Please help me to prove the following result:
Let $V(x)$ be a non constant polynomial of degree less or equal two give by $V(x)=\sum_{i=1}^d\mu_ix_i+\sum_{i=1}^d\frac{\nu_i}{2}x_i^2$. We denote $$Hess\; V=\begin{pmatrix} \nu_{1} & 0 & \ldots & 0\\ 0 & \nu_{2} & \ldots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 &\ldots & \nu_{d} \end{pmatrix} $$
its Hessian matrix.
If $det(Hess\; V)=0$, we can always find an orthonormal basis, in which $V(x)$ takes the form $$V(x)=V_0+\lambda_1x_1+\sum_{i=2}^d\frac{\nu_i}{2}x_i^2$$
where $\lambda_1>0$
Thanks
As originally stated, the result is false. The zero polynomial is of indicated form (with all $\mu_i=\nu_i = 0$) but there is no orthogonal change of variable that puts it in form $V(x)=\lambda_1 x_1$ with $\lambda_1>0$. Because the vanishing-of-the-gradient-at-zero property is invariant with respect to orthogonal change of variables.
As edited, the result is trivially true. The assumptions force $V(x)=\langle \mu,x\rangle$ with $\mu\ne0$ to be linear. There is an orthogonal matrix $O$ such that the first row of $O$ is a positive multiple of $\mu$. That matrix gives the desired change of variables.