A full-column-rank real matrix has an economy QR factorization

Let $W=\mathrm{col}X$. The columns of $X$ are a basis of $W$, so $\dim W=p$. By [F1], ${\mathbb{R}}^n=W\oplus W^{\perp }$. Since $W^{\perp }$ is a subspace of the finite-dimensional space ${\mathbb{R}}^n$, [F4] makes it finite-dimensional; [F5] then gives $n=\dim W+\dim W^{\perp }$, so $\dim W^{\perp }=n-p$. By [F2], choose an orthonormal basis of $W^{\perp }$ and index its $n-p$ vectors as $z_{p+1},\dots ,z_n$. The columns of $A=[X{z}_{p+1}\cdots z_n]$ form a basis of ${\mathbb{R}}^n$: their $X$-part spans $W$, their remaining part spans $W^{\perp }$, and the direct sum in [F1] makes the union linearly independent. Thus $A$ is invertible.

Apply [F3] to write $A=UT$, where $U$ is orthogonal and $T$ is upper triangular with positive diagonal. Let $Q$ be the first $p$ columns of $U$, let $Q_0$ be its remaining columns, and let $R$ be the leading $p\times p$ block of $T$. Because $T_{ij}=0$ when $i>j$, every entry below row $p$ in its first $p$ columns is zero. The first $p$ columns of $A$ are therefore $X=QR$. The matrix $R$ is upper triangular with positive diagonal, while $U=[Q{Q}_0]$ gives $Q^{T}Q=I_p$ and the required orthonormal completion.

Since $R$ is invertible, $X^{T}X=R^{T}R$ and $(X^{T}X{)}^{-1}=R^{-1}{R}^{-T}$. Hence $P_X=QR(R^{T}R{)}^{-1}{R}^T{Q}^T=Q{Q}^T$, because $R(R^{T}R{)}^{-1}{R}^T=I_p$.