$u_1, u_2$ is linearly independent if and only if the family $u_1 + u_2, u_1 − u_2$ is linearly independent.

Question

I have this exercise where I want to check my solutions. Can someone help me?

Let $u_1$ and $u_2$ be elements of $V$. If $1 + 1 ≠ 0$ in $K$, then the family $u_1, u_2$ is linearly independent if and only if the family $u_1 + u_2, u_1 − u_2$ is linearly independent.

So what I have done is

For $\Rightarrow$

$\alpha u_1+ \beta u_2=0 \Rightarrow \alpha = \beta=0$ so if we take for $\alpha=(a+b)$ and $\beta=(a-b)$ it becomes $(a+b) u_1+ (a-b) u_2=0 \Rightarrow (a+b) = (a-b)=0$

so $a+b=0$ and $a-b=0$ are linearly independent which means that for $a+b=0$ and $a-b=0$ we have that $a=b=0$.

So with a little bit of transformations we have $a(u_1+u_2)+b(u_1-u_2)=0 \Rightarrow a=b=0$

For $\Leftarrow$

I have $a(u_1+u_2)+b(u_1-u_2)=0 \Rightarrow u_1(a+b)+u_2(a-b)=0$ and as before with $\alpha=(a+b)$ and $\beta=(a-b)$ we have $u_1 \alpha +u_2\beta=0 \Rightarrow \alpha=\beta=0$

Everything right?

Answer 1

Your correct computations may be summarised by considering the transformation matrix, within the generated subspace, from $\,\{u_1, u_2\}\,$ to $\,\{u_1- u_2, u_1+ u_2\}\,$: $$\begin{pmatrix}1&1\\-1&1 \end{pmatrix}^{-1} \;=\; \frac12 \begin{pmatrix}1&-1\\1&1 \end{pmatrix}$$ If $1 + 1 ≠ 0$ in $K$, then determinants are non-zero, and the preceding matrix inversion is possible and correct.

Answer 2

You have the right idea but I would frame it differently to make clear you are not choosing the scalars $a,b$ in any way when showing $x,y$ are linearly independent (LI), i.e. $$ax+by=0\implies a=b=0.$$

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Thus, to show that $u_1+u_2,u_1-u_2$ LI implies $u_1,u_2$ LI, I would start with

$$au_1+bu_2=0.\quad (1)$$

We want to show $a=b=0$. This follows by writing LHS of $(1)$ as

$$\frac{1}{2}(a+b)(u_1+u_2)+\frac{1}{2}(a-b)(u_1-u_2)$$

and using the fact $u_1+u_2,u_1-u_2$ are LI. Proceed similarly for the other direction.