Given:
- $A$: $k \times n$-matrix with independent columns,
- $R$: $n \times n$-matrix,
- $A=PR$, where $P$ is a matrix.
Question: Prove that $R$ is invertible.
My proof: Since $A$ has independent columns, we know that there exists a $QR$-factorization for $A$. Let $A=QR_{1}$ be the $QR$ factorization for $A$, so $PR=QR_{1}$. Since $Q$ is an orthogonal matrix, per definition of $QR$-factorization: $Q^{T}Q=I$. So:
$PR=QR_{1}$ iff $Q^{T}PR=R_{1}$ iff $Q^{T}PRR_{1}^{-1}=I$ (per def. of $QR$ factorisation is $R_{1}$ invertible).
Now note: $Q^{T}P$ and $RR_{1}^{-1}$ are both $n \times n$ matrices and their product is equal to the identity matrix. So:
$\operatorname{Det}(Q^{T}PRR_{1}^{-1})=1$ iff $\operatorname{Det}(Q^{T}P)\operatorname{Det}(RR_{1}^{-1})=1$ this means that $\operatorname{Det}(RR_{1}^{-1}) \neq 0 $. Since $R$ and $R_{1}$ are both $n \times n$ matrices we can write: $\operatorname{Det}(RR_{1}^{-1})=\operatorname{Det}(R)\operatorname{Det}(R_{1}^{-1}) \neq 0$, so $Det(R) \neq 0$, which means that $R$ is invertible.
Are there any errors in my proof?