Problem. Prove that
for $v_1, v_2, v_3 \in \mathbb{R}^3$, if $\{v_1+v_2, v_2+v_3, v_1+v_3\}$ are linearly independent then $\{v_1, v_2, v_3\}$ are linearly independent.
What I tried:
Let $m,n,p \in \mathbb{R}$ be such that
$$mv_1+nv_2+pv_3 = 0\;(\star)$$
From the hypothesis we know that if $a,b,c \in \mathbb{R}$ with $a(v_1+v_2)+b(v_2+v_3)+c(v_1+v_3) = 0$, then $a=b=c=0$.
First, every element $\begin{pmatrix}m\\n\\p \end{pmatrix} \in \mathbb{R}^3$ can be unique written in terms of $A = \biggl\{\begin{pmatrix}1\\0\\1 \end{pmatrix},\begin{pmatrix}1\\1\\0 \end{pmatrix},\begin{pmatrix}0\\1\\1 \end{pmatrix}\biggr\}$ because $A$ is a basis in $\mathbb{R}^3$, so we can let $\begin{cases} m=a+b \\ n=b+c \\ p=a+c \end{cases}$. So, from
$$
\begin{align}
(\star) \implies (a+b)v_1 + (b+c)v_2+(a+c)v_3=0 \\
\iff av_1+bv_1+bv_2+cv_2+av_3+cv_3=0 \\
\iff a(v_1+v_3)+b(v_1+v_2)+c(v_2+v_3)=0 \\
\implies a=b=c=0 \implies m=n=p=0
\end{align}$$
$\implies \{v_1, v_2, v_3\}$ are linearly independent
Please correct me if I am wrong or not. Thanks!
Best Answer
You are correct. But we can solve this in a general setting for any vector space. Let's rename the vectors in this way: $\alpha:=v_1+v_2, \beta:=v_2+v_3, \gamma:=v_3+v_1$, then we have $v_1=\frac{1}{2}(\alpha-\beta+\gamma), v_2=\frac{1}{2}(\beta-\gamma +\alpha), v_3=\frac{1}{2}(\gamma-\alpha+\beta)$. In other words, we have $$\frac{1}{2} \begin{pmatrix} 1 & -1 & 1\\ 1 & 1 & -1\\ -1 & 1 & 1 \end{pmatrix} \begin{pmatrix} \alpha\\ \beta\\ \gamma \end{pmatrix}= \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}.$$ Let's name the above 3 by 3 matrix $A$, then the matrix A is invertible. Now if we assume that $mv_1+nv_2+pv_3=0$, then we must have
$$ \begin{pmatrix} m& n & p \end{pmatrix}A \begin{pmatrix} \alpha\\ \beta\\ \gamma \end{pmatrix}=\begin{pmatrix} m& n & p \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}=0.$$ Therefore $\begin{pmatrix} m& n & p \end{pmatrix}A \begin{pmatrix} \alpha\\ \beta\\ \gamma \end{pmatrix}=0$. Since $\alpha, \beta, \gamma$ are linearly independent we must have $\begin{pmatrix} m& n & p \end{pmatrix}A=\begin{pmatrix} 0&0&0 \end{pmatrix}$ or $A^T\begin{pmatrix} m\\ n\\ p \end{pmatrix}=\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}$. But $A^T$ is invertible so $\begin{pmatrix} m\\ n\\ p \end{pmatrix}=0$ which means $V_1, V_2 ,V_3$ are linearly independent.