# Spectral radius

In mathematics, the spectral radius of a matrix or a bounded linear operator is the supremum among the absolute values of the elements in its spectrum, which is sometimes denoted by &rho;(&middot;).

pectral radius of a matrix

Let &lambda;1, &hellip;, &lambda;"s" be the (real or complex) eigenvalues of a matrix "A" &isin; C"n" &times; "n". Then its spectral radius &rho;("A") is defined as:

:$ho\left(A\right) := max_i\left(|lambda_i|\right)$

The following lemma shows a simple yet useful upper bound for the spectral radius of a matrix:

Lemma: Let "A" &isin; C"n" &times; "n" be a complex-valued matrix, &rho;("A") its spectral radius and ||&middot;|| a consistent matrix norm; then, for each "k" &isin; N:

:$ho\left(A\right)leq |A^k|^\left\{1/k\right\}, forall k in mathbb\left\{N\right\}.$

"Proof": Let (v, &lambda;) be an eigenvector-eigenvalue pair for a matrix "A". By the sub-multiplicative property of the matrix norm, we get:

::$|lambda|^k|mathbf\left\{v\right\}| = |lambda^k mathbf\left\{v\right\}| = |A^k mathbf\left\{v\right\}| leq |A^k|cdot|mathbf\left\{v\right\}|$

:and since v &ne; 0 for each &lambda; we have

:$|lambda|^kleq |A^k|$

:and therefore

:$ho\left(A\right)leq |A^k|^\left\{1/k\right\},,square$

The spectral radius is closely related to the behaviour of the convergence of the power sequence of a matrix; namely, the following theorem holds:

Theorem: Let "A" &isin; C"n" &times; "n" be a complex-valued matrix and &rho;("A") its spectral radius; then

:$lim_\left\{k o infty\right\}A^k=0$ if and only if $ho\left(A\right)<1.$

Moreover, if &rho;("A")>1, $|A^k|$ is not bounded for increasing k values.

"Proof":

($lim_\left\{k o infty\right\}A^k = 0 Rightarrow ho\left(A\right) < 1$)

:Let (v, &lambda;) be an eigenvector-eigenvalue pair for matrix "A". Since

::$A^kmathbf\left\{v\right\} = lambda^kmathbf\left\{v\right\},$

:we have:

::

:and, since by hypothesis v &ne; 0, we must have

::$lim_\left\{k o infty\right\}lambda^k = 0$

:which implies |&lambda;| < 1. Since this must be true for any eigenvalue &lambda;, we can conclude &rho;("A") < 1.

($ho\left(A\right)<1 Rightarrow lim_\left\{k o infty\right\}A^k = 0$)

:From the Jordan Normal Form Theorem, we know that for any complex valued matrix "A" &isin; C"n" &times; "n", a non-singular matrix "V" &isin; C"n" &times; "n" and a block-diagonal matrix "J" &isin; C"n" &times; "n" exist such that:

::$A = VJV^\left\{-1\right\}$

:with

::

:where

::

:It is easy to see that

::$A^k=VJ^kV^\left\{-1\right\}$

:and, since "J" is block-diagonal,

::

:Now, a standard result on the "k"-power of an "m""i" &times; "m""i" Jordan block states that, for "k" &ge; "m""i" − 1:

::

:Thus, if &rho;("A") < 1 then |&lambda;"i"| < 1 &forall; "i", so that

::$lim_\left\{k o infty\right\}J_\left\{m_i\right\}^k=0 forall i$

:which implies

::$lim_\left\{k o infty\right\}J^k = 0.$

:Therefore,

::$lim_\left\{k o infty\right\}A^k=lim_\left\{k o infty\right\}VJ^kV^\left\{-1\right\}=V\left(lim_\left\{k o infty\right\}J^k\right)V^\left\{-1\right\}=0$

On the other side, if &rho;("A")>1, there is at least one element in "J" which doesn't remain bounded as k increases, so proving the second part of the statement.

::$square$

Theorem (Gelfand's formula, 1941)

For any matrix norm ||&middot;||, we have

:$ho\left(A\right)=lim_\left\{k o infty\right\}||A^k||^\left\{1/k\right\}.$

In other words, the Gelfand's formula shows how the spectral radius of "A" gives the asymptotic growth rate of the norm of "A""k":

:$|A^k|sim ho\left(A\right)^k$ for $k ightarrow infty.,$

"Proof": For any &epsilon; > 0, consider the matrix

::$ilde\left\{A\right\}=\left( ho\left(A\right)+epsilon\right)^\left\{-1\right\}A.$

:Then, obviously,

::$ho\left( ilde\left\{A\right\}\right) = frac\left\{ ho\left(A\right)\right\}\left\{ ho\left(A\right)+epsilon\right\} < 1$

:and, by the previous theorem,

::$lim_\left\{k o infty\right\} ilde\left\{A\right\}^k=0.$

:That means, by the sequence limit definition, a natural number "N1" &isin; N exists such that

::$forall kgeq N_1 Rightarrow | ilde\left\{A\right\}^k| < 1$

:which in turn means:

::$forall kgeq N_1 Rightarrow |A^k| < \left( ho\left(A\right)+epsilon\right)^k$

:or

::$forall kgeq N_1 Rightarrow |A^k|^\left\{1/k\right\} < \left( ho\left(A\right)+epsilon\right).$

:Let's now consider the matrix

::$check\left\{A\right\}=\left( ho\left(A\right)-epsilon\right)^\left\{-1\right\}A.$

:Then, obviously,

::$ho\left(check\left\{A\right\}\right) = frac\left\{ ho\left(A\right)\right\}\left\{ ho\left(A\right)-epsilon\right\} > 1$

:and so, by the previous theorem,$|check\left\{A\right\}^k|$ is not bounded.

:This means a natural number "N2" &isin; N exists such that

::$forall kgeq N_2 Rightarrow |check\left\{A\right\}^k| > 1$

:which in turn means:

::$forall kgeq N_2 Rightarrow |A^k| > \left( ho\left(A\right)-epsilon\right)^k$

:or

::$forall kgeq N_2 Rightarrow |A^k|^\left\{1/k\right\} > \left( ho\left(A\right)-epsilon\right).$

:Taking

::$N:=max\left(N_1,N_2\right)$

:and putting it all together, we obtain:

::$forall epsilon>0, exists Ninmathbb\left\{N\right\}: forall kgeq N Rightarrow ho\left(A\right)-epsilon < |A^k|^\left\{1/k\right\} < ho\left(A\right)+epsilon$

:which, by definition, is

::$lim_\left\{k o infty\right\}|A^k|^\left\{1/k\right\} = ho\left(A\right).,,square$

Gelfand's formula leads directly to a bound on the spectral radius of a product of finitely many matrices, namely assuming that they all commute we obtain$ho\left(A_1 A_2 ldots A_n\right) leq ho\left(A_1\right) ho\left(A_2\right)ldots ho\left(A_n\right).$

Actually, in case the norm is consistent, the proof shows more than the thesis; in fact, using the previous lemma, we can replace in the limit definition the left lower bound with the spectral radius itself and write more precisely:

::$forall epsilon>0, exists Ninmathbb\left\{N\right\}: forall kgeq N Rightarrow ho\left(A\right) leq |A^k|^\left\{1/k\right\} < ho\left(A\right)+epsilon$

:which, by definition, is

:$lim_\left\{k o infty\right\}|A^k|^\left\{1/k\right\} = ho\left(A\right)^+.$

Example: Let's consider the matrix:

whose eigenvalues are 5, 10, 10; by definition, its spectral radius is &rho;("A")=10. In the following table, the values of $|A^k|^\left\{1/k\right\}$ for the four most used norms are listed versus several increasing values of k (note that, due to the particular form of this matrix,$|.|_1=|.|_infty$):

 k $|.|_1=|.|_infty$ $|.|_F$ $|.|_2$ 1 14 15.362291496 10.681145748 2 12.649110641 12.328294348 10.595665162 3 11.934831919 11.532450664 10.500980846 4 11.501633169 11.151002986 10.418165779 5 11.216043151 10.921242235 10.351918183 $vdots$ $vdots$ $vdots$ $vdots$ 10 10.604944422 10.455910430 10.183690042 11 10.548677680 10.413702213 10.166990229 12 10.501921835 10.378620930 10.153031596 $vdots$ $vdots$ $vdots$ $vdots$ 20 10.298254399 10.225504447 10.091577411 30 10.197860892 10.149776921 10.060958900 40 10.148031640 10.112123681 10.045684426 50 10.118251035 10.089598820 10.036530875 $vdots$ $vdots$ $vdots$ $vdots$ 100 10.058951752 10.044699508 10.018248786 200 10.029432562 10.022324834 10.009120234 300 10.019612095 10.014877690 10.006079232 400 10.014705469 10.011156194 10.004559078 $vdots$ $vdots$ $vdots$ $vdots$ 1000 10.005879594 10.004460985 10.001823382 2000 10.002939365 10.002230244 10.000911649 3000 10.001959481 10.001486774 10.000607757 $vdots$ $vdots$ $vdots$ $vdots$ 10000 10.000587804 10.000446009 10.000182323 20000 10.000293898 10.000223002 10.000091161 30000 10.000195931 10.000148667 10.000060774 $vdots$ $vdots$ $vdots$ $vdots$ 100000 10.000058779 10.000044600 10.000018232

pectral radius of a bounded linear operator

For a bounded linear operator "A" and the operator norm ||&middot;||, again we have

:$ho\left(A\right) = lim_\left\{k o infty\right\}|A^k|^\left\{1/k\right\}.$

pectral radius of a graph

The spectral radius of a graph is defined to be the spectral radius of its adjacency matrix.

ee also

* Spectral gap

External links

* [http://people.csse.uwa.edu.au/gordon/planareig.html Spectral Radius of Planar Graphs]

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