# $\left \{ u,v \right \}$ is linearly independent iff $\left \{ u+v,u-v \right \}$ is linearly independent

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.

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*.

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)

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