Math 2331 { Linear Algebra

1 DiagonalizationMath 2331 Linear DiagonalizationJiwen HeDepartment of Mathematics, University of jiwenhe/math2331 Jiwen He, University of HoustonMath 2331, Linear Algebra1 / DiagonalizationDiagonalization Theorem DiagonalizationDiagonalizationMatrix Powers: ExampleDiagonalizableDiagonalization TheoremDiagonalization: ExamplesJiwen He, University of HoustonMath 2331, Linear Algebra2 / DiagonalizationDiagonalization Theorem ExamplesDiagonalizationThe goal here is to develop a useful factorizationA=PDP 1,whenAisn n. We can use this to computeAkquickly for matrixDis adiagonalmatrix ( entries off the maindiagonal are all zeros).

2 Powers of Diagonal MatrixDkis trivial to compute as the following example [5 00 4]. ComputeD2andD3. In general, what isDk,wherekis a positive integer?Jiwen He, University of HoustonMath 2331, Linear Algebra3 / DiagonalizationDiagonalization Theorem ExamplesDiagonalization (cont.)Solution:D2=[5 00 4][5 00 4]=[00]D3=D2D=[520042][5 00 4]=[00]and in general,Dk=[5k004k]Jiwen He, University of HoustonMath 2331, Linear Algebra4 / DiagonalizationDiagonalization Theorem ExamplesMatrix Powers: ExampleExampleLetA=[6 123]. Find a formula forAkgiven thatA=PDP 1whereP=[1 11 2],D=[5 00 4]andP 1=[2 1 11].

3 Solution:A2=(PDP 1)(PDP 1)=PD(P 1P)DP 1=PDDP 1=PD2P 1 AgainA3=A2A=(PD2P 1)(PDP 1)=PD2(P 1P)DP 1=PD3P 1 Jiwen He, University of HoustonMath 2331, Linear Algebra5 / DiagonalizationDiagonalization Theorem ExamplesMatrix Powers: Example (cont.)In general,Ak=PDkP 1=[1 11 2][5k004k][2 1 11]=[2 5k 4k 5k+ 4k2 5k 2 4k 5k+ 2 4k].DiagonalizableA square matrixAis said to bediagonalizableifAis similar to adiagonal matrix, ifA=PDP 1wherePis invertible andDisa diagonal He, University of HoustonMath 2331, Linear Algebra6 / DiagonalizationDiagonalization Theorem ExamplesDiagonalizableWhen isAdiagonalizable?

4 (The answer lies in examining theeigenvalues and eigenvectors ofA.)Note that[6 123][11]= 5[11],[6 123][12]= 4[12]Altogether[6 123][1 11 2]=[5 45 8]=[1 11 2][00]Equivalently,[6 123]=[1 11 2][5 00 4][1 11 2] 1 Jiwen He, University of HoustonMath 2331, Linear Algebra7 / DiagonalizationDiagonalization Theorem ExamplesDiagonalizable (cont.)In general,A[v1v2 vn]=[v1v2 vn] 10 00 2 n and if[v1v2 vn]is invertible,Aequals[v1v2 vn] 10 00 2 n [v1v2 vn] 1 Jiwen He, University of HoustonMath 2331, Linear Algebra8 / DiagonalizationDiagonalization Theorem ExamplesDiagonalization TheoremTheorem (Diagonalization)

5 Ann nmatrixAis diagonalizable if and only ifAhas nlinearly independent fact,A=PDP 1, withDa diagonal matrix, if and only ifthe columns ofParenlinearly independent eigenvectors this case, the diagonal entries ofDare eigenvalues ofAthat correspond, respectively, to the eigenvectors He, University of HoustonMath 2331, Linear Algebra9 / DiagonalizationDiagonalization Theorem ExamplesDiagonalization: ExampleExampleDiagonalize the following matrix, if 20 012 1 1 0 1 Step 1. Find the eigenvalues of (A I) = det 2 0012 1 101 = (2 )2(1 ) = ofA: = 1 and = He, University of HoustonMath 2331, Linear Algebra10 / DiagonalizationDiagonalization Theorem ExamplesDiagonalization: Example (cont.)

6 Step 2. Find three linearly independent eigenvectors of solving(A I)x=0,for each value of , we obtain the following:Basis for = 1:v1= 0 11 Basis for = 2:v2= 010 ,v3= 101 Jiwen He, University of HoustonMath 2331, Linear Algebra11 / DiagonalizationDiagonalization Theorem ExamplesDiagonalization: Example (cont.)Step 3: Construct P from the vectors in step 0 0 1 1 101 01 Step 4: Construct D from the corresponding 1 0 00 2 00 0 2 Jiwen He, University of HoustonMath 2331, Linear Algebra12 / DiagonalizationDiagonalization Theorem ExamplesDiagonalization: Example (cont.)

7 Step 5: Check your work by verifying thatAP=PDAP= 20 012 1 1 0 1 0 0 1 1 101 01 = 00 2 1 20102 PD= 0 0 1 1 101 01 1 0 00 2 00 0 2 = 00 2 1 20102 Jiwen He, University of HoustonMath 2331, Linear Algebra13 / DiagonalizationDiagonalization Theorem ExamplesDiagonalization: ExampleExampleDiagonalize the following matrix, if 2 4 60 2 20 0 4 .Since this matrix is triangular, the eigenvalues are 1= 2 and 2= 4. By solving (A I)x=0for each eigenvalue, we wouldfind the following: 1= 2 :v1= 100 , 2= 4 :v2= 511 Every eigenvector ofAis a multiple ofv1orv2which means thereare not three linearly independent eigenvectors ofAand byTheorem 5,Ais not He, University of HoustonMath 2331, Linear Algebra14 / DiagonalizationDiagonalization Theorem ExamplesDiagonalization: ExampleExampleWhy isA= 2 0 02 6 03 2 1 diagonalizable?

8 Solution:SinceAhas three eigenvalues: 1=, 2=, 3=and since eigenvectors corresponding to distinct eigenvalues arelinearly independent,Ahas three linearly independent eigenvectorsand it is therefore (6)Ann nmatrix withndistinct eigenvalues is He, University of HoustonMath 2331, Linear Algebra15 / DiagonalizationDiagonalization Theorem ExamplesDiagonalization: ExampleExampleDiagonalize the following matrix, if 200 00 20 024 12 2 0000 2 Solution:Eigenvalues: 2 and 2 (each with multiplicity 2).Solving (A I)x=0yields the following eigenspace basis for = 2 :v1= 10 60 v2= 0130 Jiwen He, University of HoustonMath 2331, Linear Algebra16 / DiagonalizationDiagonalization Theorem ExamplesDiagonalization: Example (cont.)

9 Basis for = 2 :v3= 0010 v4= 0001 {v1,v2,v3,v4}is linearly independent P= [v1v2v3v4] is invertible A=PDP 1, whereP= 1 0 0 00 1 0 0 6 3 1 00 0 0 1 andD= 20 0 00 2 0 000 2 000 0 2 .Jiwen He, University of HoustonMath 2331, Linear Algebra17 / DiagonalizationDiagonalization Theorem ExamplesDiagonalization: TheoremTheorem (7)LetAbe ann nmatrix whose distinct eigenvalues are 1,.., k p, the dimension of the eigenspace for kis lessthan or equal to the multiplicity of the eigenvalue matrixAis diagonalizable if and only if the sum of thedimensions of the distinct eigenspaces equalsn, and thishappens if and only if the dimension of the eigenspace foreach kequals the multiplicity of diagonalizable and kis a basis for the eigenspacecorresponding to kfor eachk, then the total collection ofvectors in the sets 1.

10 , pforms an eigenvector basis He, University of HoustonMath 2331, Linear Algebra18 / 18