关于最大特征值大于行列式的2×2整数矩阵计数的技术问询

Let's tackle this problem systematically. We're focusing on 2×2 matrices with integer elements from the set ( {-k, -k+1, \dots, 0, \dots, k-1, k} ), and we need to count how many of these satisfy the condition that their largest eigenvalue is greater than their determinant.

Key Definitions & Setup #

First, let's formalize the matrix:
[ M = \begin{pmatrix} a & b \ c & d \end{pmatrix} ]
where ( a,b,c,d \in {-k, ..., k} ). For this matrix:

• Trace: ( \text{tr}(M) = a + d ) (sum of diagonal elements)
• Determinant: ( \det(M) = ad - bc )
• The eigenvalues satisfy the characteristic equation: ( \lambda^2 - \text{tr}(M)\lambda + \det(M) = 0 )

The largest eigenvalue ( \lambda_{\text{max}} ) comes from solving this quadratic. We'll split our analysis into two cases: real eigenvalues (discriminant non-negative) and complex eigenvalues (discriminant negative).

Case 1: Real Eigenvalues (Discriminant ≥ 0) #

The discriminant of the characteristic equation is ( D = \text{tr}(M)^2 - 4\det(M) ). When ( D \geq 0 ), the eigenvalues are real, and the largest one is:
[ \lambda_{\text{max}} = \frac{\text{tr}(M) + \sqrt{D}}{2} ]

We need ( \lambda_{\text{max}} > \det(M) ). Let's substitute ( t = \text{tr}(M) ) and ( \Delta = \det(M) ) to simplify the inequality:
[ \frac{t + \sqrt{t^2 - 4\Delta}}{2} > \Delta ]

Rearranging and analyzing this inequality leads to two subcases:

1. When ( \Delta \leq t/2 ): The right-hand side of the rearranged inequality is non-positive, and the left-hand side (square root term) is non-negative. The inequality holds automatically (as long as ( D \geq 0 )).
2. When ( \Delta > t/2 ): We can square both sides (since both sides are positive here) and simplify to find the condition ( \Delta(\Delta - t + 1) < 0 ). Combining this with ( \Delta > t/2 ):
   • If ( t \geq 2 ): Valid ( \Delta ) values are integers in ( \lceil t/2 \rceil \leq \Delta \leq t-2 )
   • If ( t \leq 0 ): Valid ( \Delta ) values are integers in ( \lceil t/2 \rceil \leq \Delta \leq -1 )
   • If ( t = 1 ): No valid ( \Delta ) exists here (since ( \Delta ) must be integer, and the inequality leads to a contradiction)

Case 2: Complex Eigenvalues (Discriminant < 0) #

For complex eigenvalues, they come in conjugate pairs with real part ( t/2 ) and modulus ( \sqrt{\Delta} ).

• If we interpret "largest eigenvalue" as the modulus: The condition ( \sqrt{\Delta} > \Delta ) would require ( 0 < \Delta < 1 ), but ( \Delta ) is integer—so no valid matrices here.
• If we interpret it as the real part: The condition ( t/2 > \Delta ) combined with ( D < 0 ) (i.e., ( t^2 < 4\Delta )) leads to a contradiction for integer ( t ) and ( \Delta ).

In short, complex eigenvalue matrices never satisfy the condition.

Consolidated Valid Conditions #

A matrix counts towards our total if and only if:

1. It has real eigenvalues (( t^2 \geq 4\Delta )), AND
2. Either ( \Delta \leq \lfloor t/2 \rfloor ), OR (based on ( t )) the determinant falls into the valid integer ranges we outlined above.

Calculation Approach & Code Example #

To compute the count, we can iterate through all possible matrix elements, or optimize by grouping by trace and determinant. Here's a straightforward Python implementation for small ( k ):

def count_valid_matrices(k):
    count = 0
    element_range = range(-k, k + 1)
    
    # Iterate all possible diagonal elements a, d
    for a in element_range:
        for d in element_range:
            trace = a + d
            det_base = a * d  # det without bc term
            
            # Iterate all off-diagonal pairs b, c
            for b in element_range:
                for c in element_range:
                    det = det_base - b * c
                    discriminant = trace ** 2 - 4 * det
                    
                    # Skip matrices with complex eigenvalues
                    if discriminant < 0:
                        continue
                    
                    # Check if max eigenvalue > determinant
                    if det <= trace / 2:
                        count += 1
                    else:
                        # Check the secondary condition for det > t/2
                        if det * (det - trace + 1) < 0:
                            count += 1
    return count

# Example: Compute for k=1 (3^4=81 total matrices)
print(f"Valid matrices for k=1: {count_valid_matrices(1)}")

For ( k=1 ), this will return the number of valid matrices out of 81 total. For larger ( k ), you could optimize by precomputing the number of (b,c) pairs that produce each possible ( bc ) value, rather than iterating all pairs directly.

内容的提问来源于stack exchange，提问作者dektorpan