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by: Jayde Lang


Jayde Lang
Rice University
GPA 3.86


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Class Notes
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This 6 page Class Notes was uploaded by Jayde Lang on Monday October 19, 2015. The Class Notes belongs to MATH 355 at Rice University taught by Staff in Fall. Since its upload, it has received 13 views. For similar materials see /class/224939/math-355-rice-university in Mathematics (M) at Rice University.

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Date Created: 10/19/15
The question we want to answer now is the following If A is not similar to a diagonal matrix then what is the simplest matrix that A is similar to Before we can provide the answer we will have to introduce a few de nitions De nition A square matrix A is block diagonal if A has the form A1 0 0 0 A2 0 A 7 la a 11 where each A is a square matrix and the diagonals of each A lie on the diagonal of A Each 0 is a zero matrix of appropriate size Each A is called a block of A Technically every square matrix is a block diagonal matrix But we only use the terminology when there are at least two blocks in the matrix Here is an example of a typical7 block diagonal matrix 1 7 1 0 0 OOHOW OOMMM OCTJOOO 0 0 0 0 2 H0000 000033 This matrix has blocks of size 3 1 and 2 as we move down the diagonal The three blocks in this matrix are 132 21 A1 7 0 2 A26 A33 3 112 The lines are just drawn to illustrate the blocks De nition A Jordan block with value A is a square upper triangular matrix whose entries are all A on the diagonal all 1 on the entries immediately above the diagonal and 0 elsewhere A1000 0A100 00A00 JA 000A1 0000A Here7s what the Jordan blocks of size 1 2 and 3 look like A10 A 0A1 00A De nition A Jordan form matrix is a block diagonal matrix whose blocks are all Jordan blocks For example every diagonal p gtlt p matrix is a Jordan form with p 1 gtlt 1 Jordan blocks Here are some more interesting examples again lines have been drawn to illustrate the blocks 11000 0 0100 0 0 2100 0031 00 0210l l 0003 0 0 0020 0000710 l00M2l l00000 ql Now here7s the big theorem that answers our rst question Theorem 1 Let A be ap gtlt p matrip Then there is a Jordan form matricc J that is similar to A In fact we can be more speci c than that Theorem 2 Let A be a p gtlt p matrip with 5 distinct eigenvalues A1 As Let each A haue algebraic multiplicity m and geometric multiplicity Mi Then A is similar to a Jordan form matricc J1 0 0 J 0 12 15 5 39 1 where 1Mmmwm 2 For each A the number of Jordan blocks in J with value A is equal to Mi 3 A appears on the diagonal ofJ eccactly in times Further the matricc J is unique up to re ordering the Jordan blocks on the diagonal This is a pretty complicated theorem and we aren7t going to try to prove it here But we will learn a method for nding the Jordan form of a matrix A and also nding the nonsingular matrix Q such that J Q IAQ Algorithm for the Jordan Form of A H E0 00 r U 03 I Compute the distinct eigenvalues A12 As along with the associated algebraic multiplicities 77117712 m5 and geometric multiplicities M1412 Ms Treat each eigenvalue in turn For a given eigenvalue A of algebraic multiplicity m and geometric multiplicity p we start computing the E spaces and their dimensions The k th E space is Eff x A7 ADkx 0 So E is just EA and we build from there We stop when we get to an Eff that has dimension m the algebraic multiplicity of A We make a diagram of boxes as follows Compute the numbers d1 dim Ei d2 dim E i dim Ei dk dim Eff i dim Eff l Now make a diagram with d1 boxes in the rst row d2 boxes in the second row and so on For example if d1 4d2 2d3 2d4 1 then we get a diagram DEED DE DE D We are going to ll in7 the diagram with vectors as follows Start at the bottom ofthe diagram and ll the boxes in row k with linearly independent vectors that belong to Eff but not Efil Anytime you have a vector 1 in a box the box immediately above it gets lled with the vector A 7 AIM If a box is the lowest in its column and belongs to row 239 ll that box with a new vector from E3 which is linearly independent to both E34 and all the other vectors in row 239 Repeat steps 2 through 4 for each distinct eigenvalue You will get a diagram full of vectors for each one Make a matrix Q as follows For each eigenvalue consider the associated diagram The vectors in the boxes become the columns of Q as follows Start at the top of the leftmost column and use the vectors as you go down the column When you reach the end of a column go to the next column When you nish one diagram go to rst column of the next diagram This gives the matrix Q The Jordan form of A is given by J Q lAQ But the nice part of the algorithm is that you can compute J Without nding Q In fact J will have one Jordan block for each column of each diagram The value of the block is given by the eigenvalue and the size of the block is equal to the number of squares in the column You put the blocks down the diagonal of J in the same order you chose the vectors in Q Examples 1 A 2 For this matrix7 the characteristic polynomial is 1 A2 so there is one eigenvalue7 A 71 with m 2 Now7 we compute E spaces E11 Solving A Ix 0 gives 3 3 0 1 1 0 3 i3 0 0 0 0 39 1 i t i 1 Erlltlliltlll So d1 M 1 Since dim E11 lt m7 we have to compute another E space E31 Solving A 2X 0 0 0 ltAIgt20 0 so E31 span 6162 and d2 2 7 1 1 Since dim E31 m7 we dont need any more E spaces Since we have d1 17d2 17 our diagram looks like D D We put a vector in the lower box It has to be a vector in E31 that is linearly inde pendent to Ell That7s easy enough 7 how about 01 1 OF Above 111 we have to put A 01 which is i 3 i3 1 i 3 2 3 i3 0 3 39 7 Hence7 our diagram is Qlvz ll Finally7 without computing Q lAQ7 we still know what J looks like There is only one column7 so J is just one Jordan block7 of size 27 with value A 71 475 310 2A 7110 3 2 2 We skip the computation to show that A has only one eigenvalue A 2 of multiplicity 3 Computing E21 1 1 0 0 1 0 0 0 71 71 0 0 7 0 1 0 0 3 2 0 0 0 0 0 0 So E21 is spanned by the vector 0 0 1lT So d1 1 Turning to E5 we solve A 7 22x 0 So E is spanned by 71 71 OF and 0 0 1lT So d2 2 71 We need to compute the E space But computation shows that A 7 2 O the zero matrix so E is spanned by e1e2e3 So d3 1 and we can stop here Our diagram is one column of three boxes The bottom box gets lled with a vector from E that is linearly independent of The vector 01 e1 will work Above there goes 112 A 7 2111 1 7 1 3lT and above that goes 113 A 7 202 0 0 1lT Since our diagram now looks like HEB H 7 03 we get the transition matrix 0 1 1 Q 13 12 01 0 1 0 1 3 0 Again there is only one column so only one Jordan block which has value 2 and size 3 We get the Jordan form matrix J DOM OMH DHO 2 4 78 3 A 0 0 4 0 71 4 You can check that this matrix also has only the eigenvalue 2 with multiplicity 3 We compute the E spaces First for Ezl 04780 01420 A42Iloi 042 440 7 00 00 04120 0000 So E21 is spanned by the two vectors 1 0 OF and 0 2 1T Also d2 2 lt 3 so we need another E space Computing E5 we see that A 7 22 0 so E is spanned by the standard basis and d2 3 7 2 1 We can stop since E has dimension 3 Our diagram looks like where 01 is a vector in E linearly independent of We get 112 A 7 201 and we nally choose 113 E E21 linearly independent of 112 If we start by choosing vl e2 we wind up getting 4 0 1 Q 12 11 13 2 1 0 71 0 0 Finally the diagram tells us that we get 2 Jordan blocks this time Both have value 2 but one is of size 2 and one is of size 1 So 2 1 0 J 0 2 0 0 0 2 We drew the lines just to illustrate the blocks You can check in this example and in all of the previous ones that indeed J Q lAQ


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