LINEAR ALGEBRA Eigenvalues and Eigenvectors. Paul Dawkins

Transcription

1 LINEAR ALGEBRA Eigevalues ad Eigevectors Paul Dawkis

2 Table of Cotets Preface... ii Eigevalues ad Eigevectors... 3 Itroductio... 3 Review of Determiats... 4 Eigevalues ad Eigevectors... 3 Diagoalizatio Paul Dawkis i

3 Preface Here are my olie otes for my Liear Algebra course that I teach here at Lamar Uiversity. Despite the fact that these are my class otes they should be accessible to ayoe watig to lear Liear Algebra or eedig a refresher. These otes do assume that the reader has a good workig kowledge of basic Algebra. This set of otes is fairly self cotaied but there is eough Algebra type problems (arithmetic ad occasioally solvig equatios) that ca show up that ot havig a good backgroud i Algebra ca cause the occasioal problem. Here are a couple of warigs to my studets who may be here to get a copy of what happeed o a day that you missed.. Because I wated to make this a fairly complete set of otes for ayoe watig to lear Liear Algebra I have icluded some material that I do ot usually have time to cover i class ad because this chages from semester to semester it is ot oted here. You will eed to fid oe of your fellow class mates to see if there is somethig i these otes that was t covered i class.. I geeral I try to work problems i class that are differet from my otes. However, with a Liear Algebra course while I ca make up the problems off the top of my head there is o guaratee that they will work out icely or the way I wat them to. So, because of that my class work will ted to follow these otes fairly close as far as worked problems go. With that beig said I will, o occasio, work problems off the top of my head whe I ca to provide more examples tha just those i my otes. Also, I ofte do t have time i class to work all of the problems i the otes ad so you will fid that some sectios cotai problems that were t worked i class due to time restrictios. 3. Sometimes questios i class will lead dow paths that are ot covered here. I try to aticipate as may of the questios as possible i writig these otes up, but the reality is that I ca t aticipate all the questios. Sometimes a very good questio gets asked i class that leads to isights that I ve ot icluded here. You should always talk to someoe who was i class o the day you missed ad compare these otes to their otes ad see what the differeces are. 4. This is somewhat related to the previous three items, but is importat eough to merit its ow item. THESE NOTES ARE NOT A SUBSTITUTE FOR ATTENDING CLASS!! Usig these otes as a substitute for class is liable to get you i trouble. As already oted ot everythig i these otes is covered i class ad ofte material or isights ot i these otes is covered i class. 7 Paul Dawkis ii

4 Eigevalues ad Eigevectors Itroductio This is goig to be a very short chapter. The mai topic of this chapter will be the Eigevalues ad Eigevectors sectio. I this sectio we will be lookig at special situatios where give a square matrix A ad a vector x the product Ax will be the same as the scalar multiplicatio λx for some scalar, λ. This idea has importat applicatios i may areas of math ad sciece ad so we put it ito a chapter of its ow. We ll also have a quick review of determiats sice those will be required i order to due the work i the Eigevalues ad Eigevectors sectio. We ll also take a look at a applicatio that uses eigevalues. Here is a listig of the topics i this chapter. Review of Determiats I this sectio we ll do a quick review of determiats. Eigevalues ad Eigevectors Here we will take a look at the mai sectio i this chapter. We ll be lookig at the cocept of Eigevalues ad Eigevectors. Diagoalizatio We ll be lookig at diagoalizable matrices i this sectio. 7 Paul Dawkis 3

5 Review of Determiats I this sectio we are goig to do a quick review of determiats ad we ll be cocetratig almost exclusively o how to compute them. For a more i depth look at determiats you should check out the secod chapter which is devoted to determiats ad their properties. Also, we ll ackowledge that the examples i this sectio are all examples that were worked i the secod chapter. We ll start off with a quick workig defiitio of a determiat. See The Determiat Fuctio from the secod chapter for the exact defiitio of a determiat. What we re goig to give here will be sufficiet for what we re goig to be doig i this chapter. So, give a square matrix, A, the determiat of A, deoted by det ( A ), is a fuctio that associated with A a umber. That s it. That s what a determiat does. It takes a matrix ad associates a umber with that matrix. There is also some alterate otatio that we should ackowledge because we ll be usig it quite a bit. The alterate otatio is, det ( A) A. We ow eed to discuss how to compute determiats. There are may ways of computig determiats, but most of the geeral methods ca lead to some fairly log computatios. We will see oe geeral method towards the ed of this sectio, but there are some ice quick formulas that ca help with some special cases so we ll start with those. We ll be workig mostly with matrices i this chapter that fit ito these special cases. We will start with the formulas for ad 3 3 matrices. Defiitio If A a a a a det the the determiat of A is, ( A) a a aa a a a a Defiitio If a a a A a a a 3 3 a3 a3 a33 the the determiat of A is, det ( ) a a a 3 A a a a 3 a a a a a a + a a a + a a a a a a a a a a a a Okay, we said that these were ice ad quick formulas ad the formula for the matrix is fairly ice ad quick, but the formula for the 3 3 matrix is either ice or quick. Luckily there are some ice little tricks that ca help us to write dow both formulas. 7 Paul Dawkis 4

6 We ll start with the followig determiat of a matrix ad we ll sketch i two diagoals as show Note that if you multiply alog the gree diagoal you will get the first product i formula for matrices ad if you multiply alog the red diagoal you will get the secod product i the formula. Also, otice that the red diagoal, ruig from right to left, was the product that was subtracted off, while the gree diagoal, ruig from left to right, gave the product that was added. We ca do somethig similar for 3 3 matrices, but there is a differece. First, we eed to tack a copy of the leftmost two colums oto the right side of the determiat. We the have three diagoals that ru from left to right (show i gree below) ad three diagoals that ru from right to left (show i red below). As will the case, if we multiply alog the gree diagoals we get the products i the formula that are added i the formula ad if we multiply log the red diagoals we get the products i the formula that are subtracted i the formula. Here are a couple of quick examples. Example Compute the determiat of each of the followig matrices. 3 (a) A [ 9 5 Solutio] (b) B 8 [ Solutio] 7 6 (c) C 8 3 [ Solutio] 3 Solutio 3 (a) A 9 5 We do t really eed to sketch i the diagoals for matrices. The determiat is simply the product of the diagoal ruig left to right mius the product of the diagoal ruig from right to left. So, here is the determiat for this matrix. The oly thig we eed to worry about is payig attetio to mius sigs. It is easy to make a mistake with mius sigs i these computatios if you are t payig attetio. det A ( ) ( )( ) ( )( ) 7 Paul Dawkis 5

7 [Retur to Problems] (b) B 8 7 Okay, with this oe we ll copy the two colums over ad sketch i the diagoals to make sure we ve got the idea of these dow. Now, just remember to add products alog the left to right diagoals ad subtract products alog the right to left diagoals. det B ( ) ( )( )( ) ( )( )( ) ( )( )( ) ( )( )( ) ( 3)( 8)( ) ( 4)( )( ) 467 [Retur to Problems] 6 (c) C We ll do this oe with a little less detail. We ll copy the colums but ot bother to actually sketch i the diagoals this time. 6 6 ( C) det ( )( 8)( ) ( 6)( 3)( 3) ( )( )( ) ( 6)( )( ) ( )( 3)( ) ( )( 8)( 3) + + [Retur to Problems] As we ca see from this example the determiat for a matrix ca be positive, egative or zero. Likewise, as we will see towards the ed of this review we are goig to be especially iterested i whe the determiat of a matrix is zero. Because of this we have the followig defiitio. Defiitio 3 Suppose A is a square matrix. (a) If det A we call A a sigular matrix. (b) If ( ) det ( A) we call A a o-sigular matrix. So, i Example above, both A ad B are o-sigular while C is sigular. 7 Paul Dawkis 6

8 Before we proceed we should poit out that while there are formulas for larger matrices (see the Determiat Fuctio sectio for details o how to write them dow) there are ot ay easy tricks with diagoals to write them dow as we had for ad 3 3 matrices. With the statemet above made we should ote that there is a simple formula for geeral matrices of certai kids. The followig theorem gives this formula. Theorem Suppose that A is a triagular matrix with diagoal etries a, a,, a the determiat of A is, det A a a a ( ) This theorem will be valid regardless of whether the triagular matrix is a upper triagular matrix or a lower triagular matrix. Also, because a diagoal matrix ca also be cosidered to be a triagular matrix Theorem is also valid for diagoal matrices. Here are a couple of quick examples of this. Example Compute the determiat of each of the followig matrices A 3 B C Solutio Here are these determiats. det ( A) ( 5)( 3)( 4) 6 det ( B) ( 6)( ) 6 det C 6 5 ( ) ( )( )( )( ) There are several methods for fidig determiats i geeral. Oe of them is the Method of Cofactors. What follows is a very brief overview of this method. For a more detailed discussio of this method see the Method of Cofactors i the Determiats Chapter. We ll start with a couple of defiitios first. Defiitio 4 If A is a square matrix the the mior of aij, deoted by of the submatrix that results from removig the i th row ad j th colum of A. M ij, is the determiat Defiitio 5 If A is a square matrix the the cofactor of a, deoted by C, is the umber ( ) i + j ij M. Here is a quick example showig some mior ad cofactor computatios. ij ij 7 Paul Dawkis 7

9 Example 3 For the followig matrix compute the cofactors,, ad C. C C A Solutio I order to compute the cofactors we ll first eed the mior associated with each cofactor. Remember that i order to compute the mior we will remove the i th row ad j th colum of A. So, to compute M (which we ll eed for C ) we ll eed to compute the determiate of the matrix we get by removig the st row ad d colum of A. Here is that work. We ve marked out the row ad colum that we elimiated ad we ll leave it to you to verify the determiat computatio. Now we ca get the cofactor. C ( ) + M ( ) 3 ( 6) 6 Let s ow move oto the secod cofactor. Here is the work for the mior. The cofactor i this case is, C ( ) M ( ) ( ) Here is the work for the fial cofactor. C 3+ 5 ( ) M ( ) ( ) Notice that the cofactor for a give etry is really just the mior for the same etry with a + or a - i frot of it. The followig table shows whether or ot there should be a + or a - i frot of a mior for a give cofactor. 7 Paul Dawkis 8

10 C ij To use the table for the cofactor we simply go to the i th row ad j th colum i the table above ad if there is a + there we leave the mior aloe ad if there is a - there we will tack a - oto the appropriate mior. So, for C we go to the 3 rd row ad 4 th 34 colum ad see that we have a mius sig ad so we kow that C M Here is how we ca use cofactors to compute the determiat of ay matrix. Theorem If A is a matrix. (a) Choose ay row, say row i, the, A ac + ac + ac ( ) det i i i i i i (b) Choose ay colum, say colum j, the, A a C + a C + + a C ( ) det j j j j j j Here is a quick example of how to use this theorem. Example 4 For the followig matrix compute the determiat usig the give cofactor expasios. 4 A (a) Expad alog the first row. [Solutio] (b) Expad alog the third row. [Solutio] (c) Expad alog the secod colum. [Solutio] Solutio First, otice that accordig to the theorem we should get the same result i all three parts. (a) Expad alog the first row. Here is the cofactor expasio i terms of symbols for this part. det A ac + ac + ac ( ) 3 3 Now, let s plug i for all the quatities. We will just plug i for the etries. For the cofactors we ll write dow the mior ad a + or a - depedig o which sig each mior eeds. We ll determie these sigs by goig to our sig matrix above startig at the first etry i the particular row/colum we re expadig alog ad the as we move alog that row or colum we ll write dow the appropriate sig. 7 Paul Dawkis 9

11 Here is the work for this expasio det ( A) ( 4)( + ) + ( )( ) + ()( + ) ( ) ( ) ( )( ) We ll leave it to you to verify the determiat computatios. [Retur to Problems] (b) Expad alog the third row. We ll do this oe without all the explaatios. det A a C + a C + a C ( ) ( 7)( + ) + ( 5)( ) + ( )( + ) ( ) 54 ( ) + ( )( ) 54 So, the same aswer as the first part which is good sice that was supposed to happe. Notice that the sigs for the cofactors i this case were the same as the sigs i the first case. This is because the first ad third row of our sig matrix are idetical. Also, otice that we did t really eed to compute the third cofactor sice the third etry was zero. We did it here just to get oe more example of a cofactor ito the otes. [Retur to Problems] (c) Expad alog the secod colum. Let s take a look at the fial expasio. I this oe we re goig dow a colum ad otice that from our sig matrix that this time we ll be startig the cofactor sigs off with a - ulike the first two expasios. det ( A) ac + ac + a3c ( )( ) + ( 6)( + ) + ( 5)( ) ( ) 6( 7) 5( 4) 54 Agai, the same as the first two as we expected. [Retur to Problems] As this example has show it does t matter which row or colum we expad alog we will always get the same result. 7 Paul Dawkis

12 I this example we performed a cofactor expasio o a 3 3sice we could easily check the results usig the process we discussed above. Let s work oe more example oly this time we ll fid the determiat of a 4 4 matrix ad so we ll ot have ay choice but to use a cofactor expasio. Example 5 Usig a cofactor expasio compute the determiat of, A Solutio Sice the row or colum to use for the cofactor expasio was ot give i the problem statemet we get to choose which oe we wat to use. From the previous example we kow that it wo t matter which row or colum we choose.. However, havig said that otice that if there is a zero etry we wo t eed to compute the cofactor/mior for that etry sice it will just multiply out to zero. So, it looks like the secod row would be a good choice for the expasio sice it has two zeroes i it. Here is the expasio for this row. As with the previous expasios we ll explicitly give the + or - for the cofactors ad the miors as well so you ca see where everythig i the expasio is comig from. 7 5 det( A) ( )( ) 5 + ( )( + ) M + ( )( ) M3 + ( 3)( + ) We did t bother to write dow the miors M ad M 3 because of the zero etry. How we choose to compute the determiats for the first ad last etry is up to us at this poit. We could use a cofactor expasio o each of them or we could use the techique we saw above. Either way will get the same aswer ad we ll leave it to you to verify these determiats. The determiat for this matrix is, det A ( ) ( ) ( ) We ll close this review off with a sigificatly shorteed versio of Theorem 9 from Properties of Determiats sectio. We wo t eed most of the theorem, but there are two bits of it that we ll eed so here they are. Also, there are two ways i which the theorem ca be stated ow that we ve stripped out the other pieces ad so we ll give both ways of statig it here. Theorem 3 If A is a matrix the (a) The oly solutio to the system A x is the trivial solutio (i.e. x ) if ad oly if det ( A). (b) The system A x will have a o-trivial solutio (i.e. x ) if ad oly if det A. ( ) 7 Paul Dawkis

13 Note that these two statemets really are equivalet. Also, recall that whe we say if ad oly if i a theorem statemet we mea that the statemet works i both directios. For example, let s take a look at the secod part of this theorem. This statemet says that if A x has otrivial solutios the we kow that we ll also have det A. O the other had, it also says that if det ( A ) ( ) the we ll also kow that the system will have o-trivial solutios. This theorem will be key to allowig us to work problems i the ext sectio. This is the the review of determiats. Agai, if you eed a more detailed look at either determiats or their properties you should go back ad take a look at the Determiat chapter. 7 Paul Dawkis

14 Eigevalues ad Eigevectors As oted i the itroductio to this chapter we re goig to start with a square matrix A ad try to determie vectors x ad scalars λ so that we will have, Ax λx I other words, whe this happes, multiplyig x by the matrix A will be equivalet of multiplyig x by the scalar λ. This will ot be possible for all vectors, x, or will it be possible for all scalars λ. The goal of this sectio is to determie the vectors ad scalars for which this will happe. So, let s start off with the followig defiitio. Defiitio Suppose that A is a matrix. Also suppose that x is a o-zero vector from ad that λ is ay scalar (this ca be zero) so that, Ax λx We the call x a eigevector of A ad λ a eigevalue of A. We will ofte call x the eigevector correspodig to or associated with λ ad we will ofte call λ the eigevalue correspodig to or associated with x. Note that eigevalues ad eigevectors will always occur i pairs. You ca t have a eigevalue without a eigevector ad you ca t have a eigevector without a eigevalue. Example Suppose eigevalue 4 because 6 6 A 4 the 8 x is a eigevector with correspodig Ax 4 λ 4 4 x Okay, what we eed to do is figure out just how we ca determie the eigevalues ad eigevectors for a give matrix. This is actually easier to do that it might at first appear to be. We ll start with fidig the eigevalues for a matrix ad oce we have those we ll be able to fid the eigevectors correspodig to each eigevalue. Let s start with Ax λx ad rewrite it as follows, Ax λix Note that all we did was isert the idetity matrix ito the right side. Doig this will allow us to further rewrite this equatio as follows, λix Ax λi A x ( ) Now, if λ is goig to be a eigevalue of A, this system must have a o-zero solutio, x, sice we kow that eigevectors associated with λ caot be the zero vector. However, Theorem 3 7 Paul Dawkis 3

15 from the previous sectio or more geerally Theorem 8 from the Fudametal Subspaces sectio tells us that this system will have a o-zero solutio if ad oly if det λi A ( ) So, eigevalues will be scalars, λ, for which the matrix λi A will be sigular, i.e. det ( λi A). Let s get a couple more defiitios out of the way ad the we ll work some examples of fidig eigevalues. Defiitio Suppose A is a matrix the, det λi A is called the characteristic equatio of A. Whe computed it will be a th degree polyomial i λ of the form, p ( λ) λ + c λ + + cλ+ c called the characteristic polyomial of A. ( ) Note that the coefficiet of λ is (oe) ad that is NOT a typo i the defiitio. This also guaratees that this polyomial will be of degree. Also, from the Fudametal Theorem of Algebra we ow kow that there will be exactly eigevalues (possibly icludig repeats) for a matrix A. Note that because the Fudametal Theorem of Algebra does allow for the possibility of repeated eigevalues there will be at most distict eigevalues for a matrix. Because a eigevalue ca repeat itself i the list of all eigevalues we d like a way to differetiate betwee eigevalues that repeat ad those that do t repeat. The followig defiitio will do this for us. Defiitio 3 Suppose A is a matrix ad that λ, λ,, λ is the complete list of all the eigevalues of A icludig repeats. If λ occurs exactly oce i this list the we call λ a simple eigevalue. If λ occurs k times i the list we say that λ has multiplicity of k. Okay, let s fid some eigevalues we ll start with some matrices. Example Fid all the eigevalues for the give matrices. 6 6 (a) A 4 [ Solutio] 4 (b) A 3 5 [ Solutio] 7 (c) A 4 3 [ Solutio] Solutio 6 6 (a) A 4 We ll do this oe with a little more detail tha we ll do the other two. First we ll eed the matrix λi A. 7 Paul Dawkis 4

16 λ 6 6 λ 6 6 λi A λ 4 λ+ 4 Next we eed the determiat of this matrix, which gives us the characteristic polyomial. det λi A λ 6 λ+ 4 6 λ λ 8 ( ) ( )( ) ( ) Now, set this equal to zero ad solve for the eigevalues. λ λ 8 λ 4 λ+ λ, λ 4 ( )( ) So, we have two eigevalues ad sice they occur oly oce i the list they are both simple eigevalues. [Retur to Problems] (b) 4 A 3 5 Here is the matrix λi A ad its characteristic polyomial. λ + 4 λi A det ( λi A) λ + 5 λ λ We ll leave it to you to verify both of these. Now, set the characteristic polyomial equal to zero ad solve for the eigevalues. λ + 9λ+ 4 λ+ 7 λ+ λ 7, λ ( )( ) Agai, we get two simple eigevalues. [Retur to Problems] (c) 7 A 4 3 Here is the matrix λi A ad its characteristic polyomial. λ 7 λi A det ( λi A) λ 3 λ λ We ll leave it to you to verify both of these. Now, set the characteristic polyomial equal to zero ad solve for the eigevalues. λ λ+ 5 λ 5 λ 5 ( ) I this case we have a eigevalue of multiplicity two. Sometimes we call this kid of eigevalue a double eigevalue. Notice as well that we used the otatio λ, to deote the fact that this was a double eigevalue. [Retur to Problems] Now, let s take a look at some 3 3 matrices., 7 Paul Dawkis 5

17 Example 3 Fid all the eigevalues for the give matrices. 4 (a) A 6 5 [ Solutio] (b) A 3 [ Solutio] (c) A [ Solutio] 4 (d) A 3 [ Solutio] Solutio 4 (a) A 6 5 As with the previous example we ll do this oe i a little more detail tha the remaiig two parts. First, we ll eed λ,, λ 4 λ 4 λi A λ 6 λ+ 6 λ 5 5 λ Now, let s take the determiat of this matrix ad get the characteristic polyomial for A. We ll use the trick that we reviewed i the previous sectio to take the determiat. You could also use cofactors if you prefer that method. The result will be the same. λ 4 λ 4 ( λi A) det λ+ 6 λ+ 6 5 λ 5 ( 4)( 6) 5( 6) λ λ λ+ λ+ 3 λ λ λ Next, set this equal to zero. 3 λ + λ 9λ 3 Now, most of us are t that great at fid the roots of a cubic polyomial. Luckily there is a way to at least get us started. It wo t always work, but if it does it ca greatly reduce the amout of work that we eed to do. 7 Paul Dawkis 6

18 Suppose we re tryig to fid the roots of a equatio of the form, λ + c λ + + cλ+ c where the c i are all itegers. If there are iteger solutios to this (ad there may NOT be) the we kow that they must be divisors of c. This wo t give us ay iteger solutios, but it will allow us to write dow a list of possible iteger solutios. The list will be all possible divisors of. c I this case the list of possible iteger solutios is all possible divisors of -3. ±, ±, ± 3, ± 5, ± 6, ±, ± 5, ± 3 Now, that may seem like a lot of solutios that we ll eed to check. However, it is t quite that bad. Start with the smaller possible solutios ad plug them i util you fid oe (i.e. util the polyomial is zero for oe of them) ad the stop. I this case the smallest oe i the list that works is -. This meas that λ ( ) λ + must be a factor i the characteristic polyomial. I other words, we ca write the characteristic polyomial as, 3 λ + λ 9λ 3 ( λ+ ) q ( λ ) where q ( λ ) is a quadratic polyomial. We fid q ( λ ) by performig log divisio o the characteristic polyomial. Doig this i this case gives, 3 λ + λ 9λ 3 λ+ λ + λ 3 ( )( ) At this poit all we eed to do is fid the solutios to the quadratic ad icely eough for us that factors i this case. So, puttig all this together gives, λ+ λ+ 6 λ 5 λ, λ 6, λ 5 ( )( )( ) 3 So, this matrix has three simple eigevalues. [Retur to Problems] (b) A 3 Here is λi A ad the characteristic polyomial for this matrix. λ λi A λ + det ( λi A) λ λ 6λ λ + 3 Now, i this case the list of possible iteger solutios to the characteristic polyomial are, ±, ±, ± 4, ± 5, ±, ± Agai, if we start with the smallest itegers i the list we ll fid that - is the first iteger 7 Paul Dawkis 7

19 ( ) solutio. Therefore, λ λ+ must be a factor of the characteristic polyomial. Factorig this out of the characteristic polyomial gives, 3 λ λ 6λ λ+ λ 3λ ( )( ) Fially, factorig the quadratic ad settig equal to zero gives us, λ+ λ 5 λ, λ 5 ( ) ( ), 3 So, we have oe double eigevalue ( λ, ) ad oe simple eigevalue ( λ 3 5 ). [Retur to Problems] (c) A Here is λi A ad the characteristic polyomial for this matrix. λ 3 λi A λ det( λi A) λ 3λ λ We have a very small list of possible iteger solutios for this characteristic polyomial. ±, ± The smallest iteger that works i this case is - ad we ll leave it to you to verify that the complete factored form is characteristic polyomial is, 3 λ 3λ ( λ+ ) ( λ ) ad so we ca see that we ve got two eigevalues λ, (a multiplicity eigevalue) ad λ 3 (a simple eigevalue). [Retur to Problems] (d) 4 A 3 Here is λi A ad the characteristic polyomial for this matrix. λ 4 3 λi A λ 3 det( λi A) λ 9λ + 7λ 7 λ Okay, i this case the list of possible iteger solutios is, ±, ± 3, ± 9, ± 7 7 Paul Dawkis 8

20 The smallest iteger that will work i this case is 3. We ll leave it to you to verify that the factored form of the characteristic polyomial is, 3 λ 9λ + 7λ 7 ( λ 3) 3 ad so we ca see that if we set this equal to zero ad solve we will have oe eigevalue of multiplicity 3 (sometimes called a triple eigevalue), λ,,3 3 [Retur to Problems] As you ca see the work for these get progressively more difficult as we icrease the size of the matrix, for a geeral matrix larger tha 3 3 we d eed to use the method of cofactors to determie the characteristic polyomial. There is oe case kid of matrix however that we ca pick the eigevalues right off the matrix itself without doig ay work ad the size wo t matter. Theorem Suppose A is a etries,, a,, a. a triagular matrix the the eigevalues will be the diagoal Proof : We ll give the proof for a upper triagular matrix, ad leave it to you to verify the proof for a lower triagular ad a diagoal matrix. We ll start with, a a a a a A a Now, we ca write dow λi A, λ a a a λ a a λi A λ a Now, this is still a upper triagular matrix ad we kow that the determiat of a triagular matrix is just the product of its mai diagoal etries. The characteristic polyomial is the, det λi A λ a λ a λ a ( ) ( )( ) ( ) Settig this equal to zero ad solvig gives the eigevalues of, λ a λ a λ a 7 Paul Dawkis 9

21 Example 4 Fid the eigevalues of the followig matrix A Solutio Sice this is a lower triagular matrix we ca use the previous theorem to write dow the eigevalues. It will simply be the mai diagoal etries. The eigevalues are, λ 6 λ 4 λ λ λ We ca ow fid the eigevalues for a matrix. We ext eed to address the issue of fidig their correspodig eigevectors. Recall give a eigevalue, λ, the eigevector(s) of A that correspod to λ will be the vectors x such that, Ax λx OR λi A x ( ) Also, recall that λ was chose so that λi A was a sigular matrix. This i tur guarateed that we would have a o-zero solutio to the equatio above. Note that i doig this we do t just guaratee a o-zero solutio, but we also guaratee that we ll have ifiitely may solutios to the system. We have oe quick defiitio that we eed to take care of before we start workig problems. Defiitio 4 The set of all solutios to ( I A) correspodig to λ. λ x is called the eigespace of A Note that there will be oe eigespace of A for each distict eigevalue ad so there will be aywhere from to eigespaces for a matrix depedig upo the umber of distict eigevalues that the matrix has. Also, otice that we re really just fidig the ull space for a system ad we ve looked at that i several sectios i the previous chapter. Let s take a look at some eigespaces for some of the matrices we foud eigevalues for above. Example 5 For each of the followig matrices determie the eigevectors correspodig to each eigevalue ad determie a basis for the eigespace of the matrix correspodig to each eigevalue. 6 6 (a) A [ 4 Solutio] 7 (b) A [ 4 3 Solutio] Solutio We determied the eigevalues for each of these i Example above so refer to that example for the details i fidig them. For each eigevalue we will eed to solve the system, λi A x ( ) 7 Paul Dawkis

22 to determie the geeral form of the eigevector. Oce we have that we ca use the geeral form of the eigevector to fid a basis for the eigespace. (a) 6 6 A 4 We kow that the eigevalues for this matrix are λ ad λ 4. Let s first fid the eigevector(s) ad eigespace for λ. Referrig to Example for the formula for λi A ad pluggig λ ito this we ca see that the system we eed to solve is, 8 6 x x We ll leave it to you to verify that the solutio to this system is, x t x t Therefore, the geeral eigevector correspodig to λ is of the form, t x t t The eigespaces is all vectors of this form ad so we ca see that a basis for the eigespace correspodig to λ is, v Now, let s fid the eigevector(s) ad eigespace for λ 4. Pluggig λ 4 ito the formula for λi A from Example gives the followig system we eed to solve, 6 x 8 x The solutio to this system is (you should verify this), x 8t x t The geeral eigevector ad a basis for the eigespace correspodig to λ 4 is the, 8t 8 8 x t & t v Note that if we wated our hads o specific eigevalues for each eigevector the basis vector for each eigespace would work. So, if we do that we could use the followig eigevectors (ad their correspodig eigevalues) if we d like. 7 Paul Dawkis

23 8 λ v, λ 4 v Note as well that these eigevectors are liearly idepedet vectors. [Retur to Problems] (b) 7 A 4 3 From Example we kow that λ, 5 is a double eigevalue ad so there will be a sigle eigespace to compute for this matrix. Usig the formula for λi A from Example ad pluggig λ, 5 ito this gives the followig system that we ll eed to solve for the eigevector ad eigespace. x 4 x The solutio to this system is, x t x t The geeral eigevector ad a basis for the eigespace correspodig λ, 5 is the, t x t & t v I this case we get oly a sigle eigevector ad so a good eigevalue/eigevector pair is, λ, 5 v [Retur to Problems] We did t look at the secod matrix from Example i the previous example. You should try ad determie the eigespace(s) for that matrix. The work will follow what we did i the first part of the previous example sice there are two simple eigevalues. Now, let s determie the eigespaces for the matrices i Example 3 Example 6 Determie the eigevectors correspodig to each eigevalue ad a basis for the eigespace correspodig to each eigevalue for each of the matrices from Example 3 above. Solutio The work fidig the eigevalues for each of these is show i Example 3 above. Also, we ll be doig this with less detail tha those i the previous example. I each part we ll use the formula for λi A foud i Example 3 ad plug i each eigevalue to fid the system that we eed to solve for the eigevector(s) ad eigespace. 7 Paul Dawkis

24 (a) 4 A 6 5 The eigevalues for this matrix are λ, λ 6 ad λ 3 5 so we ll have three eigespaces to fid. Startig with λ we ll eed to fid the solutio to the followig system, 5 x 9 5 x x t x t x 3 t x 3 The geeral eigevector ad a basis for the eigespace correspodig to λ is the, 5t t t 5 & 9 x v 5 t Now, let s take a look at λ 6. Here is the system we eed to solve, x x x x t 5 6 x 3 x 3 Here is the geeral eigevector ad a basis for the eigespace correspodig to λ 6. x t t & v Fially, here is the system for λ 3 5. x 3 x x t x t x x 3 The geeral eigevector ad a basis for the eigespace correspodig to λ 3 5 is the, t 3 3 t t & 3 x v3 t t 7 Paul Dawkis 3

25 Now, we as with the previous example, let s write dow a specific set of eigevalue/eigevector pairs for this matrix just i case we happe to eed them for some reaso. We ca get specific eigevectors usig the basis vectors for the eigespace as we did i the previous example λ v 5, λ 6, λ3 5 v 3 v You might wat to verify that these three vectors are liearly idepedet vectors. (b) A 3 The eigevalues for this matrix are λ, ad λ 3 5 so it looks like we ll have two eigespaces to fid for this matrix. We ll start with λ,. Here is the system that we eed to solve ad it s solutio x x x t x x3 t x 3 The geeral eigevector ad a basis for the eigespace correspodig λ, is the, t x t & v t Note that eve though we have a double eigevalue we get a sigle basis vector here. Next, the system for λ 3 5 that we eed to solve ad it s solutio is, 3 8 x 7 x x 8t x x 3 8 x 3 The geeral eigevector ad a basis for the eigespace correspodig to λ 3 5 is, 8t 8 8 x t & v t A set of eigevalue/eigevector pairs for this matrix is, t 7 Paul Dawkis 4

26 8 λ, v, λ3 5 v Ulike the previous part we oly have two eigevectors here eve though we have three eigevalues (if you iclude repeats ayway). (c) A As with the previous part we ve got two eigevalues, λ, ad λ 3 ad so we ll agai have two eigespaces to fid here. We ll start with λ,. Here is the system we eed to solve, x x x s t x s x 3 x 3 The geeral eigevector correspodig to λ, is the, s t x s t + s t Now, the eigespace is spaed by the two vectors above ad sice they are liearly idepedet we ca see that a basis for the eigespace correspodig to λ, is, v v Here is the system for λ 3 that we eed to solve, x x x t x t x 3 t x 3 The geeral eigevector ad a basis for the eigespace correspodig to λ 3 is, t x t t & v3 t t 7 Paul Dawkis 5

27 Okay, i this case if we wat to write dow a set of eigevalue/eigevector pairs we ve got a slightly differet situatio that we ve see to this poit. I the previous example we had a eigevalue of multiplicity two but oly got a sigle eigevector for it. I this case because the eigespace for our multiplicity two eigevalue has a dimesio of two we ca use each basis vector as a separate eigevector ad so the eigevalue/eigevector pairs this time are, λ v, λ, λ3 v v3 Note that we listed the eigevalue of - twice, oce for each eigevector. You should verify that these are all liearly idepedet. (d) 4 A 3 I this case we had a sigle eigevalue, λ,,3 3 so we ll have a sigle eigespace to fid. Here is the system ad its solutio for this eigevalue. x x x t x s x3 t x 3 The geeral eigevector correspodig to λ,,3 3 is the, t x s t + s t As with the previous example we ca see that the eigespace is spaed by the two vectors above ad sice they are liearly idepedet we ca see that a basis for the eigespace correspodig to λ,,3 3 is, v v Note that the two vectors above would also make a ice pair of eigevectors for the sigle eigevalue i this case. Okay, let s go back ad take a look at the eigevector/eigevalue pairs for a secod. First, there are reasos for watig these as we ll see i the ext sectio. O occasio we really do wat specific eigevectors ad we geerally wat them to be liearly idepedet as well as we ll see. Also, we saw two examples of eigevalues of multiplicity. I oe case we got a sigle eigevector ad i the other we got two liearly idepedet eigevectors. This will always be the case, if λ is a eigevalue of multiplicity k the there will be aywhere from to k liearly idepedet eigevectors depedig upo the dimesio of the eigespace. 7 Paul Dawkis 6

28 How there is oe type of eigevalue that we ve bee avoidig to this poit ad you may have oticed that already. Let s take a look at the followig example to see what we ve bee avoidig to this poit. Example 7 Fid all the eigevalues of Solutio First we ll eed the matrix I A 6 5 A 8 6 λ ad the we ll use that to fid the characteristic equatio. λ 6 5 λi A 8 λ + 6 det λi A λ 6 λ λ + 4 ( ) ( )( ) From this we ca see that the eigevalues will be complex. I fact they will be, λ i λ i So we got a complex eigevalue. We ve ot see ay of these to this poit ad this will be the oly oe we ll look at here. We have avoided complex eigevalues to this poit o very importat reaso. Let s recall just what a eigevalue is. A eigevalue is a scalar such that, 6 5 ( i) 8 6 x x Ca you see the problem? I order to talk about the scalar multiplicatio of the right we eed to be i a complex vector space! Up to this poit we ve bee workig exclusively with real vector spaces ad remember that i a real vector space the scalars are all real umbers. So, i order to work with complex eigevalues we would eed to be workig i a complex vector space ad we have t looked at those yet. So, sice we have t looked at complex vector spaces yet we will be workig oly with matrices that have real eigevalues. Note that this does t mea that complex eigevalues are ot importat. There are some very importat applicatios to complex eigevalues i various areas of math, egieerig ad the scieces. Of course, there are also times where we would igore complex eigevalues. We will leave this sectio out with a couple of ice theorems. Theorem Suppose that λ is a eigevalue of the matrix A with correspodig eigevector x. k k The if k is a positive iteger λ is a eigevalue of the matrix A with correspodig eigevector x. Proof : The proof here is pretty simple. 7 Paul Dawkis 7

29 k k A x A A A k λ A k λ A λ λ k k k ( x) ( λx) ( Ax) ( λx) ( Ax) ( λx) k λ x k k So, from this we ca see that λ is a eigevalue of A with correspodig eigevector x. Theorem 3 Suppose A is a repeats). The, matrix with eigevalues λ, λ,, λ (possibly icludig det A λ λ λ. A λ + λ + + λ. (a) The determiat of A is ( ) (b) The trace of A is ( ) tr We ll prove part (a) here. The proof of part (b) ivolves some fairly messy algebra ad so we wo t show it here. Proof of (a) : First, recall that eigevalues of A are roots of the characteristic polyomial ad hece we ca write the characteristic polyomial as, det λi A λ λ λ λ λ λ ( ) ( )( ) ( ) Now, plug i λ ito this ad we get, det A λ λ λ λ λ ( ) ( )( ) ( ) ( ) λ Fially, from Theorem of the Properties of Determiat sectio we kow that det det ( A) ( ) ( A) So, pluggig this i gives, det ( ) A λ λ λ 7 Paul Dawkis 8

30 Diagoalizatio I this sectio we re goig to take a look at a special kid of matrix. We ll start out with the followig defiitio. Defiitio Suppose that A is a square matrix ad further suppose that there exists a ivertible matrix P (of the same size as A of course) such that P AP is a diagoal matrix. I such a case we call A diagoalizable ad say that P diagoalizes A. The followig theorem will ot oly tell us whe a matrix is diagoalizable, but the proof will tell us how to costruct P whe A is diagoalizable. Theorem Suppose that A is a matrix, the the followig are equivalet. (a) A is diagoalizable. (b) A has liearly idepedet eigevectors. Proof : We ll start by provig that ( a) ( b). So, assume that A is diagoalizable ad so we kow that a ivertible matrix P exists so that P AP is a diagoal matrix. Now, let p, p,, p be the colums of P ad suppose that D is the diagoal matrix we get from P AP, i.e. D P AP. So, both P ad D have the followig forms, λ λ P [ ] D p p p λ Also ote that because P is a ivertible matrix Theorem 8 from the Fudametal Subspaces sectio tells us that the colums of P will form a basis for ad hece must be liearly idepedet. Therefore,, p,, p are a set of liearly idepedet colums vectors. Now, if we rewrite A p D P AP we arrive at AP PD or, [ p p p ] [ p p p ] λ λ λ Theorem from the Matrix Arithmetic sectio tell us that the j th colum of PD is P[j th colum of D] ad so the j th colum of PD is othig more tha λ jp j. The same theorem tells us that j th colum of AP is A[j th colum of P] or Ap j. Now, sice we have AP PD the colums of both sides must be equal ad so we must have, Ap λ p Ap λ p Ap λ p 7 Paul Dawkis 9

31 So, the diagoal etries from D, λ, λ,, λ are the eigevalues of A ad their correspodig eigevectors are the colums of P,, p,, p. Also as we oted above these p are a set of liearly idepedet vectors which is what we were asked to prove. We ow eed to prove ( b) ( a) ad we ve doe most of the work for this i the previous part. Let s start by assumig that the eigevalues of A are eigevectors are, p,, p are liearly idepedet. p λ, λ,, Now, form a matrix P whose colums are, p,, p. So, P has the form, p [ ] P p p p λ ad that their associated Now, as we oted above the colums of AP are give by Ap Ap Ap However, usig the fact that p, p,, p are the eigevectors of A each of these colums ca be writte as, Ap λ p Ap λ p Ap λ p Therefore, AP ca be writte as, AP [ Ap Ap Ap] λp λp λ p However, as we saw above, the matrix o the right ca be writte as PD where D the followig diagoal matrix, λ λ D λ So, we ve maaged to show that by defiig P as above we have AP PD. Fially, sice the colums of P are liearly idepedet vectors i we kow that they will form a basis for ad so by Theorem 8 from the Fudametal Subspaces sectio we kow that P must be ivertible ad hece we have, P AP D where D is a ivertible matrix. Therefore A is diagoalizable. Let s take a look at a couple of examples. 7 Paul Dawkis 3

32 Example Fid a matrix P that will diagoalize each of the followig matrices. 4 (a) A 6 5 (b) A Solutio Okay, provided we ca fid 3 liearly idepedet eigevectors for each of these we ll have a pretty easy time of this sice we kow that that the colums of P will the be these three eigevectors. Nicely eough for us, we did exactly this i the Example 6 of the previous sectio. At the time is probably seemed like there was o reaso for writig dow specific eigevectors for each eigevalue, but we did it for the problems i this sectio. So, i each case we ll just go back to Example 6 ad pull the eigevectors from that example ad form up P. (a) This was part (a) from Example 6 ad so P is, P 5 We ll leave it to you to verify that we get, P AP (b) This was part (c) from Example 6 so P is, P Agai, we ll leave it to you to verify that, P AP Paul Dawkis 3

33 Example Neither of the followig matrices are diagoalizable (a) A 3 4 (b) A 3 Solutio To see that either of these are diagoalizable simply go back to Example 6 i the previous sectio to see that either matrix has 3 liearly idepedet eigevectors. I both cases we have oly two liearly idepedet eigevectors ad so either matrix is diagoalizable. For referece purposes. Part (a) of this example matches part(b) of Example 6 ad part (b) of this example matches part (d) of Example 6. We did t actually do ay of the work here for these problems so let s summarize up how we eed to go about fidig P, provided it exists of course. We first fid the eigevalues for the matrix A ad the for each eigevalue fid a basis for the eigespace correspodig to that eigevalue. The set of basis vectors will the serve as a set of liearly idepedet eigevectors for the eigevalue. If, after we ve doe this work for all the eigevalues we have a set of eigevectors the A is diagoalizable ad we use the eigevectors to form P. If we do t have a set of eigevectors the A is ot diagoalizable. Actually, we should be careful here. I the above statemet we assumed that if we had eigevectors that they would be liearly idepedet. We should always verify this of course. There is also oe case were we ca guaratee that we ll have liearly idepedet eigevectors. Theorem If v, v,, vk are eigevectors of A correspodig to the k distict eigevalues λ, λ,, λ the they form a liearly idepedet set of vectors. k Proof : We ll prove this by assumig that v, v,, vk are i fact liearly depedet ad from this we ll get a cotradictio ad we we ll see that v, v,, vk must be liearly idepedet. So, assume that v, v,, vk form a liearly depedet set. Now, sice these are eigevectors we kow that they are all o-zero vectors. This meas that the set { v } must be a liearly idepedet set. So, we kow that there must be a liearly idepedet subset of, v v k { v, v,, vp} v v,,. So, let p be the largest iteger such that set. Note that we must have p < k v is a liearly idepedet because we are assumig that,,, are liearly depedet. Therefore we kow that if we take the ext vector v ad add it to our liearly idepedet vectors, { v, v,, v p}, the set { v, v,, vp, p+ } depedet set. p+ v k v will be a liearly 7 Paul Dawkis 3

34 So, if we kow that { v, v,, vp, vp+ } scalars c, c,, c p, c p +, ot all zero so that, is a liearly depedet set we kow that there are c v + c v + + c v + c + v + p p p p () Now, multiply this by A to get, cav + cav + + c Av + c Av p p p+ p+ We kow that the vi are eigevectors of A correspodig to the eigevalue λ i ad so we kow that Av λiv i. Usig this gives us, cλ v + c λ v + + c λ v + c λ v () p p p p+ p+ p+ Next, multiply both sides of () by λ p + to get, cλ p+ v+ cλ p+ v + + cpλp+ vp + cp+ λp+ vp+ ad subtract this from (). Doig this gives, c λ λ v + c λ λ v + + c λ λ v + c λ λ v ( p+ ) ( p+ ) p( p p+ ) p p+ ( p+ p+ ) c( λ λ p+ ) v+ c( λ λ p+ ) v + + cp( λ p λ p+ ) p Now, recall that we assumed that { v, v,, vp} + v were a liearly idepedet set ad so the coefficiets here must all be zero. Or, c λ λ c λ λ c λ λ ( p+ ) ( p+ ) p( p p+ ) However the eigevalues are distict ad so the oly way all these ca be zero is if, c c cp Pluggig these values ito () gives us but, v p+ c + v + p p is a eigevector ad hece is ot the zero vector ad so we must have c + p So, what have we show to this poit? We ll we ve just see that the oly possible solutio to cv+ cv + + cpvp + c p + vp + is c c cp c p + This however would mea that the set { v, v,, vp, vp+ } is liearly idepedet ad we assumed that at least some of the scalars were ot zero. Therefore, this cotradicts the fact that p 7 Paul Dawkis 33

35 we assumed that this set was liearly depedet. Therefore our origial assumptio that, v form a liearly depedet set must be wrog. k We ca the see that v, v,, v k form a liearly idepedet set. v, v, We ca use this theorem to quickly idetify some diagoalizable matrices. Theorem 3 Suppose that A is a matrix ad that A has distict eigevalues, the A is diagoalizable. Proof : By Theorem we kow that the eigevectors correspodig to each of the eigevectors ar e a liearly idepedet set ad the by Theorem above we kow that A will be diagoalizable. We ll close this sectio out with a ice theorem about powers of diagoalizable matrices ad the iverse of a ivertible diagoalizable matrix. Theorem 4 Suppose that A is a diagoalizable matrix ad that P AP D the, (a) If k is ay positive iteger we have, k k A PD P (b) If all the diagoal etries of D are o-zero the A is ivertible ad, A PD P Proof : (a) We ll give the proof for k ad leave it to you to geeralize the proof for larger values of k. Let s start with the followig. ( ) D P AP ( P AP)( P AP) P A( PP ) AP P A( I ) AP P A P So, we ca see that, D P A P We ca fiish this off by multiply the left of this equatio by P ad the right by P to arrive at, A PD P (b) First, we kow that if the mai diagoal etries of a diagoal matrix are o-zero the the diagoal matrix is ivertible. Now, all that we eed to show that, A( PD P ) I This is easy eough to do. All we eed to do is plug i the fact that from part (a), usig k, we have, A PDP So, do the followig. A PD P PDP PD P PD P P D P P DD P PP I let s ( ) ( ) ( ) ( ) So, we re doe. 7 Paul Dawkis 34

36 7 Paul Dawkis 35