Qr Decomposition 176817

1. **State the problem:** Find the QR-decomposition of the matrix $$A=\begin{bmatrix}1 & 2 & 1 \\ 1 & 1 & 1 \\ 0 & 3 & 1\end{bmatrix}$$ using the Gram-Schmidt process. 2. **Recall the QR-decomposition:** Any matrix $$A$$ can be decomposed as $$A = QR$$ where $$Q$$ is an orthogonal matrix (columns are orthonormal vectors) and $$R$$ is an upper triangular matrix. 3. **Gram-Schmidt process:** Given column vectors $$a_1, a_2, a_3$$ of $$A$$, we find orthonormal vectors $$q_1, q_2, q_3$$ as follows: - $$u_1 = a_1$$ - $$q_1 = \frac{u_1}{\|u_1\|}$$ - $$u_2 = a_2 - \text{proj}_{q_1}(a_2) = a_2 - (q_1^T a_2) q_1$$ - $$q_2 = \frac{u_2}{\|u_2\|}$$ - $$u_3 = a_3 - \text{proj}_{q_1}(a_3) - \text{proj}_{q_2}(a_3) = a_3 - (q_1^T a_3) q_1 - (q_2^T a_3) q_2$$ - $$q_3 = \frac{u_3}{\|u_3\|}$$ 4. **Calculate:** - $$a_1 = \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}$$ - $$\|a_1\| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2}$$ - $$q_1 = \frac{1}{\sqrt{2}} \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}$$ - $$a_2 = \begin{bmatrix}2 \\ 1 \\ 3\end{bmatrix}$$ - Compute $$q_1^T a_2 = \frac{1}{\sqrt{2}}(2 + 1 + 0) = \frac{3}{\sqrt{2}}$$ - $$u_2 = a_2 - (q_1^T a_2) q_1 = \begin{bmatrix}2 \\ 1 \\ 3\end{bmatrix} - \frac{3}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix} = \begin{bmatrix}2 \\ 1 \\ 3\end{bmatrix} - \frac{3}{2} \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix} = \begin{bmatrix}2 - 1.5 \\ 1 - 1.5 \\ 3 - 0\end{bmatrix} = \begin{bmatrix}0.5 \\ -0.5 \\ 3\end{bmatrix}$$ - $$\|u_2\| = \sqrt{0.5^2 + (-0.5)^2 + 3^2} = \sqrt{0.25 + 0.25 + 9} = \sqrt{9.5}$$ - $$q_2 = \frac{1}{\sqrt{9.5}} \begin{bmatrix}0.5 \\ -0.5 \\ 3\end{bmatrix}$$ - $$a_3 = \begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}$$ - Compute $$q_1^T a_3 = \frac{1}{\sqrt{2}}(1 + 1 + 0) = \frac{2}{\sqrt{2}} = \sqrt{2}$$ - Compute $$q_2^T a_3 = \frac{1}{\sqrt{9.5}}(0.5 \cdot 1 + (-0.5) \cdot 1 + 3 \cdot 1) = \frac{1}{\sqrt{9.5}}(0.5 - 0.5 + 3) = \frac{3}{\sqrt{9.5}}$$ - $$u_3 = a_3 - (q_1^T a_3) q_1 - (q_2^T a_3) q_2 = \begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix} - \sqrt{2} \cdot \frac{1}{\sqrt{2}} \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix} - \frac{3}{\sqrt{9.5}} \cdot \frac{1}{\sqrt{9.5}} \begin{bmatrix}0.5 \\ -0.5 \\ 3\end{bmatrix} = \begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix} - \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix} - \frac{3}{9.5} \begin{bmatrix}0.5 \\ -0.5 \\ 3\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix} - \begin{bmatrix}0.158 \\ -0.158 \\ 0.947\end{bmatrix} = \begin{bmatrix}-0.158 \\ 0.158 \\ 0.053\end{bmatrix}$$ - $$\|u_3\| = \sqrt{(-0.158)^2 + 0.158^2 + 0.053^2} = \sqrt{0.025 + 0.025 + 0.003} = \sqrt{0.053}$$ - $$q_3 = \frac{1}{\sqrt{0.053}} \begin{bmatrix}-0.158 \\ 0.158 \\ 0.053\end{bmatrix}$$ 5. **Form matrices:** $$Q = \begin{bmatrix} q_1 & q_2 & q_3 \end{bmatrix}$$ $$R = Q^T A$$ 6. **Final answer:** $$Q = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{0.5}{\sqrt{9.5}} & \frac{-0.158}{\sqrt{0.053}} \\ \frac{1}{\sqrt{2}} & \frac{-0.5}{\sqrt{9.5}} & \frac{0.158}{\sqrt{0.053}} \\ 0 & \frac{3}{\sqrt{9.5}} & \frac{0.053}{\sqrt{0.053}} \end{bmatrix}$$ $$R = \begin{bmatrix} \sqrt{2} & \frac{3}{\sqrt{2}} & \sqrt{2} \\ 0 & \sqrt{9.5} & \frac{3}{\sqrt{9.5}} \\ 0 & 0 & \sqrt{0.053} \end{bmatrix}$$ This completes the QR-decomposition of matrix $$A$$ using Gram-Schmidt.