Orthogonal vectors in case of Lorentz Metric Let us consider $\mathbb{R}^4$ equipped with the Lorentz metric
$$\eta(X,Y)=x^0y^0-(x^1y^1+x^2y^2+x^3y^3)$$
Let $X\in\mathbb{R}^4$ a time-like vector, that is $\eta(X,X)>0$. I want to show that all orthogonal vectors $Y\in\mathbb{R}^4$ are zero or space-like vectors.
It is not difficult to prove that the zero-vector is orthogonal to the time-like vector. To show the case of a space-like vector I assume Y is time-like. Hence $\eta(Y,Y)>0$. It follows
$$0<\eta(X,X)+\underbrace{2\eta(X,Y)}_{=0}+\eta(Y,Y)=\eta(X+Y,X+Y),\quad X\not=0$$
Putting $X=-Y$ we get the contradiction $0<0$. Hence Y must be a space-like vector.
Is the reasoning correct?
 A: Suppose $X$ is timelike, so that $\eta(X,X)>0$. Without loss of generality, suppose $\eta(X,X)=a\ $ for some $positive$ constant $a\in\mathrm{R}^{+}$ (positive since its timelike).
Now $choose$ a coordinate system in which $x^{0}=\sqrt{a}\ $ and $\ x^{1}=x^{2}=x^{3}=0$ (you can always do this since the quantity $\eta(X,X)$ is invariant of the coordinate system being used). You can check that $\eta(X,X)=a\ $ in this coordinate system.
Now suppose $Y$ is a vector with components $y^{0}$, $y^{1}$, $y^{2}$, $y^{3}$, and is orthogonal to $X$. This means that $\eta(X,Y)=0$. What conditions does this give you on the coordinates of $Y$?
A: No. The reasoning is incorrect. The vector $X$ has been fixed from the get go, so it is "illegal" to later on say "set $X = -Y$." 
One correct way to prove the assertion is as follows. 
Write the given time-like vector $X$ in components $(X_0, X_1, X_2, X_3)$, we know that since it is timelike the components satisfy
$$ (X_0)^2 > (X_1^2 + X_2^2 + X_3^2) $$
Now suppose $Y$ is timelike, so that
$$ (Y_0)^2 > (Y_1^2 + Y_2^2 + Y_3^2) $$
we compute
$$ \eta(X,Y) = X_0 Y_0 - (X_1 Y_1 + X_2 Y_2 + X_3 Y_3) $$
By Cauchy's inequality we know
$$ |X_1 Y_1 + X_2 Y_2 + X_3 Y_3| \leq \sqrt{ X_1^2 + X_2^2 + X_3^2} \sqrt{Y_1^2 + Y_2^2 + Y_3^2} $$
but by the timelikeness we then see
$$ |X_1 Y_1 + X_2 Y_2 + X_3 Y_3| < |X_0 Y_0| $$
which implies 
$$ \eta(X,Y) \neq 0 $$
This concludes the proof that two time-like vectors can not be orthogonal in the Minkowski metric. 
