Let $V$ be a $k$-vector space equipped with an anisotropic bilinear form $B\colon V\otimes V\to k$, e.g. with $k=\mathbb{R}$ we can let $B$ be any inner product on $V$.
Let $L$ be the tautological line bundle over $\mathbb{P}(V)$, and let $E$ be the vector bundle over $\mathbb{P}(V)$ for which
$$E_{[x]} = \{ y\in V: B(x,y)=0\}.$$ Let $T$ be the trivial bundle $T=V\times \mathbb{P}(V)$.
Given local sections $u\in \Gamma_L(U)$ and $v\in \Gamma_E(U)$ we obtain $u+v\in \Gamma_T(U)$. Conversely, given $w\in \Gamma_T(U)$ we obtain $u\in \Gamma_L(U)$ and $v\in \Gamma_E(U)$ by $B$-orthogonal projection. Since $B$ is bilinear the map $w\mapsto(u,v)$ is algebraic, and since $B$ is anisotropic we have $w=u+v$. Thus it seems we have an algebraic isomorphism
$$T\cong L\oplus E.$$ But $\mathbb{P}(V)$ is complete, so by Theorem 3 of Atiyah's On the Krull-Schmidt theorem with application to sheaves, vector bundles over $\mathbb{P}(V)$ should have unique decomposition into indecomposables. Clearly $T$ is a direct sum of $\dim(V)$ copies of the trivial line bundle, so this seems to lead to the false conclusion that $L$ is trivial.
Where am I making a mistake? Vector spaces over algebraically closed fields don't admit anisotropic bilinear forms, so a potential resolution would be if Atiyah's result only applies to varieties over algebraically closed fields, but this doesn't seem to be the case.