orthogonal basis - counter example I need a counter example for this:

Let V be inner product space, B={u1,u2,...,un) a basis of V. Suppose there are scalars a1,a2,...,an. 

If



then B is orthonormal basis of V.

I cant find an counter example.
Any hint?
Pick $u_1=\{1,1\}$, $u_2=\{1,0\}$, $\alpha_1=-2$ and $\alpha_2=1$.
It seems that $\langle u_i,u_j\rangle=\delta_{ij}$, thus $B$ is indeed  an orthonormal base: 1) Putting $\alpha_j=\delta_{ij}$ shows that $\Vert u_i\Vert^2=1$. Putting $\alpha_i=1=\alpha_j$ and $\alpha_k=0$ for $k\not=i,j$ if $i\not=j$ results in $\Vert u_i+u_j\Vert^2=2$. Therefore $2=\langle u_i+u_j,u_i+u_j\rangle=\langle u_i,u_i\rangle+2\langle u_i,u_j\rangle+\langle u_j,u_j\rangle=2+2\langle u_i,u_j\rangle$ and thus $\langle u_i,u_j\rangle=0$.

EDIT: This of course only holds when one may choose the coefficients arbitrarily. Otherwise there are, as shown, counterexamples.