. We wish to show that $\lambda_1=0$ and $\lambda_2=0$.

And, again, your strategy would need to be about rewriting $\lambda_1 u +\lambda_2 v$ as some linear combination of $(u+v)$ and $(u-v)$ and equating the coefficients of that sum to zero - and then doing more algebra to get to your conclusion.

Your second proof seems a bit closer to complete - basically, your $\gamma$'s take the place of what I'm using $\lambda$'s for - but your proof fails because you define the $\gamma$'s to be equal to something whereas the coefficients of $u$ and $v$ are the things that are given to you (i.e. you're proving a "for all" statement about them) - hence you may not exercise any influence over their values. The solution, in your notation, is just to write that $\lambda$'s in terms of the $\gamma$'s.

(*You may need to be careful about the base field, if you're doing this generally - things don't work out right in fields of characteristic $2$, where the statement is false. If this note is confusing, you don't need to worry about it)

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  • $\begingroup$ Thank you very much, you have really helped me to understand why my proof was incorrect! $\endgroup$
    – luisegf
    Jan 2, 2021 at 4:33

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