if u v and w are vectors that are linearly independent will u-v v-w and u-w be linearly independent?
So you are given $\displaystyle a \vec{u} + b \vec{v} + c \vec{w} = 0 \Rightarrow a = b = c = 0 $.
Then you want to show that $\displaystyle x(\vec{u} - \vec{v}) + y(\vec{v}- \vec{w}) + z(\vec{u} - \vec{w}) = 0 \Rightarrow x = y = z = 0$ is true or false.
Expand out: $\displaystyle x \vec{u} - x \vec{v} + y \vec{v} - y \vec{w} + z \vec{u} - z \vec{w} = 0 $.
Then $\displaystyle (x+z) \vec{u} + (y-x) \vec{v} + (-y-z) \vec{w} = 0 $
But that doesn't imply that $\displaystyle x = y = z = 0 $. Only that $\displaystyle a = b = c = 0 $, where $\displaystyle a = x+z, \ b = y-x, \ c = -y-z $. So they are linearly dependent.
$\displaystyle A = \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 2 & 1 \\ 0 & 3 & 2 & 1 \\ 0 & 1 & -1 & 1 \end{bmatrix} $
$\displaystyle \det A = -1 \begin{bmatrix} 0 & 0 & 2 \\ 0 & 3 & 2 \\ 0 & 1 & -1 \end{bmatrix} = -1(-2)(0) =0 $
Linearly dependent.
i believe they are dependent if the Wronskian is zero.
recall that a set of two or more vectors is linearly dependent if and only if at least one of the vectors in the set can be expressed as the linear combination of finitely many other vectors in the set.
now, note that:
$\displaystyle u - v = u - v + w - w = (u - w) - (v - w)$
$\displaystyle v - w = v - w + u - u = (u - w) - (u - v)$
and
$\displaystyle u - w = u - w + v - v = (u - v) + (v - w)$
so we see that each of the vectors can be expressed as a linear combination of the other two, thus the vectors u - v, v - w, and u - w form a linearly dependent set of vectors