$u,v,w$ are distinct vectors show the following is also a basis

Question

Let $u,v,w$ be distinct vectors of vector space over $\mathbb C$, such that $\{u,v,w\}$ is a basis of $V$.

Show that $\{u−(1+i)v, u+v+w, −2iu\}$ is also a basis.

Answer 1

$\beta = \{u,v,w\}$ Is a basis of $V$

Is $\beta'=\{u−(1+i)v, u+v+w, −2iu\}$ also a basis of $V$?

First see to the definition of a basis, it is a set of vectors belonging to $V$, this set of vectors generate $V$ when lineally combined and are linearly independent.

Define $\beta' $ as $\beta' = \{u',v',w'\}$

The items on both basis are taken from the same vectorial space, so we can prove that $\{u',v',w'\}$ are linearly independent (or not) and therefore $\beta'$ will (or wont) be a basis.

lets get the equations for $\{u',v',w'\}$

$$u' = u−(1+i)v\\ v' = u+v+w\\ w'=−2iu $$

Now the matrice corresponding to this system:

$$\left(\begin{matrix} 1&-(1+i)&0\\ 1&1&1\\ -2i&0&0\\ \end{matrix}\right)$$

From this we make zeros to get this other matrice:

$$\left(\begin{matrix} 1&-(1+i)&0\\ 0&2+i&1\\ 0&0&(-2+6i)/5\\ \end{matrix}\right)$$

Which has range 3, then $\{u',v',w'\}$ are linearly independent and are the same number of vectors as in $\beta$ , then $\beta'$ generates $V$, and is a free system, then it is a basis.