TITLE: Examples of non trivial vector bundles
QUESTION [8 upvotes]: Once you see the notion of vector bundle, next thing you want to see are examples of   non trivial vector bundles.
Here, I want to collect such examples with justification of one or two lines saying why this vector bundle is non trivial.
Please add one per answer.

REPLY [6 votes]: You get an example for every non-orientable smooth manifold $M$:
A smooth $n$-dimensional manifold $M$ is  orientable  iff there exists a nowhere vanishing $n$-form  i.e. a nowhere vanishing  section  of the bundle $\Lambda^n(T^*M)$  whose fiber at $p$ is the vectorspace of all multlinear alternating maps from $(T_pM)^n$ to $\mathbb R$. Since this bundle is 1-dimensional the existence of a nowhere vanishing section is equivalent to the trviality of the bundle. 
As an example if $n$ is even then $\mathbb{RP^n}$ is not orientable. The reason : 
Any $n$-form $\omega$ on $\mathbb{RP^n}$ can be pulledback to a $n$-form $\bar\omega$ on $S^n$ via the quotient map $q:S^n\rightarrow\mathbb{RP^n}$. Then after identifying $TS^n\subset S^n\times\mathbb{R^{n+1}}$ there is $f\in C^\infty(S^n)$ such that $\bar\omega_p(X_1,...,X_n)=f(p)\det(X_1,...,X_n,p)$. As $q$ cannot distinguish antipodal points $\bar\omega_{p}(X_1,...,X_n)=\bar\omega_{-p}(-X_1,...,-X_n)$. In particular setting $X_i=e_i$, $p=e_{n+1}$ yields $f(-p)=-f(p)$ so as $S^n$ is connected $f$ has a zero so $\bar\omega$ vanishes somewhere and hence so does $\omega$.