线性方程组特征向量（不动点）求解：大矩阵高效计算方法问询

云上老周

2026-5-19

Let's tackle this problem head-on—you're absolutely right that structured high-order matrices like this never require tedious row reduction. The key is leveraging their symmetry and special structure to skip the grunt work. Let's dive in.

Understanding the Problem First

First, a non-trivial fixed point means we’re looking for $\vec{x} \neq \vec{0}$ such that $A\vec{x} = \vec{x}$. Substitute the given matrix $A = \frac{1}{5}M$ (where $M$ is your 6x6 matrix), and this simplifies to:
$$\frac{1}{5}M\vec{x} = \vec{x} \implies M\vec{x} = 5\vec{x}$$
So we’re really finding the eigenvectors of $M$ corresponding to the eigenvalue 5—that’s our core fixed point condition.

Breaking Down Matrix $M$'s Structure

Take a close look at $M$: every row has exactly five 1s and one 0, with the 0 shifting left across rows (row 1 has a 0 in column 6, row 2 in column 5, ..., row 6 in column 1). We can rewrite $M$ as a combination of two simple matrices:
$$M = J - K$$
Where:

$J$ is the 6x6 all-ones matrix (every entry is 1),
$K$ is a permutation matrix with 1s only at positions $(1,6), (2,5), (3,4), (6,1), (5,2), (4,3)$—it swaps pairs of opposite rows/columns.

We already know the eigenvalues and eigenvectors for both $J$ and $K$, so we can combine these properties to analyze $M$ without any row reduction.

Finding the Eigenvalue 5 Eigenvectors

Let’s start with the most obvious observation:

The all-ones vector $\vec{1} = [1,1,1,1,1,1]^T$. When we multiply $M$ by $\vec{1}$, each entry is the sum of the corresponding row of $M$, which is 5. So:
$$M\vec{1} = 5\vec{1}$$
That’s our first (and only, as we’ll confirm) linearly independent eigenvector for eigenvalue 5.

To confirm this is the only one:

The all-ones matrix $J$ has eigenvalues 6 (with eigenvector $\vec{1}$) and 0 (a 5-dimensional eigenspace consisting of all vectors whose entries sum to 0).
The permutation matrix $K$ has eigenvalues 1 (a 3-dimensional eigenspace: vectors where $x_1=x_6, x_2=x_5, x_3=x_4$) and -1 (a 3-dimensional eigenspace: vectors where $x_1=-x_6, x_2=-x_5, x_3=-x_4$).

Since $M = J - K$, its eigenvalues are $\lambda_J - \lambda_K$ for corresponding shared eigenvectors:

For $\vec{1}$, $\lambda_J=6$ and $\lambda_K=1$ (since $K\vec{1}=\vec{1}$), so $\lambda_M=6-1=5$.
For all other eigenvectors, $\lambda_J=0$, so $\lambda_M=0-1=-1$ or $0-(-1)=1$—neither equals 5.

This means the eigenspace for $\lambda=5$ is 1-dimensional, so all non-trivial fixed points are scalar multiples of $\vec{1}$.

Why the Shortcut Works (and When to Use It)

The reference solution’s efficiency comes from leaning into structural clues instead of brute-force computation. Here’s when you can apply this kind of approach:

Constant row sums: If every row of your matrix adds up to the same value, the all-ones vector is automatically an eigenvector with that sum as the eigenvalue.
Decomposable into simpler matrices: If your matrix is a combination of known matrices (all-ones, permutation, identity), you can reuse their eigenvalue/eigenvector properties instead of starting from scratch.
Symmetry/permutation structure: Matrices with cyclic, mirrored, or swapped row/column patterns often have eigenvectors that match that symmetry (like equal/opposite paired entries, or all-ones).
Symmetric matrices: These have orthogonal eigenvectors, so once you find one eigenvector, you can restrict your search to vectors orthogonal to it for other solutions.

Final Result

All non-trivial fixed points of the system are scalar multiples of the all-ones vector:
$$\vec{x} = c \begin{bmatrix}1\1\1\1\1\1\end{bmatrix}, \quad c \in \mathbb{R}, c \neq 0$$

内容的提问来源于stack exchange，提问作者Aesir