全为正数的矩阵

hbghlyj

math19b_2011 lecture34.pdf

Perron Frobenius theorem: If all entries of a $n \times n$ matrix $A$ are positive, then it has a unique maximal eigenvalue. Its eigenvector has positive entries.

在Perron Frobenius theorem的证明中用到了$T$是压缩映射,但是没有证明$T$是压缩映射,只证明了$T$是$X$到$X$的映射

The matrix $A$ defines a map $T (v) = Av/|Av|$ on $X$ because the entries of the matrix are nonnegative. Because they are positive, $TX$ is contained in the interior of $X$. This map is a contraction, there exists $0 < k < 1$ such that $d(T x, T y) ≤ kd(x, y)$ where $d$ is the geodesic sphere distance.

hbghlyj

今天在people.maths.ox.ac.uk/ritter/algebraic-topology/sheet0.pdf 问题 3 中看到了类似的问题：
Recall the Brouwer fixed point theorem:
Let $\mathbb{D}^n=\left\{x \in \mathbb{R}^n:\|x\| \leqslant 1\right\}, f: \mathbb{D}^n \rightarrow \mathbb{D}^n$ continuous $\Rightarrow \exists p \in \mathbb{D}^n, f(p)=p$.

Use this to prove that any real $n \times n$ matrix with (strictly) positive entries has a real eigenvalue $\lambda>0$ with eigenvector $\left(v_1, \dots, v_n\right)$ st. $v_i \geqslant 0 \forall i$.

hbghlyj

Brouwer不动点定理和Banach不动点定理的区别在于Banach不动点定理保证唯一性，而Brouwer不动点定理不保证唯一性。

对于这个问题，不能使用 Banach 不动点定理，因为映射 $T$ 不一定是压缩！数值反例参阅此帖子。所以 1# 证明是不正确的！事实上，此帖子的作者在同一个 PDF 笔记上也有同样的疑问，并且也怀疑映射 $T$ 并不总是压缩！

hbghlyj

Czhang271828 发表于 2023-6-10 09:57
$S$ 是单位球面, $\Pi$ 是第一卦限, 那么 $T$ 将 $\overline{S\cap \Pi}$ 映射至其内部. 由于 $\overline{S\cap \Pi}$ 是紧集, 从而...

注意Brouwer不动点定理只给出存在性，而不给出唯一性。
@Czhang271828 只证明了 Perron-Frobenius 定理的一部分。
那么如何证明唯一性来完成证明？

hbghlyj

hbghlyj 发表于 2024-10-24 09:33
那么如何证明唯一性来完成证明？

Wikipedia

Multiplicity one

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This section proves that the Perron–Frobenius eigenvalue is a simple root of the characteristic polynomial of the matrix. Hence the eigenspace associated to Perron–Frobenius eigenvalue r is one-dimensional. The arguments here are close to those in Meyer.^[12]

Given a strictly positive eigenvector v corresponding to r and another eigenvector w with the same eigenvalue. (The vectors v and w can be chosen to be real, because A and r are both real, so the null space of A-r has a basis consisting of real vectors.) Assuming at least one of the components of w is positive (otherwise multiply w by −1). Given maximal possible α such that u=v- α w is non-negative, then one of the components of u is zero, otherwise α is not maximum. Vector u is an eigenvector. It is non-negative, hence by the lemma described in the previous section non-negativity implies strict positivity for any eigenvector. On the other hand, as above at least one component of u is zero. The contradiction implies that w does not exist.

Case: There are no Jordan cells corresponding to the Perron–Frobenius eigenvalue r and all other eigenvalues which have the same absolute value.

If there is a Jordan cell, then the infinity norm (A/r)^k_∞ tends to infinity for k → ∞ , but that contradicts the existence of the positive eigenvector.

Given r = 1, or A/r. Letting v be a Perron–Frobenius strictly positive eigenvector, so Av=v, then:

$ \|v\|_{\infty }=\|A^{k}v\|_{\infty }\geq \|A^{k}\|_{\infty }\min _{i}(v_{i}),~~\Rightarrow ~~\|A^{k}\|_{\infty }\leq \|v\|/\min _{i}(v_{i}) $ So A^k_∞ is bounded for all k. This gives another proof that there are no eigenvalues which have greater absolute value than Perron–Frobenius one. It also contradicts the existence of the Jordan cell for any eigenvalue which has absolute value equal to 1 (in particular for the Perron–Frobenius one), because existence of the Jordan cell implies that A^k_∞ is unbounded. For a two by two matrix:

$ J^{k}={\begin{pmatrix}\lambda &1\\0&\lambda \end{pmatrix}}^{k}={\begin{pmatrix}\lambda ^{k}&k\lambda ^{k-1}\\0&\lambda ^{k}\end{pmatrix}}, $

hence J^k_∞ = |k + λ| (for |λ| = 1), so it tends to infinity when k does so. Since J^k = C⁻¹ A^kC, then A^k ≥ J^k/ (C⁻¹ C ), so it also tends to infinity. The resulting contradiction implies that there are no Jordan cells for the corresponding eigenvalues.

Combining the two claims above reveals that the Perron–Frobenius eigenvalue r is simple root of the characteristic polynomial. In the case of nonprimitive matrices, there exist other eigenvalues which have the same absolute value as r. The same claim is true for them, but requires more work.

hbghlyj

Definition:
Primitive matrix:
A nonnegative square matrix 𝐴 where 𝐴^𝑘 is positive for some positive integer 𝑘.

Czhang271828

hbghlyj 发表于 2024-10-24 17:33
注意Brouwer不动点定理只给出存在性，而不给出唯一性。
@Czhang271828 只证明了 Perron-Frobenius 定理的一 ...

最大特征值肯定唯一. 如果有两个线性无关的特征向量, 其线性组合也是最大特征值的特征向量. 存在某个线性组合, 使得特征向量同时有严格正和 $\leq 0$ 的项. 矛盾.

hbghlyj

Czhang271828 发表于 2024-10-24 14:57
使得特征向量同时有严格正和 $\leq 0$ 的项. 矛盾.

要得出矛盾需要一个引理：

Lemma: given a positive (or more generally irreducible non-negative) matrix A and v as any non-negative eigenvector for A, then it is necessarily strictly positive and the corresponding eigenvalue is also strictly positive.

Proof. One of the definitions of irreducibility for non-negative matrices is that for all indexes i,j there exists m, such that (A^m)_ij is strictly positive. Given a non-negative eigenvector v, and that at least one of its components say j-th is strictly positive, the corresponding eigenvalue is strictly positive, indeed, given n such that (Aⁿ)_ii >0, hence: rⁿv_i = Aⁿv_i ≥ (Aⁿ)_iiv_i >0. Hence r is strictly positive. The eigenvector is strict positivity. Then given m, such that (A^m)_ij >0, hence: r^mv_j = (A^mv)_j ≥ (A^m)_ijv_i >0, hence v_j is strictly positive, i.e., the eigenvector is strictly positive.