# Condition for existence of an orthonormal matrix whose column space is orthogonal to the column spac

Condition for existence of an orthonormal matrix whose column space is orthogonal to the column space of another matrix
While I was reading a statistics paper, I came across one statement that I don't understand (I just have basic linear algebra knowledge).
Assume (in the context of regressions), we have a $n×p$ data matrix X, assuming that X is invertible and n>p. The paper states
"$U\in {\mathbb{R}}^{n×p}$ is an orthonormal matrix whose column space is orthogonal to that of X s.t. ${U}^{T}X=0$": such matrix exists if $n\ge 2p$. I don't understand where the last statement comes from.
I know that the nullspace of X has dimension n−rank(X)=n−p in full rank case and U is the orthonormal basis of the null space of X. But I don't get the link why U only exists, if $n\ge p+rank\left(X\right)$, i.e. $n\ge 2p$.
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billyfcash5n
Note: it is not standard to refer to the matrix X as "invertible" unless X is a square matrix. Presumably, "X is invertible" refers in this case to the fact that X has full rank. In this case, because X is $n×p$ with n>p, this means that X has rank p (i.e. has "full column rank").
As you say, "$U\in {\mathbb{R}}^{n×p}$ is an orthonormal matrix whose column space is orthogonal to that of X". The fact that the column space is orthogonal to that of X is equivalent to the statement that ${U}^{T}X=0$. Because U is $n×p$ with orthonormal columns, the dimension of its column space is p. Because the column space of U is orthogonal to that of X, the column space of U must be a subspace of the orthogonal complement to the column space of X. The column space of X is a p-dimensional subspace of ${\mathbb{R}}^{n}$, which means that its orthogonal complement has dimension n−p.
Putting all this together leads us to the following conclusion:$\mathrm{col}\left(U\right)\subseteq \mathrm{col}\left(X{\right)}^{\perp }\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{dim}\left(\mathrm{col}\left(U\right)\right)\le \mathrm{dim}\left(\mathrm{col}\left(X{\right)}^{\perp }\right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{0ex}{0ex}}p\le n-p\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{0ex}{0ex}}2p\le n.$