Quadratic Canonical 5Ebae7

1. **State the problem:** Reduce the quadratic form $$2x_1^2 + x_2^2 + x_3^2 + 2x_1x_2 - 2x_1x_3 - 4x_2x_3$$ to its canonical form using an orthogonal transformation. 2. **Write the quadratic form in matrix form:** $$Q(\mathbf{x}) = \mathbf{x}^T A \mathbf{x}$$ where $$\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$$ and $$A = \begin{bmatrix} 2 & 1 & -1 \\ 1 & 1 & -2 \\ -1 & -2 & 1 \end{bmatrix}$$ Note: The matrix is symmetric, as required for quadratic forms. 3. **Find eigenvalues of matrix $$A$$:** Solve $$\det(A - \lambda I) = 0$$. Calculate: $$\det \begin{bmatrix} 2-\lambda & 1 & -1 \\ 1 & 1-\lambda & -2 \\ -1 & -2 & 1-\lambda \end{bmatrix} = 0$$ Expanding the determinant: $$ (2-\lambda) \left((1-\lambda)(1-\lambda) - (-2)(-2)\right) - 1 \left(1(1-\lambda) - (-2)(-1)\right) -1 \left(1(-2) - (1-\lambda)(-1)\right) = 0 $$ Simplify inside: $$ (2-\lambda) \left((1-\lambda)^2 - 4\right) - 1 \left(1-\lambda - 2\right) - 1 \left(-2 + 1 - \lambda\right) = 0 $$ $$ (2-\lambda) \left((1-\lambda)^2 - 4\right) - ( -1 - \lambda ) - ( -1 - \lambda ) = 0 $$ $$ (2-\lambda) \left((1-\lambda)^2 - 4\right) + 1 + \lambda + 1 + \lambda = 0 $$ $$ (2-\lambda) \left((1-\lambda)^2 - 4\right) + 2 + 2\lambda = 0 $$ Calculate $$(1-\lambda)^2 - 4 = 1 - 2\lambda + \lambda^2 - 4 = \lambda^2 - 2\lambda - 3$$ So: $$ (2-\lambda)(\lambda^2 - 2\lambda - 3) + 2 + 2\lambda = 0 $$ Multiply: $$ (2)(\lambda^2 - 2\lambda - 3) - \lambda(\lambda^2 - 2\lambda - 3) + 2 + 2\lambda = 0 $$ $$ 2\lambda^2 - 4\lambda - 6 - \lambda^3 + 2\lambda^2 + 3\lambda + 2 + 2\lambda = 0 $$ Combine like terms: $$ -\lambda^3 + (2\lambda^2 + 2\lambda^2) + (-4\lambda + 3\lambda + 2\lambda) + (-6 + 2) = 0 $$ $$ -\lambda^3 + 4\lambda^2 + 1\lambda - 4 = 0 $$ Multiply both sides by $$-1$$: $$ \lambda^3 - 4\lambda^2 - \lambda + 4 = 0 $$ 4. **Find roots of the cubic:** Try rational roots $$\pm1, \pm2, \pm4$$. Test $$\lambda=1$$: $$1 - 4 - 1 + 4 = 0$$ So $$\lambda=1$$ is a root. Divide polynomial by $$(\lambda - 1)$$: Using polynomial division or synthetic division: $$\lambda^3 - 4\lambda^2 - \lambda + 4 = (\lambda - 1)(\lambda^2 - 3\lambda - 4)$$ Solve quadratic: $$\lambda^2 - 3\lambda - 4 = 0$$ Use quadratic formula: $$\lambda = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm 5}{2}$$ So: $$\lambda_2 = 4, \quad \lambda_3 = -1$$ 5. **Eigenvalues are:** $$\lambda_1 = 1, \quad \lambda_2 = 4, \quad \lambda_3 = -1$$ 6. **Find eigenvectors for each eigenvalue:** For $$\lambda=1$$ solve $$(A - I)\mathbf{v} = 0$$: $$\begin{bmatrix} 1 & 1 & -1 \\ 1 & 0 & -2 \\ -1 & -2 & 0 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$ From first row: $$v_1 + v_2 - v_3 = 0$$ From second row: $$v_1 - 2 v_3 = 0 \Rightarrow v_1 = 2 v_3$$ From third row: $$-v_1 - 2 v_2 = 0 \Rightarrow -2 v_3 - 2 v_2 = 0 \Rightarrow v_2 = -v_3$$ Substitute into first row: $$2 v_3 + (-v_3) - v_3 = 0$$ $$0=0$$ (consistent) Eigenvector: $$\mathbf{v}_1 = \begin{bmatrix} 2 v_3 \\ -v_3 \\ v_3 \end{bmatrix} = v_3 \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}$$ Normalize: $$\|\mathbf{v}_1\| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$$ So normalized eigenvector: $$\mathbf{u}_1 = \frac{1}{\sqrt{6}} \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}$$ 7. Repeat for $$\lambda=4$$: $$(A - 4I) = \begin{bmatrix} -2 & 1 & -1 \\ 1 & -3 & -2 \\ -1 & -2 & -3 \end{bmatrix}$$ Solve $$(A - 4I)\mathbf{v} = 0$$. From first row: $$-2 v_1 + v_2 - v_3 = 0$$ From second row: $$v_1 - 3 v_2 - 2 v_3 = 0$$ From third row: $$-v_1 - 2 v_2 - 3 v_3 = 0$$ From first row: $$v_2 = 2 v_1 + v_3$$ Substitute into second row: $$v_1 - 3(2 v_1 + v_3) - 2 v_3 = 0$$ $$v_1 - 6 v_1 - 3 v_3 - 2 v_3 = 0$$ $$-5 v_1 - 5 v_3 = 0 \Rightarrow v_1 = -v_3$$ Substitute $$v_1 = -v_3$$ into expression for $$v_2$$: $$v_2 = 2(-v_3) + v_3 = -2 v_3 + v_3 = -v_3$$ Check third row: $$-(-v_3) - 2(-v_3) - 3 v_3 = v_3 + 2 v_3 - 3 v_3 = 0$$ Consistent. Eigenvector: $$\mathbf{v}_2 = v_3 \begin{bmatrix} -1 \\ -1 \\ 1 \end{bmatrix}$$ Normalize: $$\|\mathbf{v}_2\| = \sqrt{(-1)^2 + (-1)^2 + 1^2} = \sqrt{3}$$ Normalized eigenvector: $$\mathbf{u}_2 = \frac{1}{\sqrt{3}} \begin{bmatrix} -1 \\ -1 \\ 1 \end{bmatrix}$$ 8. For $$\lambda = -1$$: $$(A + I) = \begin{bmatrix} 3 & 1 & -1 \\ 1 & 2 & -2 \\ -1 & -2 & 2 \end{bmatrix}$$ Solve $$(A + I)\mathbf{v} = 0$$. From first row: $$3 v_1 + v_2 - v_3 = 0$$ From second row: $$v_1 + 2 v_2 - 2 v_3 = 0$$ From third row: $$-v_1 - 2 v_2 + 2 v_3 = 0$$ Add second and third rows: $$(v_1 + 2 v_2 - 2 v_3) + (-v_1 - 2 v_2 + 2 v_3) = 0 + 0 = 0$$ This is always true, so two equations are dependent. From first row: $$v_3 = 3 v_1 + v_2$$ Substitute into second row: $$v_1 + 2 v_2 - 2(3 v_1 + v_2) = 0$$ $$v_1 + 2 v_2 - 6 v_1 - 2 v_2 = 0$$ $$-5 v_1 = 0 \Rightarrow v_1 = 0$$ Then from first row: $$v_3 = 3(0) + v_2 = v_2$$ Eigenvector: $$\mathbf{v}_3 = \begin{bmatrix} 0 \\ v_2 \\ v_2 \end{bmatrix} = v_2 \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}$$ Normalize: $$\|\mathbf{v}_3\| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}$$ Normalized eigenvector: $$\mathbf{u}_3 = \frac{1}{\sqrt{2}} \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}$$ 9. **Form orthogonal matrix $$P$$ with columns $$\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3$$:** $$P = \begin{bmatrix} \frac{2}{\sqrt{6}} & \frac{-1}{\sqrt{3}} & 0 \\ \frac{-1}{\sqrt{6}} & \frac{-1}{\sqrt{3}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} \end{bmatrix}$$ 10. **Canonical form:** $$Q(\mathbf{x}) = \mathbf{y}^T D \mathbf{y} = \lambda_1 y_1^2 + \lambda_2 y_2^2 + \lambda_3 y_3^2 = y_1^2 + 4 y_2^2 - y_3^2$$ where $$\mathbf{y} = P^T \mathbf{x}$$. **Final answer:** The quadratic form reduces to canonical form $$y_1^2 + 4 y_2^2 - y_3^2$$ via the orthogonal transformation $$\mathbf{y} = P^T \mathbf{x}$$ with $$P$$ as above.