Take a real matrix QR decomposition as $A_{m\times n}= QR$ where $Q$ is an $m\times m$ orthogonal matrix and $R$ is an $m \times n$ upper triangular matrix.
How can I go about proving the following norm equivalences:
$\alpha \|A\|_1 \leq \|R\|_1 \leq \beta\|A\|_1$
$\alpha \|A\|_{\infty} \leq \|R\|_{\infty} \leq \beta\|A\|_{\infty}$
What I've tried is:
$\|A\|_1 = \|QR\|_1$
And by the inequality: $\|A\|_1 =\|QR\|_1 \leq \|Q\|_1\|R\|_1$
Taking $\alpha = \frac{1}{\|Q\|_1}$, we would end up with:
$\alpha\|A\|_1 \leq \|R\|_1$
On the other hand we'd have: $\|IR\|_1 = \|Q^TQR\|_1 \leq \|Q^T\|_1\|QR\|_1$ and taking $\beta = \|Q^T\|$ yields:
$\alpha \|A\|_1 \leq \|R\|_1 \leq \beta\|A\|_1$
Since inequality arguments work for both 1-norm and $\infty$-norm this would finish it off.
However, I don't know if I can find explicit constants for a general case for $\|Q\|_1$, $\|Q\|_\infty$.
I feel like they could be $n$ or something like this.
Hints would be very appreciated.
For an Euclidean unit vector $v$ you get by the Cauchy-Schwarz inequality that $$ |v_1|+...+|v_n|\le\sqrt{1+...+1}\sqrt{|v_1|^2+...+|v_n|^2}=\sqrt{n} $$ As the row and column sum norms of an orthogonal matrix are maxima over such sums, you get that both $\|Q\|_1\le\sqrt{n}$ and $\|Q\|_\infty\le\sqrt{n}$.