1. **Problem Statement:** (a) Check if the matrix $$A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & -1 \\ 1 & -1 & -1 \end{bmatrix}$$ is diagonalizable. If yes, find diagonal matrix $$D$$ and invertible matrix $$P$$ such that $$P^{-1}AP = D$$. (b) Given the linear functional $$f: \mathbb{C}^3 \to \mathbb{C}$$ defined by $$f((z_1,z_2,z_3)) = z_1 - i z_2 + i z_3$$, find $$(y_1,y_2,y_3) \in \mathbb{C}^3$$ such that $$f((z_1,z_2,z_3)) = \langle (z_1,z_2,z_3), (y_1,y_2,y_3) \rangle$$. --- 2. **Diagonalizability Check:** We find eigenvalues by solving $$\det(A - \lambda I) = 0$$. $$A - \lambda I = \begin{bmatrix} 1-\lambda & 1 & 1 \\ 1 & 1-\lambda & -1 \\ 1 & -1 & -1-\lambda \end{bmatrix}$$ Calculate determinant: $$\det(A - \lambda I) = (1-\lambda) \begin{vmatrix} 1-\lambda & -1 \\ -1 & -1-\lambda \end{vmatrix} - 1 \begin{vmatrix} 1 & -1 \\ 1 & -1-\lambda \end{vmatrix} + 1 \begin{vmatrix} 1 & 1-\lambda \\ 1 & -1 \end{vmatrix}$$ Calculate minors: $$= (1-\lambda)((1-\lambda)(-1-\lambda) - (-1)(-1)) - 1(1(-1-\lambda) - 1(-1)) + 1(1(-1) - 1(1-\lambda))$$ Simplify: $$= (1-\lambda)((1-\lambda)(-1-\lambda) - 1) - 1(-1-\lambda + 1) + 1(-1 - (1-\lambda))$$ $$= (1-\lambda)((1-\lambda)(-1-\lambda) - 1) - 1(-\lambda) + 1(-1 - 1 + \lambda)$$ $$= (1-\lambda)((1-\lambda)(-1-\lambda) - 1) + \lambda + (\lambda - 2)$$ Calculate $$(1-\lambda)(-1-\lambda) = -1 - \lambda + \lambda + \lambda^2 = -1 + \lambda^2$$ So: $$= (1-\lambda)(-1 + \lambda^2 - 1) + \lambda + \lambda - 2 = (1-\lambda)(\lambda^2 - 2) + 2\lambda - 2$$ Expand: $$= (1)(\lambda^2 - 2) - \lambda(\lambda^2 - 2) + 2\lambda - 2 = \lambda^2 - 2 - \lambda^3 + 2\lambda + 2\lambda - 2$$ Simplify: $$= -\lambda^3 + \lambda^2 + 4\lambda - 4$$ Set equal to zero: $$-\lambda^3 + \lambda^2 + 4\lambda - 4 = 0$$ Multiply both sides by -1: $$\lambda^3 - \lambda^2 - 4\lambda + 4 = 0$$ Try rational roots: possible roots are $$\pm1, \pm2, \pm4$$. Test $$\lambda=1$$: $$1 - 1 - 4 + 4 = 0$$ so $$\lambda=1$$ is a root. Divide polynomial by $$(\lambda - 1)$$: $$\lambda^3 - \lambda^2 - 4\lambda + 4 = (\lambda - 1)(\lambda^2 - 4)$$ Factor further: $$\lambda^2 - 4 = (\lambda - 2)(\lambda + 2)$$ Eigenvalues are $$\lambda = 1, 2, -2$$. 3. **Find eigenvectors:** For $$\lambda=1$$: $$(A - I)\mathbf{v} = 0$$ $$\begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & -1 \\ 1 & -1 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$ From first row: $$y + z = 0 \Rightarrow z = -y$$ Second row: $$x - z = 0 \Rightarrow x = z = -y$$ Third row: $$x - y - 2z = 0 \Rightarrow -y - y - 2(-y) = 0 \Rightarrow -y - y + 2y = 0$$ which is true. Eigenvector for $$\lambda=1$$ is $$\mathbf{v}_1 = (-y, y, -y) = y(-1,1,-1)$$. For $$\lambda=2$$: $$(A - 2I)\mathbf{v} = 0$$ $$\begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & -1 \\ 1 & -1 & -3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = 0$$ From first row: $$-x + y + z = 0$$ Second row: $$x - y - z = 0$$ Add first and second rows: $$(-x + y + z) + (x - y - z) = 0 \Rightarrow 0 = 0$$ From first row: $$-x + y + z = 0 \Rightarrow y + z = x$$ Third row: $$x - y - 3z = 0$$ Substitute $$x = y + z$$: $$(y + z) - y - 3z = 0 \Rightarrow z - 3z = 0 \Rightarrow -2z = 0 \Rightarrow z = 0$$ Then $$x = y + 0 = y$$. Eigenvector is $$\mathbf{v}_2 = (y, y, 0) = y(1,1,0)$$. For $$\lambda = -2$$: $$(A + 2I)\mathbf{v} = 0$$ $$\begin{bmatrix} 3 & 1 & 1 \\ 1 & 3 & -1 \\ 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = 0$$ From first row: $$3x + y + z = 0$$ Second row: $$x + 3y - z = 0$$ Third row: $$x - y + z = 0$$ Add first and second rows: $$(3x + y + z) + (x + 3y - z) = 0 \Rightarrow 4x + 4y = 0 \Rightarrow x + y = 0 \Rightarrow y = -x$$ From third row: $$x - y + z = 0 \Rightarrow x - (-x) + z = 0 \Rightarrow 2x + z = 0 \Rightarrow z = -2x$$ Eigenvector is $$\mathbf{v}_3 = (x, -x, -2x) = x(1, -1, -2)$$. 4. **Construct matrices $$P$$ and $$D$$:** $$P = \begin{bmatrix} -1 & 1 & 1 \\ 1 & 1 & -1 \\ -1 & 0 & -2 \end{bmatrix}$$ (columns are eigenvectors) $$D = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -2 \end{bmatrix}$$ 5. **Linear functional representation:** Given $$f((z_1,z_2,z_3)) = z_1 - i z_2 + i z_3$$ and inner product $$\langle (z_1,z_2,z_3), (y_1,y_2,y_3) \rangle = z_1 \overline{y_1} + z_2 \overline{y_2} + z_3 \overline{y_3}$$. We want: $$z_1 - i z_2 + i z_3 = z_1 \overline{y_1} + z_2 \overline{y_2} + z_3 \overline{y_3}$$ Equate coefficients: $$\overline{y_1} = 1, \quad \overline{y_2} = -i, \quad \overline{y_3} = i$$ Take conjugates: $$y_1 = 1, \quad y_2 = i, \quad y_3 = -i$$ **Final answer:** (a) Matrix $$A$$ is diagonalizable with $$P = \begin{bmatrix} -1 & 1 & 1 \\ 1 & 1 & -1 \\ -1 & 0 & -2 \end{bmatrix}, \quad D = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -2 \end{bmatrix}$$ (b) The vector $$\mathbf{y} = (1, i, -i)$$ satisfies $$f((z_1,z_2,z_3)) = \langle (z_1,z_2,z_3), \mathbf{y} \rangle$$.