Matrix Theory, Math6304 Lecture Notes from November 27, 2012 taken by Charles Mills

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1 Matrix Theory, Math6304 Lecture Notes from November 27, 202 take by Charles Mills Last Time (9/20/2) Gelfad s formula for spectral radius Gershgori s circle theorem Warm-up: Let s observe what Gershgori s circle theorem tells us about the eigevalues of the matrix A = 0 2 By observig the o-diagoal row sums, Gershgori tells us that there are two eigevalues, both of which lie iside the uio of a circle of radius cetered at the poit x =ad the poit x =2. So Gershgori tells us that all of the eigevalues of A lie withi a circle of a radius cetered at the poit x =. This however is ot much of a isight sice the matrix is already i upper triagular form, ad we ca therefore read off the two eigevalues of ad 2. I this lecture we will develop a ew way of usig Gershgori that will give us the spectrum of this matrix ad i fact, of ay upper triagular matrix. 5.5 Improvig Gershgori Before we proceed we eed to prove a theorem we have see before, but have ot yet prove Theorem. If ad are two orms o C,thetheyareequivalet,thatis, c>0 such that x x c x c for each x C.

2 Proof. By trasitivity it is eough to show that every orm is equivalet to. We begi with x = x j e j x j e j (max j x j ) = c x e j So we have obtaied a upper boud. O the other had, assume there is o >0 such that x x for all x, the for every k N we ca fid x k C such that x k k > x k ad we ca assume that x k is ormalized, x =by homogeeity of both sides. So each x k lies iside the polydisk D with D = {z C : z }. Now x k is a sequece i a compact set, so we ca take a coverget subsequece x kj x D. The triagle iequality, the upper boud we proved ad the assumptio o the orms of the sequece give x = x x kj + x kj x x kj + x kj c x x kj + k j. By cotiuity of the max orm i the etries we have the covergece x kj x 0, so each of these terms goes to zero as j,thusx =0. This cotradicts the cotiuity of the max orm sice x =lim j x kj =. Thus we have obtaied the boudedess ad cotiuity from the theorem. Questio: Ca we improve Gershgori to get better estimates o the spectrum? aswer this questio we will eed the help of a lemma. I order to Lemma. If A M ad S is ivertible, the A has eigevalue λ iff S AS has eigevalue λ. Proof. If Ax = λx ad x = 0,defiey = S x,the S ASy = S ASS x = S Ax = λs x = λy Thus we have obtaied oe directio of the implicatio i the lemma. Repeatig this process with B = S AS ad A = S BS gives the other directio. 2

3 Now we have all of the tools we eed i order to prove the followig geeralizatio of the Gershgori circle theorem Theorem. Let [A i,j ]=A M, {p j }, p j > 0, adr j (A, p) = p j l= p l A j,l. The if λ is a eigevalue of A, wehave λ G j (A, p) where G j = {z C : z A j,j R j (A, p)} Proof. The precedig lemma allows us to replace A with S AS ad the perform the usual strategy for derivig Gershgori circles. Choose S = D adiagoalmatrixwithd j,j = p j. The [D AD] j,l = p j A j,l p l ad R j (A, p) =R j (D AD) = l==j p j A j,l p l. Now by applyig Gershgori we obtai that all eigevalues lie withi G j(a, p). Now goig back to the warm-up from earlier, take 0 D = 0 2 By the previous theorem we ca compute D AD ad the apply Gershgori. D 0 AD = 0 A = 0 2 This holds for ay >0. Thuswecataketheitersectiooverallsuch ad ow Gershgori gives us the exact spectrum of the matrix A, amely{, 2}. Nextwelookatsomecosequeces of Gershgori. 5.6 Cosequeces of Gershgori Defiitio. Let A =[A i,j ] M,theA is diagoally domiat if for all j, A j,j l==j A j,l, adstrictlydiagoallydomiatiftheiequalityisstrict. Now as a direct cosequece of Gershgori, we get 2 theorems o ivertibility Theorem. Let A M.IfA is strictly diagoally domiat, the A is ivertible. Moreover, if A = A ad A j,j > 0 for all j, the all of the eigevalues of A are strictly positive. Proof. By Gershgori, the spectrum of A is iside the uio of disks are cetered at the diagoal values ad have radii that are strictly smaller tha the magitude of the correspodig diagoal etry. Thus, the (complex) triagle iequality shows that oe of these disks cotais the origi. Thus zero is ot a eigevalue, ad therefore A is ivertible. Now if A = A ad A j,j > 0 for 3

4 each j, thealloftheeigevaluesarerealadarefouditheitersectioofthediskswiththe real lie. Usig the real triagle iequality the shows that the eigevalues are cotaied i G j (A) R R + \{0}, so they are strictly positive. Now we ca pose aother questio. Ca we relax the codictio o strict diagoal domiatio ad still have some critero for determiig whether or ot a matrix is ivertible from Gershgori? The aswer is yes if we make a additioal assumptio o A Theorem. If A M is diagoally domiat, has o zero etries, ad if there is oe i {, 2,...,} such that A i,i > A i,j the A is ivertible. Proof. Assume A is ot ivertible. Take x = 0with Ax =0.Theforayi, A i,i x i = A i,i x i = A i,j x j A i,j x j A i,i x A i,j x j Now we ca pick a i such that equality holds o the first ad last step by chosig the i such that x i = x.thisforcesequalitytoholdthroughout,adweobtaithat A i,j x j = A i,j x j So we see that if A i,j =0,the x j = x,butthisiswhatweassumedforallofa. So x j = x for all j. O the other had if we chose i such that A i,i > A i,j, the A i,i x = A i,i x i = A i,j x j A i,j x j < A i,i x A i,j x j 4

5 So we see that the assumptio of strict domiatio for oe row is cotradicted. Thus we ca coclude that such a matrix has o otrivial vectors such that Ax =0.ThusA is ivertible. 5