Are linear combinations of vectors that form a basis still a basis? V is a subspace of $R^n$ and $B={u,v,w}$ is a basis for V and
$x = 2u-3v+w$
$y = u+v$
$z = -v-w$
Explain why ${x,y,z}$ are a basis for V
Can I say that:
Any linear combinations of vectors that form a basis are still a basis?
 A: you can see that $$\pmatrix{x\\y\\z} = \pmatrix{2&-3&1\\1&1&0\\0&-1&-1}
\pmatrix{u\\v\\w}$$ and you need to show that the coefficient matrix in invertible. you can do that either by row reducing it see that you have three pivots or you could find the determinant not zero.
A: Here's the elementary approach:
You want to show that $x,y,z$ are linearly independent, so assume
$$ \lambda x + \mu y + \nu z = 0$$
for some scalars $\lambda, \mu, \nu$ and conclude that $\lambda=\mu=\nu=0$.
Substitute your knowledge about $x,y,z$ into the equation to get
\begin{align*}
0 &= \lambda x + \mu y + \nu z \\
  &= \lambda (2u - 3v + w)
 + \mu (u+v) + \nu (-v-w) \\
&= (2\lambda + \mu) u + (-3\lambda+\mu-\nu)v + (\lambda-\nu)w.
\end{align*}
Since $u,v,w$ are linearly independent we conclude that
\begin{align*}
2\lambda + \mu &= 0, \\
-3\lambda+\mu-\nu &= 0, \\
\lambda - \nu &= 0.
\end{align*}
Solving this linear system of equations yields the unique solution $\lambda=\mu=\nu=0$, so $x,y,z$ are linearly independent.
A: No. Consider the linear combination of the canonical basis (0,1)+(1,0) and (1,0)+(0,1). They are linearly dependent and hence do not form a basis for R^2
