Prove that for vectors $v_1,...,v_n$ in $\mathbb C^n$, $\{v_1,...,v_n\}$ is a basis for $\mathbb C^n$ iff its conjugate is a basis for $\mathbb C^n$ Prove that for vectors $v_1,...,v_n$ in $\mathbb C^n$, $\{v_1,...,v_n\}$ is a basis for $\mathbb C^n$ if and only if $\{\bar v_1,..., \bar v_n\}$ is a basis for $\mathbb C^n$.
I know intuitively that the conjugate of $a+bi$, $a-bi$ is orthogonal to $a+bi$. Since they are orthogonal, then the span of $v_1,...,v_n$ and $\bar v_1,...,\bar v_n$ is the same, and $v_i \in \mathbb C^n$ and $\bar v_i\in\mathbb C^n$. 
Please drop hints on how I can prove this formally.
Edit: not all conjugates are orthogonal, thanks Abramos.
 A: Hints: for $\;a_i\in\Bbb C\;$ :
$$\sum_{i=1}^n a_iv_i=0\iff 0=\overline 0=\overline{\left(\sum_{i=1}^n a_iv_i\right)}=\sum_{i=1}^n\overline{(a_iv_i)}=\sum_{i=1}^n\overline{a_i}\overline{v_i} $$
So we get that $\;\forall\,i\;,\;a_i=0\;\iff\;\forall\,i\;,\;\overline{a_i}=0\;$
A: This should be very straightforward - just write out the usual arguments for proving that a set of vectors is (a) linearly independent and (b) spans the whole space, and use the fact that the conjugate satisfies the properties:
$$\overline{z} = 0 \iff z = 0\text{ and }\overline{z_1 + \dots z_n} = \overline{z_1} + \dots \overline{z_n} \text{ and } \overline{az} = \overline{a}\,\overline{z}$$
So, for example, if there were a linear combination equal to zero:
$$a_1v_1 + \dots a_nv_n = 0$$
then we would have 
$$\overline{a_1v_1 + \dots a_nv_n} = 0$$
and then 
$$\overline{a_1v_1} + \dots \overline{a_nv_n} = 0$$
etc.
A: What I say below may be more than is wanted, and it may only be USD $ \$ 0.02 $ worth, but it's my $ \$ 0.02 $, so what the hell, here goes:
If the $v_i \in \Bbb C^n$ span $\Bbb C^n$, then for any vector $w \in C^n$ we also have $\bar w \in C^n$, so there exist coeffiecients $b_i \in \Bbb C$ such that
$\bar w = \sum_i b_i v_i, \tag{1}$
whence
$w = \sum_i \bar b_i \bar v_i, \tag{2}$
which shows that the $\bar v_i$ also span $\Bbb C^n$.  The argument for linear independence of the $\bar v_i$ over $\Bbb C$ is of course the same as given by Old John and DonAntonio in their answers:  if
$\sum_i b_i \bar v_i = 0, \tag{3}$
then by the august factoid that $\bar 0 = 0$ we have
$\sum_i \bar b_i v_i = 0, \tag{4}$
which shows that the $\bar b_i = 0$ by linear independence of the $v_i$.  Thus the $b_i = 0$ implying the $\bar v_i$ are linearly independent,and hence form a basis for $\Bbb C^n$.
Of course the whole argument can be run in reverse to show that the $\bar v_i$ form a basis implies the $v_i$ do.  QED
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
A: A different approach would be to note that $\mathbb{C}^n \cong \mathbb{R}^{2n}$ and then write down the image of a basis under the canonical isomorphism. After that you can work with real basis in order to prove the statement.
