$X$ is a complex compact algebraic manifold. Let $[V]\in [X,Gr_C(k,\infty)]$ be a vector bundle. Is this even a holomorphic vector bundle? $X$ is a complex compact algebraic manifold. Let $[V]\in [X,Gr_C(k,\infty)]$ be a  vector bundle representing the corresponding class in homotopy class $[X,Gr_C(k,\infty)]$ where $k$ where $Gr_C(k,\infty)$ is $k-$plane grassmanian over complex number. 
$\textbf{Q:}$ This $V$ may not be a holomorphic rank $k$ vector bundle. Now consider all representatives $[V]$. Is there a way to tell whether $V$ possibly possesses a holomorphic vector bundle structure? 
$\textbf{Q':}$ Let $V_1,V_2\in [V]$ be two holomorphic vector bundles in the same class. Now in smooth category $V_1\cong V_2$. However, there is one complex vector bundle structure on each of them. Should these 2 complex structure even agree? In other words, if $V_i$ give rise to identical homotopy maps $f$, then should $f$ detect difference in non biholomorphic holomorphic vector bundles $V_1,V_2$?
 A: I don't know of any simple way to detect whether a vector bundle admits a holomorphic stucture in general.  For line bundles, though, there is a very nice criterion coming from the exponential sequence.  Recall there is a short exact sequence of sheaves $$0\to\mathbb{Z}\to\mathcal{O}_X\to\mathcal{O}_X^\times\to 0$$ where the first map is the inclusion of the constant functions with integer values and the second map is $f\mapsto\exp(2\pi i f)$.  This induces a long exact sequence in cohomology $$H^1(X,\mathcal{O}_X)\to H^1(X,\mathcal{O}_X^\times)\stackrel{c_1}\to H^2(X,\mathbb{Z})\to H^2(X,\mathcal{O}_X).$$  There is a natural isomorphism between $H^1(X,\mathcal{O}_X^\times)$ and the Picard group $\operatorname{Pic}(X)$ of isomorphism classes of holomorphic line bundles on $X$, and it turns out that the connecting homomorphism labelled $c_1$ above is exactly the map taking a holomorphic line bundle to its first Chern class.  On the other hand, topological (or smooth) line bundles are completely classified by their first Chern classes (each element of $H^2(X,\mathbb{Z})$ is the first Chern class of a topological line bundle which is unique up to isomorphism).  So, a topological line bundle admits a holomorphic structure iff it is in the image of the map $c_1$ above, or equivalently if it is in the kernel of the map $H^2(X,\mathbb{Z})\to H^2(X,\mathcal{O}_X)$.  By Hodge theory this map can be identified with the map $H^2(X,\mathbb{Z})\to H^2(X,\mathbb{C})\to H^{0,2}(X)$ where the second map is the projection given by the Hodge decomposition.  An integral cohomology class is fixed by complex conjugation, so its projection to $H^{0,2}(X)$ is trivial iff its projection to $H^{2,0}(X)$ is trivial.  So, a topological line bundle admits a holomorphic structure iff its first Chern class has Hodge type $(1,1)$.
This also gives examples of holomorphic line bundles which are topologically but not holomorphically isomorphic: those just correspond to $c_1$ failing to be injective.  In particular, when $X$ is a curve and we identify $H^2(X,\mathbb{Z})$ with $\mathbb{Z}$, then $c_1$ is just the map taking a holomorphic line bundle to its degree.  So, any two holomorphic line bundles of the same degree are topologically isomorphic, but if $X$ has genus greater than $0$ then it has non-isomorphic holomorphic line bundles of the same degree.
For vector bundles of higher rank it is still true that all the Chern classes of a holomorphic vector bundle have Hodge type $(p,p)$.  I don't know whether this a sufficient criterion for a vector bundle to admit a holomorphic structure though (I suspect it is not).
