Is the set $S$ of vectors linearly independent?

Question

$S = \{ u, v, w \}$ is a subset of $\mathbb{R}^3$, where it is known that there is an invertible $3 \times 3$ matrix $D$ such that $uD = (1,2,3)$, $vD = (4,5,6)$, and $w D = (5,7,9)$.

Is the set $S$ of vectors linearly independent?

Below is what I did:

$wD = (5,7,9) = (1,2,3) + (4,5,6) = uD + vD$
Since $wB$ is a redundant vector, $w$ is a redundant vector as well. Therefore, set of vectors that contain $w$ is linearly dependent.

Is there anything wrong with my explanation ?

Answer 1

You’ve shown that the set $\{uD,vD,wD\}$ is linearly dependent, but have failed to justify why this implies that $\{u,v,w\}$ is also linearly dependent. As a counterexample, suppose that $u$, $v$ and $w$ are the standard basis vectors of $\mathbb R^3$ and $$D=\begin{bmatrix}1&2&3\\4&5&6\\5&7&9\end{bmatrix}.$$ Then $uD=(1,2,3)$, $vD=(4,5,6)$ and $wD=(5,7,9)$, but $u$, $v$ and $w$ are linearly independent.

The key is that $D$ is invertible. You can therefore say that $$0=0D^{-1}=(uD+vD-wD)D^{-1}=uDD^{-1}+vDD^{-1}-wDD^{-1}=u+v-w.$$

Answer 2

A nonsingular $n×n$ matrix is a change of basis matrix (from the basis consisting of the columns to the standard basis). Hence it always takes a basis to a basis. Since this did not happen when applied to $\{u,v,w\}$ (as you have shown), it isn't a basis...