1. **Problem Statement:** Find matrix $A$ for the quadratic form $2xy - 2yz + 2zx$. Then find eigenvalues and normalized eigenvectors. Finally, diagonalize matrix $A$, convert it to its canonical form, and determine the index, rank, signature, and nature of $A$. 2. **Form the matrix $A$: ** The quadratic form is $Q = 2xy - 2yz + 2zx$. For variables $(x,y,z)$, the quadratic form can be written as $Q = \begin{bmatrix} x & y & z \end{bmatrix} A \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ where $A$ is symmetric. Since $Q$ has terms: - $2xy$ means $A_{12} + A_{21} = 2$ - $-2yz$ means $A_{23} + A_{32} = -2$ - $2zx$ means $A_{13} + A_{31} = 2$ Diagonal entries are 0 because no $x^2, y^2,$ or $z^2$ terms. Thus, $$ A = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & -1 \\ 1 & -1 & 0 \end{bmatrix} $$ 3. **Find eigenvalues:** Solve characteristic polynomial $\det(A - \lambda I) = 0$. $$ A - \lambda I = \begin{bmatrix} -\lambda & 1 & 1 \\ 1 & -\lambda & -1 \\ 1 & -1 & -\lambda \end{bmatrix} $$ Calculate determinant: $$ \det(A - \lambda I) = -\lambda \begin{vmatrix} -\lambda & -1 \\ -1 & -\lambda \end{vmatrix} - 1 \begin{vmatrix} 1 & -1 \\ 1 & -\lambda \end{vmatrix} + 1 \begin{vmatrix} 1 & -\lambda \\ 1 & -1 \end{vmatrix} $$ Compute minors: $$ = -\lambda (\lambda^2 - 1) - 1( -\lambda - (-1)) + 1( -1 - (-\lambda)) $$ $$ = -\lambda (\lambda^2 - 1) - ( -\lambda + 1) + ( -1 + \lambda ) $$ $$ = -\lambda^3 + \lambda - ( -\lambda + 1) + ( -1 + \lambda ) $$ $$ = -\lambda^3 + \lambda + \lambda - 1 - 1 + \lambda $$ $$ = -\lambda^3 + 3 \lambda - 2 $$ Solve $-\lambda^3 + 3\lambda - 2 = 0$ or equivalently: $$ \lambda^3 - 3\lambda + 2 = 0 $$ Trial roots: $\lambda=1:$ $$ 1 - 3 + 2 = 0 $$ so $\lambda=1$ is root. Divide polynomial by $(\lambda -1)$: $$ \lambda^3 - 3\lambda + 2 = (\lambda -1)(\lambda^2 + \lambda -2) $$ Solve quadratic: $$ \lambda^2 + \lambda -2 = 0 $$ $$ \lambda = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm 3}{2} $$ So, $$ \lambda = 1, \quad \lambda = 1, \quad \lambda = -2 $$ Eigenvalues are $\boxed{\lambda_1 = 1, \lambda_2 = 1, \lambda_3 = -2}$. 4. **Find normalized eigenvectors:** - For $\lambda=1$, solve $(A - I)\mathbf{v} = \mathbf{0}$. $$ A - I = \begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & -1 \\ 1 & -1 & -1 \end{bmatrix} $$ From row 1: $-v_1 + v_2 + v_3 =0$ or $v_1 = v_2 + v_3$ From row 2: $v_1 - v_2 - v_3 =0$ same as above. Only one independent equation, so eigenvectors form two-dimensional subspace. Choose $v_2 = s$, $v_3 = t$, then $v_1 = s + t$. Two basis eigenvectors: For $(s,t)=(1,0)$: $\mathbf{v}_1 = (1,1,0)$ For $(s,t)=(0,1)$: $\mathbf{v}_2 = (1,0,1)$ Normalize: $$ |\mathbf{v}_1| = \sqrt{1^2 + 1^2 + 0} = \sqrt{2} $$ $$ \mathbf{u}_1 = \frac{1}{\sqrt{2}}(1,1,0) $$ $$ |\mathbf{v}_2| = \sqrt{1^2 + 0 + 1^2} = \sqrt{2} $$ $$ \mathbf{u}_2 = \frac{1}{\sqrt{2}}(1,0,1) $$ - For $\lambda = -2$, solve $(A + 2I)\mathbf{v} = 0$: $$ A + 2I = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix} $$ Solve system: Row 1: $2v_1 + v_2 + v_3=0$ Row 2: $v_1 + 2 v_2 - v_3 =0$ Row 3: $v_1 - v_2 + 2 v_3=0$ Add Row2 + Row3: $$ (v_1 + 2v_2 - v_3) + (v_1 - v_2 + 2v_3) = 0 $$ $$ 2v_1 + v_2 + v_3= 0$$ same as Row1. From Row1: $$ 2v_1 + v_2 + v_3=0 $$ From Row2: $$ v_1 + 2v_2 - v_3=0 $$ Add Row1 and Row2: $$ (2v_1 + v_2 + v_3) + (v_1 + 2v_2 - v_3) = 3v_1 + 3 v_2 = 0 $$ $$ v_1 + v_2 =0 \implies v_2 = -v_1 $$ From Row1: $$ 2v_1 + (-v_1) + v_3= 0 \implies v_1 + v_3=0 \implies v_3= -v_1 $$ Eigenvector: $$ \mathbf{v}_3 = (v_1, -v_1, -v_1) = v_1 (1,-1,-1) $$ Normalize: $$ |\mathbf{v}_3| = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{3} $$ $$ \mathbf{u}_3 = \frac{1}{\sqrt{3}} (1,-1,-1) $$ 5. **Diagonalize $A$:** Form matrix $P = [\mathbf{u}_1 \ \mathbf{u}_2 \ \mathbf{u}_3]$: $$ P = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{3}} \\ 0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{3}} \end{bmatrix} $$ Then, $$ D = P^{-1} A P = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -2 \end{bmatrix} $$ 6. **Canonical form:** The canonical quadratic form is $$ Q = y_1^2 + y_2^2 - 2 y_3^2 $$ where $\mathbf{y} = P^{-1} \mathbf{x}$. 7. **Index, rank, signature, nature:** - Rank = number of nonzero eigenvalues = 3. - Signature = (number positive eigenvalues, number negative eigenvalues, zero eigenvalues) = (2,1,0). - Index = number of positive eigenvalues = 2. - Since 1 negative eigenvalue, matrix is indefinite. **Final answers:** $$ A = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & -1 \\ 1 & -1 & 0 \end{bmatrix}, \quad \text{Eigenvalues } = \{1,1,-2\} $$ Normalized eigenvectors: $$ \mathbf{u}_1 = \frac{1}{\sqrt{2}}(1,1,0), \quad \mathbf{u}_2 = \frac{1}{\sqrt{2}}(1,0,1), \quad \mathbf{u}_3 = \frac{1}{\sqrt{3}}(1,-1,-1) $$ Diagonal form: $$ D = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -2 \end{bmatrix} $$ Index = 2, Rank = 3, Signature = (2,1,0), Nature = indefinite matrix.