Showing Dependence of Linear Combinations of Vectors This is an exercise problem in the book Linear Algebra by Larry Smith, $3^{rd}$ Edition. Given 4 vectors A, B, C and D in a vector Space V, we must show that the following 4 vectors in V are linear dependent:
$v_1 = A + B + C + D, v_2 = 2A + 2B + C - D, v_3 = A - B + C$ and $v_4 = A - C + D$
These vectors $\{v_1,v_2,v_3,v_4\}$ are linear combinations of the above 4. Also, there is no information about the dependence of the vectors A, B C and D.
I tried using the relation $a_1v_1+a_2v_2+a_3v_3+a_4v_4=0$ and solving for the co-efficients $a_1,a_2,a_3,a_4$ but that did not work.
Thank you for the help.
 A: The statement is false. Suppose that $V=\Bbb R^4$, that $A=(1,0,0,0)$, that $B=(0,1,0,0)$, that $C=(0,0,1,0)$, and that $D=(0,0,0,1)$. Then

*

*$A+B+C+D=(1,1,1,1)$;

*$2A+2B+C-D=(2,2,1,-1)$;

*$A-B+C=(1,-1,1,0)$;

*$A-C+D=(1,0,-1,0)$.

And these four vectors are linearly independent, since$$\begin{vmatrix}1 & 1 & 1 & 1 \\ 2 & 2 & 1 & -1 \\ 1 & -1 & 1 & 0 \\ 1 & 0 & -1 & 1\end{vmatrix}=-13\ne0.$$
A: Let $w_1=A,...,w_4 = D$ for notational convenience.
We can write $v_k = \sum_i [M]_{ik} w_i$, where $M = \begin{bmatrix} 1 & 2 & 1 & 1 \\ 1 & 2 & -1 & 0 \\ 1 & 1 & 1 & -1 \\ 1 & -1 & 0 & 1\end{bmatrix}$.
Since $\det M  = -13 $ we see that $M$ is invertible.
It is straightforward to check that $w_i = \sum_k [M^{-1}]_{ki} v_k$.
In particular, we see that if $\sum_k \alpha_k v_k = 0$ then $\sum_i \beta_i w_i = 0$ where $\beta = M \alpha$.
Similarly
if $\sum_i \beta_i w_i = 0$ then $\sum_k \alpha_k v_k = 0$  where $\beta = M \alpha$.
Hence if the $v_k$ are linearly dependent then so are the $w_i$ and vice bersa.
