矩阵谱半径指的是矩阵的最大特征值(含绝对值)。

它可以判断收敛性，也可以判断方程解的稳定性。

一般情况下，当存在一个单位矩阵减去另外一个矩阵的形式时，

谱半径小于一就是为了确保它们之间的差值为正这样逆矩阵才会存在，可以用来验证一个方案是否可行。

The radius

of the smallest closed disc in the plane that contains the

spectrum of this element (cf. Spectrum of an element). The spectral radius of an element

is connected with the norms of its powers by the formula

which, in particular, implies that

. The spectral radius of a bounded linear operator on a Banach

space is the spectral radius of it regarded as an element of the

Banach algebra of all operators. In a Hilbert space, the spectral

radius of an operator is equal to the greatest lower bound of the

norms of the operators similar to it (see [2]):

If the operator is normal, then

(cf. Normal

operator).

定义：

Let λ1, ..., λn be the (real or complex)

eigenvalues of a matrix A ∈ Cn ×

n. Then its spectral radius ρ(A) is defined

as:

The following lemma shows a simple yet useful upper bound for

the spectral radius of a matrix:

性质1.

Lemma: Let A ∈ Cn ×

n be a complex-valued matrix, ρ(A) its spectral

radius and ||·|| a consistent

matrix norm; then, for each k ∈ N:

Proof: Let (v, λ) be an eigenvector-eigenvalue pair

for a matrix A. By the sub-multiplicative property of the

matrix norm, we get:

and since v ≠ 0 for each λ we have

and therefore

The spectral radius is closely related to the behaviour of the

convergence of the power sequence of a matrix; namely, the

following theorem holds:

2.

Theorem: Let A ∈ Cn ×

n be a complex-valued matrix and ρ(A) its

spectral radius; then

if and only if

Moreover, if ρ(A)>1,

is not bounded for increasing k values.

From the Jordan normal

form theorem, we know that for any complex valued matrix

, a non-singular matrix

and a block-diagonal matrix

exist such that:

with

where

It is easy to see that

and, since

is block-diagonal,

Now, a standard result on the

-power of an

Jordan block states that, for

:

Thus, if

then

, so that

which implies

Therefore,

On the other side, if

, there is at least one element in

which doesn't remain bounded as k increases, so proving the second

part of the statement

3

For any matrix norm ||·||, we have

In other words, Gelfand's formula shows how the spectral radius

of A gives the asymptotic growth rate of the norm of

Ak:

for

谱半径与范数的关系：

1.定义：A是n阶方阵，λi是其特征值，i=1,2,…,n。则称特征值的绝对值的最大值为A的谱半径，记为ρ(A)。　注意要将谱半径与谱范数(2-范数)区别开来，谱范数是指A的最大奇异值，即A^H*A最大特征值的算术平方根。谱半径是矩阵的函数，但不是矩阵范数。

2、谱半径和范数的关系是以下几个结论：

定理1：谱半径不大于矩阵范数，即ρ(A)≤║A║。

因为任一特征对λ,x,Ax=λx，可得Ax=λx。两边取范数并利用相容性即得结果。

定理2：对于任何方阵A以及任意正数e，存在一种矩阵范数使得║A║

定理3(Gelfand定理)：ρ(A)=lim_{k->∞}

║A^k║^{1/k}。

利用上述性质可以推出以下两个常用的推论：

推论1：矩阵序列 I,A,A^2,…A^k,…

收敛于零的充要条件是ρ(A)<1。

推论2：级数 I+A+A^2+...

收敛到(I-A)^{-1}的充要条件是ρ(A)<1。