Linear Algebra and Its Applications · Problem 3.4.30

Orthogonality — Gilbert Strang

Mathematics · qr-elimination-connection

Problem

If A=QRA = QR then ATA=RTR=A^T A = R^T R = \_\_\_\_ triangular times \_\_\_\_ triangular. Gram-Schmidt on AA corresponds to elimination on ATAA^T A. Compare A=[100−1100−1100−1]withATA=[2−10−12−10−12].A = \begin{bmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & -1 \end{bmatrix} \quad \text{with} \quad A^T A = \begin{bmatrix} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{bmatrix}. For ATAA^T A, the pivots are 22, 32\frac{3}{2}, 43\frac{4}{3} and the multipliers are −12-\frac{1}{2} and −23-\frac{2}{3}. (a) Using those multipliers in AA, show that column 1 of AA and B=B = column 2−122 - \frac{1}{2}(column 1) and C=C = column 3−233 - \frac{2}{3}(column 2) are orthogonal. (b) Check that ∥column 1∥2=2\|\text{column 1}\|^2 = 2, ∥B∥2=32\|B\|^2 = \frac{3}{2}, and ∥C∥2=43\|C\|^2 = \frac{4}{3}, using the pivots.

Browse the problem library