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?
you can see that $$\pmatrix{x\\y\\z} = \pmatrix{2&amp;-3&amp;1\\1&amp;1&amp;0\\0&amp;-1&amp;-1}
\pmatrix{u\\v\\w}$$ 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.
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