How can I prove $dim(span(v_{1},v_{2}...v_{n})) = dim(span((\overline{v_{1}},\overline{v_{2}}...\overline{v_{n}}))$? Let $v_{1},v_{2}...v_{n}$ be a set of vectors in a vector space over $\mathbb{C}$.
How can I prove $dim(span(v_{1},v_{2}...v_{n})) = dim(span((\overline{v_{1}},\overline{v_{2}}...\overline{v_{n}}))$ ?
I'm using this result as a lemma to prove $rank(A) = rank(A^*)$, so any help is appreciated. 
 A: Take a basis for one of the spans. The complex conjugates of the basis vectors make up a basis for the other span.
Namely if $\bar v$ is any vector in the conjugated span, its conjugate $v$ is in the original span and is therefore a linear combination of the basis vectors. The same linear combination (except with conjugated coefficients) of the conjugated basis vectors produces $\bar v$.
A: Arrange $v_1,v_2,...,v_n$ in the rows of a matrix $A$. Let $v_i\in R^m$
$dim(span(v_{1},v_{2},...,v_{n}))=m-dim(nul(A))$
Then, define $B=\overline{A}$
$dim(span(\overline{v_{1}},\overline{v_{2}},...,\overline{v_{n}}))=m-dim(nul(B))$
If a vector $x_i$ is in the null space of $A$, then $\overline{x_i}$ is in the nul space of $B$. So, having $x_1,x_2,...,x_k$ as a basis for the null space of $A$, the copmplex conjugate of each, which is $\overline{x_1},\overline{x_2},...,\overline{x_k}$, is in the null space of $B$. Then it is required to show that $\overline{x_1},\overline{x_2},...,\overline{x_k}$ are linearly independent, which can be done using contradiction. In short, if they are not independent, then $x_1,x_2,...,x_k$ are not independent either. Finally, $dim(nul(B))=dim(nul(A))$ and consequently,
$dim(span(v_{1},v_{2},...,v_{n}))=dim(span(\overline{v_{1}},\overline{v_{2}},...,\overline{v_{n}}))$
