Linear dependency of vectors $a,b,c,d$ Let $V$ be a vector space and $a,b,c,d\in V$, where the set $\{a,d\}$ is linearly independent and sets $\{a, b+c+d\}$ and $\{d,a+b+c\}$ are linearly dependent.
We need to show that the set $\{b,a+c+d\}$ is linearly dependent.
So far I've tried to express the vectors $a$ and $d$ using $b+c+d$ and $a+b+c$, respectively, and I tried to express $b$ in the form of $b=\alpha (a+c+d)$, for some $\alpha \in\mathbb{F}$, but that didn't work.
One other thing I think follows from here, is that the set $\{b+c+d, a+b+c\}$ is linearly independent, but that also doesn't seem to help me with my goal.
 A: By assumption $b+c+d=ra$ and $a+b+c=sd$ for $r,s\in K$, the field. Taking the difference gives $(s+1)d-(r+1)a=0$, so that $r=s=-1$, because $a$ and $d$ are linearly independent.
So $a+b+c+d=0$ and therefore $b$ and $a+c+d$ are linearly dependent.
A: Since $\{a,d\}$ is linearly independent, neither $a$ nor $d$ can be zero.  Therefore
$$b+c+d=\lambda a\ ,\quad a+b+c=\mu d$$
for some scalars $\lambda,\mu$, and so
$$b+c=\lambda a-d=\mu d -a\ .$$
Since $a,d$ are independent, any vector in their span can be written uniquely as a linear combination of $a,d$.  Hence
$$\lambda=\mu=-1$$
Therefore
$$a+c+d=-b$$
and so $\{b, a+c+d\}$ is dependent.
A: First note that neither $a$ nor $b$ can be $0$. From the dependence of $\{a,b+c+d\}$ we get $a=\alpha (b+c+d)$ for some $\alpha$.  Similarly, $d=\beta (a+b+c)$ for some $\beta$.  This gives $\beta a=\alpha \beta (b+c+d)$ and $\alpha d =\alpha  \beta(a+b+c)$ . Subtraction gives $\beta a -\alpha d= \alpha \beta (d-a)$. Use indepednece of $\{a,d\}$ to see that $-\beta =\alpha \beta$ and $\alpha= -\alpha \beta$.  This gives $\alpha =-1$ and $\beta =-1$. But then $a=\alpha (b+c+d)=-(b+c+d)$ or $a+b+c+d=0$. Clearly this implies linear dependence of $b$ and $a+c+d$.
