Given l.i. integer vectors $x^1,\dots,x^{n-1} \in \mathbb Z^n$ can we find $x^n \in Z^n$ orthogonal to all other $x^i$?

Question

Consider some nonzero vector $x^1=(x_1,x_2) \in \mathbb Z^2$ with integer components. Then by defining $$x^2 = (x_2, -x_1)$$ we can find a vector with integer components orthogonal to $x^1$. I am wondering whether this is true more generally.

Question: If $x^1,\dots,x^{n-1} \in \mathbb Z^n$ are linear independent vectors with integer components, can we always find another vector $x^n \in \mathbb Z^n$ with integer components that is orthogonal to all the $x^i,$ $i=1,\dots,n-1$?

If not, what conditions on $x^1,\dots,x^{n-1}$ would make sure that we can find such a $x^n$?

Answer 1

Sure. Requiring orthogonality to $x^1,\dots,x^{n-1}$ can be looked at as $n-1$ linear equations in $n$ rational unknowns with rational coefficients. When the vectors you want orthogonality to are linearly independent, this system has a line of solutions in $\mathbb{Q}^n$, say $\{ c x^* : c \in \mathbb{Q} \}$ for a particular nontrivial solution $x^*$ in $\mathbb{Q}^n$.

Now take $c$ to be any common multiple of the denominators of the components of $x^*$ (with the convention that the "denominator" of an integer is $1$) and you'll get a solution in $\mathbb{Z}^n$.

Answer 2

Here is a general approach based on a standard mnemonic for the vector cross product in $\Bbb R^3$:

Take your $n-1$ vectors, write their components as the columns of an $n\times (n-1)$ matrix, and add an $n$th column consisting of the $n$ basis vectors $\vec e_1, \vec e_2,\ldots, \vec e_n$ (so you get a "matrix" with $n-1$ columns consisting of integers and one column consisting of vectors). Then take the "determinant" of this square "matrix". If all the entries were integers to start with, the resulting determinant will be a linear combination of the basis vectors with only integer coefficients.

For $n=2$, you get your "swap the elements and change a sign" result. For $n=3$ you get the vector cross product.

Brief explanation of what's going on: given $n$ vectors with entries written into a matrix as above, the determinant of this matrix gives the signed hypervolume of the parallelepiped spanned by the $n$ vectors. Given your $n-1$ vectors, you can use this to construct a linear function $\Bbb R^n\to \Bbb R$ ("insert the final vector and calculate the volume").

Any linear function $\Bbb R^n\to \Bbb R$ is given by a dot product by some vector. My description above gives that vector for this function. If the $n-1$ given vectors are linearly independent, then this associated linear function is not the zero function. Thus the associated vector is not the zero vector. Finally, the function evaluates to zero on each of the given $n-1$ vectors (the volume becomes zero), so the associated vector is orthogonal to each of the given vectors.