Uniqueness of the QR-factorization

I'm preparing a talk about the QR-factorization.

I alredy have proved the existence of it: If $A \in \mathbb{R}^{n \times m} $ there exists an orthogonal matrix $Q \in \mathbb{R}^{n \times n}$ and an upper triangular $R \in \mathbb{R}^{n \times m}$ with $A=QR$.

There is a mistake in the following theorem, see my answer below.

Now I want to prove the following theorem: If $A \in \mathbb{R}^{n \times m} $ with $rank(A)=m$ has the QR-factorization $A=QR$ where $R$ has positive diagonal entries, then the Q and R are unique.

But all literature use proves with another factorisation. So i dont now how to handle, maybe sombody can help, thanks.

Once you have one $QR$ factorization, say $A=Q_1R_1$, then it is easy to produce another one by defining $Q_2=Q_1B$ and $R_2=B^{-1}R_1$. But for $Q_2$ and $R_2$ to be orthogonal and upper triangular, respectively, $B$ must be orthogonal and diagonal. That means it can only have $\pm 1$ as elements on the diagonal. If $R_1$ already has positive diagonal entries, then $B$ must be the identity.

There is an mistake in the Theorem of the uniqueness.

For example: If $A= \begin{pmatrix} 1&0\\ 0&1\\ 0&0\\ \end{pmatrix} $. We have $A= \begin{pmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{pmatrix} \begin{pmatrix} 1&0\\ 0&1\\ 0&0\\ \end{pmatrix} $ and $A=\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&-1 \end{pmatrix} \begin{pmatrix} 1&0\\ 0&1\\ 0&0\\ \end{pmatrix} $. Two different QR-factrorisations.

You can fix that Theorem, by additional demanding of m=n. Then the prove above works, the problem in was the null space of R.