Proof for $\langle a,b\rangle=\langle a-b,a+b\rangle $ check.

Question

I want to provide the following proof, but both directions feel a bit.... identical. I also wonder if my assumptions are correct, especially that we can pick any two real numbers and write them as $\lambda + \mu$ and $\lambda - \mu$.

We wish to prove for vectors $a,b$ that the span: $$\langle a,b\rangle =\langle a-b,a+b\rangle $$ We will prove this via mutual inclusion. Notice that any element $x\in\langle a,b\rangle $ can be written as $x= (\lambda+ \mu) a + (\lambda - \mu) b$ for some $\lambda, \mu \in \mathbb{R}$. We can rewrite this to $x=\lambda(a+b) + \mu(a-b)$. Thus $x \in \langle a-b,a+b\rangle $ and $\langle a,b\rangle \subseteq \langle a-b, a+b\rangle $.

For the other inclusion, consider $\tau, \sigma \in \mathbb{R}$ and $y\in \langle a-b,a+b\rangle $ then $y= \tau (a-b) + \sigma (a+b)=(\tau+ \sigma)a + (\tau-\sigma) b$. So $y \in \langle a,b\rangle $. We conclude $\langle a-b,a+b\rangle \subseteq \langle a,b\rangle $, consequently we haven proven $\langle a,b\rangle =\langle a-b,a+b\rangle $. As desired $ \square$

Answer 1

Your proof is correct, but you need to check some details/rewrite some things to make the proof's structure make more sense.

For one, $ x \in <a, b> $ means $ x = \lambda a + \mu b $ for $ \lambda, \mu \in \mathbb R $. This is not of the form you've written. You want to show that $ (\lambda, \mu) = (\alpha + \beta, \alpha - \beta) $ for some $ \alpha, \beta \in \mathbb R $. Then the proof of your first inclusion is correct.

For the second inclusion, the $ \lambda $ and $ \mu $ depend on $ y $, so you should introduce them after you write $ y $. Specifically, I would write it like this:

Let $ y \in <a - b, a + b> $. Then there are $ \lambda, \mu \in \mathbb R $ such that $ y = \lambda (a - b) + \mu (a + b) $.

I can also suggest another proof. $ <a, b> $ is the smallest subspace containing $ a $ and $ b $. Similarly, $ <a - b, a + b> $ is the smallest subspace containing $ a - b $ and $ a + b $. Thus, it suffices to show that $ a, b \in <a - b, a + b> $ and $ a - b, a + b \in <a, b> $, which are immediate.

Answer 2

You want to prove that for any $\lambda,\mu$,

$$\lambda a+\mu b=\sigma (a+b)+\tau(a-b)=(\sigma+\tau)a+(\sigma-\tau)b.$$ has a solution in $\sigma$ and $\tau$.

This is certainly true because the system

$$\begin{cases}\sigma+\tau=\lambda,\\\sigma-\tau=\mu\end{cases}$$ has a solution.