I'm trying to prove the following statement :
Suppose $V$ is an inner product space, its dimension is 4, and $U$ is a subspace of $V$ spanned by the basis $\{u_1, u_2\}$. Then there is a scalar $t$ such that $u_2-tu_1$ is perpendicular to $u_1$ (That is, $\langle u_2-tu_1, u_1\rangle = 0$)
As far as I understand the only way I can prove this is to show that $u_2 -tu_1 = 0$ but its impossible since $u_2$ and $u_1$ form a basis to $U$ (That is, there isn't a scalar $t$ such that $u_2=tu_1$ which implies $u_2-tu_1\ne 0$).
Am I missing something? Is it possible that I'm trying to prove a false statement?
Well, $$ t = \frac{\langle u_1,u_2 \rangle}{\langle u_1,u_1 \rangle} $$ since the elements of a basis are nonzero. Therefore you have found the scalar $t$ with the required property.