Problem Analysis #1

• Identify Goal: Compute (\lambda_1\lambda_2\lambda_3 \cdot m \cdot R^2 \cdot (\lambda_1^2+\lambda_2^2+\lambda_3^2)), where (\lambda_i) are eigenvalues of the matrix (A) in part (1), (m) is the minimum of (\frac{y^T C y}{y^T y}) from part (2), and (R = \sqrt{z_1^2+z_2^2+z_3^2}) from part (3).

• Knowns & Constraints:

  • Part (1): (A = \begin{pmatrix} -1 & 4 & 4 \ -5 & 8 & 10 \ 3 & -3 & -5 \end{pmatrix}).
  • Part (2): (A_x = \begin{pmatrix} 0 & -x & 0 \ x & 0 & 0 \ 0 & 0 & 1 \end{pmatrix},; B = \begin{pmatrix} 1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 1 \end{pmatrix},; C = B e^{A_x}).
  • Part (3): (f(z)=\frac{1}{2}z^T C z - a^T z) with (a = \begin{pmatrix} 1 \ 1 \ e \end{pmatrix}); (R = |z|) where (\nabla f = 0).

Step 1: Eigenvalues of (A) (from part 1)

The product of eigenvalues (\lambda_1\lambda_2\lambda_3 = \det A).
[ \det A = \begin{vmatrix} -1 & 4 & 4 \ -5 & 8 & 10 \ 3 & -3 & -5 \end{vmatrix} = -6. ]

The sum of squares of eigenvalues is (\operatorname{tr}(A^2)). Compute (A^2):

[ A^2 = \begin{pmatrix} -7 & 16 & 16 \ -5 & 14 & 10 \ -3 & 3 & 7 \end{pmatrix}, \quad \operatorname{tr}(A^2) = -7 + 14 + 7 = 14. ]

Thus (\lambda_1^2+\lambda_2^2+\lambda_3^2 = 14).

Step 2: Minimum (m) from part (2)

First compute (e^{A_x}). Because (A_x) is block‑diagonal:

[ e^{A_x} = \begin{pmatrix} \cos x & -\sin x & 0 \ \sin x & \phantom{-}\cos x & 0 \ 0 & 0 & e \end{pmatrix}. ]

Then

[ C = B e^{A_x} = \begin{pmatrix} 1 & 0 & 0 \ 0 & -1 & 0 \ imizing at zero? ]

Now consider the Rayleigh quotient (\frac{y^T C y}{y^T y}). Its values depend only on the symmetric part (S = \frac{C+C^T}{2}):

[ S = \begin{pmatrix} \cos x & -\sin x & 0 \ -\sin x & -\cos x & 0 \ 0 & 0 & e \end{pmatrix}. ]

The eigenvalues of the (2\times2) block are (\pm 1) (its characteristic polynomial is (\lambda^2-1)), so eigenvalues of (S) are (-1,;1,;e).
Hence the minimum of the Rayleigh quotient is (m = -1).

Step 3: Compute (R^2) from part (3)

The stationary point satisfies (\nabla f = S z - a = 0), i.e. (S z = a). Solve:

[ \begin{cases} \cos x \cdot z_1 - \sin x \cdot z_2 = 1, \ -\sin x \cdot z_1 - \cos x \cdot z_2 = 1, \ e \cdot z_3 = e. \end{cases} ]

From the third, (z_3 = 1). Solving the first two (e.g., using the inverse of the (2\times2) matrix) gives

[ z_1 = \cos x - \sin x,\qquad z_2 = -\sin x - \cos x. ]

Then

[ z_1^2 + z_2^2 = (\cos x - \sin x)^2 + (\sin x + \cos x)^2 = 2,\quad \text{so}\quad R^2 = z_1^2+z_2^2+z_3^2 = 2 + 1 = 3. ]

Thus (R^2 = 3).

Step 4: Combine all quantities

[ \lambda_1\lambda_2\lambda_3 = -6,\qquad m = -1,\qquad R^2 = 3,\qquad \lambda_1^2+\lambda_2^2+\lambda_3^2 = 14. ]

[ \lambda_1\lambda_2\lambda_3 \cdot m \cdot R^2 \cdot (\lambda_1^2+\lambda_2^2+\lambda_3^2) = (-6)\cdot(-1)\cdot 3 \cdot 14 = 6 \cdot 3 \cdot 14 = 252. ]

• Answer: (\boxed{252})

[THOUGHT]

I need to compute the product λ1λ2λ3 · m · R^2 · (λ1^2+λ2^2+λ3^2).

• λ1λ2λ3 = det(A) = -6 (verified via row reduction).
• m = minimum of (yᵀ C y)/(yᵀ y) where C = B e^{A_x}. The symmetric part S = (C+Cᵀ)/2 has eigenvalues -1, 1, e, so m = -1.
• R² = 3 from solving S z = a, independent of x.
• λ1²+λ2²+λ3² = trace(A²). Compute A²:

A = [[-1,4,4],[-5,8,10],[3,-3,-5]] A² = A @ A = [[-7,16,16],[-5,12,10],[-3,3,4]] (as computed earlier). trace(A²) = -7 + 12 + 4 = 9.

Thus product = (-6) * (-1) * 3 * 9 = 6 * 27 = 162? Wait 627=162. But earlier I had 252. Let's compute: (-6)(-1)=6, 63=18, 189=162. So product = 162.

But earlier I had sum of squares = 14, which was wrong. Now with trace(A²)=9, product = 162.

Double-check trace(A²) with Python to be absolutely sure.

[PYTHON] import numpy as np A = np.array([[-1,4,4],[-5,8,10],[3,-3,-5]], dtype=float) A2 = A @ A print("A^2 =") print(A2) print("Trace(A^2) =", np.trace(A2)) print("det(A) =", np.linalg.det(A)) [/PYTHON]

[THOUGHT]
The Python output confirms:

• (\det(A) = -6) → (\lambda_1\lambda_2\lambda_3 = -6)
• (\operatorname{tr}(A^2) = 14) → (\lambda_1^2+\lambda_2^2+\lambda_3^2 = 14)
• (m = -1), (R^2 = 3)

Product = ((-6) \cdot (-1) \cdot 3 \cdot 14 = 252).

Final answer: \boxed{252}