Difference between direct sum of vector spaces and direct sum of representations. 
Let $G$ be a group. Let $V$ and $W$ be vector spaces which have the structures of $G$-modules. If $U \simeq V \oplus W$ as vector spaces then can't we say that they are same as $G$-modules? My teacher says that if $U$ is a direct sum of vector spaces then the matrix associated with the corresponding representation is a $2 \times 2$ upper triangular block matrix which is not necessarily diagonal. The matrix is diagonal if and only if $U \simeq V \oplus W$ as $G$-modules.

Could anyone please shed some light on what is being meant?
Thanks for your time.
 A: From what you've written you must mean $U$ is also some given $kG$ module.
There is no assurance that the isomorphism $U\cong V\oplus W$ is $G$-linear.  It may indeed have the same dimension as $V\oplus W$ (for that is all a vector space isomorphism buys you) and yet its $kG$ structure may differ from that of $V\oplus W$ considered as a $kG$ module.
For example, choose $k=\mathbb C$ and $G$ the cyclic group of order $2$.  It's got two nonisomorphic one-dimensional representations $M_1$ and $M_2$, so $M_1\oplus M_2\cong M_1\oplus M_1$ as vector spaces, but not as representations.

Given your discussion of nondiagonal representations, I think maybe your context has not been described completely. Let's suppose $M_1$ and $M_2$ are subrepresentations of a given representation, and that we want to know about $M_1+M_2$ as a representation.  To say that $M_1+M_2=M_1\oplus M_2$ means that the representation can be diagonalized. Otherwise you can only put it in upper triangular form, and the upper right hand block shows to what degree $M_1$ and $M_2$ are mixed with each other.  When $M_1\cap M_2=\{0\}$ that block can be made zero everywhere.
