# 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

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!

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.

You can also use contradiction. Suppose $$\{v_1,v_2,v_3\}$$ is not linearly independent. WLOG we can write $$v_1=av_2+bv_3,$$

where $$a\ne 0$$ or $$b\ne 0$$.

It means that

$$\{v_1+v_2,v_1+v_3,v_2+v_3\}=\{(1+a)v_2+v_3,v_2+(1+b)v_3,v_2+v_3\}$$

See that this last set of vectors is not linearly independent.

Consider the vector space $$V$$ generated by $$\{v_1,v_2,v_3\}$$ and the linear transformation $$T: V\to V$$defined by $$v_1\mapsto v_1+v_2,\ v_2\mapsto v_2+v_3;\ v_3\mapsto v_1+v_3.$$ The matrix of $$T$$ in the basis $$\{v_1,v_2,v_3\}$$

$$\begin{pmatrix} 1 &0 &1 \\ 1& 1 &0 \\ 0 &1 &1 \end{pmatrix}$$

which is invertible. Hence $$T$$ maps the basis $$\{v_1,v_2,v_3\}$$ to a basis $$\{v_1+v_2,v_2+v_3,v_1+v_3\},$$ and so the vectors are indeed linearly independent.

Given that

$$c_1(v_1+v_2)+c_2(v_2+v_3)+c_3(v_1+v_3)=0$$ holds iff $$c_1=c_2=c_3=0$$

$$\implies (c_1+c_3)v_1+(c_1+c_2)v_2+(c_2+c_3)v_3=0$$ holds iff $$c_1=c_2=c_3=0$$

$$\implies \color{red}{\alpha v_1+\beta v_2+\gamma v_3=0\ holds\ iff\ \alpha=\beta=\gamma=0}$$

where $$\alpha=c_1+c_3,\beta=c_1+c_2, \gamma=c_2+c_3$$

Proving the contrapositive, if $$\{v_1, v_2, v_3\}$$ are linearly dependent, then their span has dimension less than 3, so any three vectors in their span are linearly dependent, hence so are $$\{v_1+v_2, v_2+v_3, v_1+v_3\}$$.

Of course, the acceptable proofs in a class depend on what you are allowed to use. Here I used that (1) the dimension of the span of a nonempty, finite set of vectors is the cardinality of the set if and only if they are linearly independent, and (2) the maximum cardinality of any linearly independent set of vectors from a finite dimensional vector space is the dimension of the space.

Another approach using determinants: Let's notice that we can show the tuple as a matrix $$\left\{ v_{1}+v_{2},v_{2}+v_{3},v_{1}+v_{3}\right\} =\left[\begin{array}{ccc} v_{1} & 0 & v_{1}\\ v_{2} & v_{2} & 0\\ 0 & v_{3} & v_{3} \end{array}\right]$$ Because the columns are independent we know that the determinant is non zero. $$\left|\begin{array}{ccc} v_{1} & 0 & v_{1}\\ v_{2} & v_{2} & 0\\ 0 & v_{3} & v_{3} \end{array}\right|=v_{1}v_{2}v_{3}+v_{1}v_{2}v_{3}=2v_{1}v_{2}v_{3}\ne0$$ Let us now work with $$A=\left[\begin{array}{ccc} v_{1} & 0 & 0\\ 0 & v_{2} & 0\\ 0 & 0 & v_{3} \end{array}\right]\Rightarrow\left|A\right|=\left|\begin{array}{ccc} v_{1} & 0 & 0\\ 0 & v_{2} & 0\\ 0 & 0 & v_{3} \end{array}\right|=v_{1}v_{2}v_{3}\ne0$$

• What is the multiplication $v_1v_2v_3$ here? Commented Jul 22, 2022 at 14:56