When can we cancel vector bundles from tensor products? Let $E,F,G$ be algebraic vector bundles over $\mathbb P_{\mathbb C}^n$. My general question is:
Assume $E\otimes G \cong F\otimes G$, under what conditions can one conclude that $E\cong F$? 
Some easy answers (if I am not mistaken): one can when $n=1$ or  when $G$ is a line bundle. At this point I am mostly interested in the case when $E$ is a direct sum of line bundles, but any comments/reference/solutions/analogues about other cases would be appreciated.  
 A: It seems, that in the case in which E and F are direct sums of line bundles (and G is non-zero!), you can reconstruct E and F knowing that $E \otimes G \simeq F \otimes G$: this simply imitates the proof of the fact that vector bundles on $\mathbb{P}^1$ are sums of line bundles. Indeed, since the reconstruction is fine in the case in which G is a line bundle, we may replace E and F by E(e) and F(e) for any integer e: hence, exchanging if necessary E and F, we may suppose that E has a section and E(-1),F(-1) do not have sections.  Let g be the integer such that G(g) has a section and G(g-1) does not. Thus we have $E \otimes G(g) \simeq F \otimes G(g)$; by considering global sections, we deduce that the multiplicity of the number of trivial direct summands in E is the same as the multiplicity in F. Remove the copies of $\mathcal{O}$ from both E and F and repeat.
A: The answer to your question is yes, if $F$ is a direct sum of line bundles. So, let us assume that and its rank is $m$. And as you observed, we will assume that we are working over $\mathbb{P}^2$. Let $0\to F_0\to F_1\to E\to 0$ be the minimal resolution of $E$, where $F_i$ are direct sum of line bundles with rank of $F_0=n$ so that rank of $F_1=n+m$. Tensoring with $G$, we get $0\to G\otimes F_0\to G\otimes F_1\to G\otimes F\to 0$, the last by the assumption. Taking cohomologies, letting $H_*^0(G)=M,H_*^1(G)=N$, we get an exact sequence $N^n\to N^{n+m}\to N^m$, which for length considerations (length of $N$ is finite) can easily seen to be exact on the left (and right). In particular, we have surjectivity of global sections and thus an exact sequence $0\to M^n\to M^{n+m}\to M^m\to 0$. This splits (I have forgotten whose theorem it is, but I think Ihave read it recently in some comment by Graham Leuschke) which is impossible since the first map has all entries in the maximal ideal unless $n=0$.
A: According to this preprint, over a connected proper algebraic variety $X$ there is a universal reductive group $G$ such that isomorphism classes of vector bundles of rank $n$ are in bijection with isomorphism classes of $n$-dimensional representations of $G$. Furthermore the component group is $\pi_1(X)$.
So the cancellation property you seek holds for any simply-connected proper algebraic variety, because it holds for representations of any connected reductive group. We can see this by characters - a representation is uniquely determined by its character, and because the group is connected, there are no zero-divisors in the ring of characters.
This certainly includes $\mathbb P^n$.
A: Well, I am by no means an algebraic geometer and maybe this is not even helpful, but anyway:
If you have a bundle $G^\perp$, such that $G \oplus G^\perp \simeq \underline{\mathbb{C}^n}$ is trivial and $E \otimes G^\perp$ is isomorphic to $F \otimes G^\perp$, then you at least get $E^n \simeq F^n$, which is kind of analogous to your remark concerning line bundles. But I better stop mumbling trivialities now.
