We are given a $3 \times 3$ real symmetric matrix $A$ with eigenvalues $\lambda_1 = 2$, $\lambda_2 = 3$, and $\lambda_3 = 5$. The quadratic form associated with $A$ is defined as:

$Q(x) = x^T A x$

where $x$ is a vector in $\mathbb{R}^3$. Additionally, we know that:

• $\|x\| = 1$, meaning $x$ is a unit vector,
• $x_1 + x_2 + x_3 = 0$, which is an additional constraint on $x$.

Step 1: Understanding the quadratic form and the eigenvalues of $A$

Since $A$ is a real symmetric matrix, we can diagonalize it. The quadratic form $Q(x)$ can be expressed as:

$Q(x) = x^T A x = \sum_{i=1}^{3} \lambda_i (v_i^T x)^2$

where $\lambda_1, \lambda_2, \lambda_3$ are the eigenvalues of $A$, and $v_1, v_2, v_3$ are the corresponding orthonormal eigenvectors of $A$. Since $A$ is symmetric, it has an orthonormal basis of eigenvectors.

Step 2: Analyzing the constraint $x_1 + x_2 + x_3 = 0$

The condition $x_1 + x_2 + x_3 = 0$ suggests that $x$ lies in a subspace orthogonal to the vector $\mathbf{1} = (1, 1, 1)^T$. To proceed, we will find the vector $\mathbf{1}$ and project the space onto the subspace orthogonal to it.

Eigenvalue decomposition and subspaces

• The matrix $A$ has eigenvalues $\lambda_1 = 2$, $\lambda_2 = 3$, and $\lambda_3 = 5$, with their corresponding eigenvectors.
• The condition $x_1 + x_2 + x_3 = 0$ restricts $x$ to the plane orthogonal to the vector $(1, 1, 1)^T$, which has an eigenvalue $\lambda_3 = 5$.

The maximum value of $Q(x)$ occurs when $x$ aligns with the eigenvector corresponding to the largest eigenvalue, subject to the constraint $x_1 + x_2 + x_3 = 0$. Thus, we need to find the maximum value of $Q(x)$ when $x$ is a unit vector and $x$ lies in the plane orthogonal to $(1, 1, 1)^T$.

Step 3: The maximum value of $Q(x)$

Given the eigenvalues $\lambda_1 = 2$, $\lambda_2 = 3$, and $\lambda_3 = 5$, the maximum value of $Q(x)$ will be achieved by the eigenvalue corresponding to the largest eigenvalue in the subspace orthogonal to $(1, 1, 1)^T$.

The eigenvalue $\lambda_3 = 5$ corresponds to the direction along $(1, 1, 1)^T$, but since we are restricting $x$ to the plane orthogonal to this vector, the largest eigenvalue in the remaining space is $\lambda_2 = 3$.

Thus, the maximum value of $Q(x)$ occurs when $x$ is aligned with the eigenvector corresponding to $\lambda_2 = 3$, and the value is:

$Q(x) = 3$

Final Answer:

The maximum value of $Q(x)$ is $\boxed{3}$.

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Would you like further clarification or details on any part of the solution? Here are some related questions:

1. How does the condition $x_1 + x_2 + x_3 = 0$ affect the eigenvectors of $A$?
2. How would the solution change if $A$ had different eigenvalues?
3. What is the significance of the matrix being symmetric in this problem?
4. How can we generalize the method to higher-dimensional spaces?
5. How would we solve for $Q(x)$ if the matrix $A$ were not diagonalizable?

Tip: For any real symmetric matrix, the quadratic form is always maximized or minimized along its eigenvectors.