Hi,
zu (a)
Die Norm der j-ten Spalte von \( Q A \) berechnet sich zu
$$ \| (QA)^{(j)} \|_2^2 = \sum_{i=1}^n (Q A )_{i,j}^2 = \sum_{i=1}^n \sum_{k=1}^n \sum_{l=1}^n Q_{i,k} A_{k,j} Q_{i,l} A_{l,j} $$ weil \( Q \) orthogonal ist folgt $$ \sum_{i=1}^n Q_{i,k} Q_{i,l} = \delta_{k,l} \Rightarrow \\\| (QA)^{(j)} \|_2^2 = \sum_{i=1}^n (Q A )_{i,j}^2 = \sum_{k=1}^n \sum_{l=1}^n \delta_{k,l} A_{k,j} A_{l,j} = \sum_{k=1}^n A_{k,j}^2 = \| A^{(j)} \|^2 $$
zu (b)
$$ | \det(A) | = | \det(Q R) | = | \det(Q) | \cdot | \det(R) | = \left| \prod_{j=1}^n R_{j,j} \right| \le \left| \prod_{j=1}^n \sqrt{ \sum_{i=1}^n R_{i,j}^2 } \right| = \left| \prod_{j=1}^n \left \| R^{(j)} \right \|_2 \right| $$
Wegen $$ \left\| R^{(j)} \right\|_2 = \left\| Q R^{(j)} \right\|_2 = \left\| A^{(j)} \right\|_2 $$ folgt
$$ | \det(A) | \le \left| \prod_{j=1}^n \left \| A^{(j)} \right \|_2 \right| $$
