Finding Eigenvalues and Eigenvectors

Introduction

Link 1 and link 2 give good tutorials on how to get the maximum and minimum eigenvalues and eigenvectors by iterative processes.  However, I have noticed that eigenvalue converges to its final value much more quickly than the eigenvector.  In this short document I will demonstrate how to get eigenvectors that are in agreement with their eigenvalues.

Math

            It is obvious that eigenvectors can be multiplied by any constant and they still are perfectly good solutions of the eigenvector equation.  There I choose to make the first element of any eigenvector unity (1) and scale the remainder of the elements with respect to this first element.  To demonstrate the procedure for getting the remaining elements, I will assume a 3x3 matrix but it will work for any square matrix:

( m 11 m 12 m 13 m 21 m 22 m 23 m 31 m 32 m 33 )( 1 v 2 v 3 )=λ( 1 v 2 v 3 ) MathType@MTEF@5@5@+= feaahiart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaabmaabaqbae qabmWaaaqaaiaad2gadaWgaaWcbaGaaGymaiaaigdaaeqaaaGcbaGa amyBamaaBaaaleaacaaIXaGaaGOmaaqabaaakeaacaWGTbWaaSbaaS qaaiaaigdacaaIZaaabeaaaOqaaiaad2gadaWgaaWcbaGaaGOmaiaa igdaaeqaaaGcbaGaamyBamaaBaaaleaacaaIYaGaaGOmaaqabaaake aacaWGTbWaaSbaaSqaaiaaikdacaaIZaaabeaaaOqaaiaad2gadaWg aaWcbaGaaG4maiaaigdaaeqaaaGcbaGaamyBamaaBaaaleaacaaIZa GaaGOmaaqabaaakeaacaWGTbWaaSbaaSqaaiaaiodacaaIZaaabeaa aaaakiaawIcacaGLPaaadaqadaqaauaabeqadeaaaeaacaaIXaaaba GaamODamaaBaaaleaacaaIYaaabeaaaOqaaiaadAhadaWgaaWcbaGa aG4maaqabaaaaaGccaGLOaGaayzkaaGaeyypa0Jaeq4UdW2aaeWaae aafaqabeWabaaabaGaaGymaaqaaiaadAhadaWgaaWcbaGaaGOmaaqa baaakeaacaWG2bWaaSbaaSqaaiaaiodaaeqaaaaaaOGaayjkaiaawM caaaaa@5E3C@  

(1.1)

We want to solve equation (1.1) for v2 and v3.  To do that we form the matrix product for the lower two rows times our eigenvector and subtract out λ from the resulting diagonal elements.

( m 22 −λ m 23 m 32 m 33 −λ )( v 2 v 3 )=−( m 21 m 31 ) MathType@MTEF@5@5@+= feaahiart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaabmaabaqbae qabiGaaaqaaiaad2gadaWgaaWcbaGaaGOmaiaaikdaaeqaaOGaeyOe I0Iaeq4UdWgabaGaamyBamaaBaaaleaacaaIYaGaaG4maaqabaaake aacaWGTbWaaSbaaSqaaiaaiodacaaIYaaabeaaaOqaaiaad2gadaWg aaWcbaGaaG4maiaaiodaaeqaaOGaeyOeI0Iaeq4UdWgaaaGaayjkai aawMcaamaabmaabaqbaeqabiqaaaqaaiaadAhadaWgaaWcbaGaaGOm aaqabaaakeaacaWG2bWaaSbaaSqaaiaaiodaaeqaaaaaaOGaayjkai aawMcaaiabg2da9iabgkHiTmaabmaabaqbaeqabiqaaaqaaiaad2ga daWgaaWcbaGaaGOmaiaaigdaaeqaaaGcbaGaamyBamaaBaaaleaaca aIZaGaaGymaaqabaaaaaGccaGLOaGaayzkaaaaaa@557E@  

(1.2)

We can solve for v2 and v3 by inverting the 2x2 matrix as shown below:

 

MathType@MTEF@5@5@+= feaahiart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOaa@35E5@

 

   ( m 22 −λ m 23 m 32 m 33 −λ ) −1 ( m 22 −λ m 23 m 32 m 33 −λ )( v 2 v 3 )=−   ( m 22 −λ m 23 m 32 m 33 −λ ) −1 ( m 21 m 31 ) MathType@MTEF@5@5@+= feaahiart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaabaaaaaaaaape GaaiiOaiaacckapaWaaeWaaeaafaqabeGacaaabaGaamyBamaaBaaa leaacaaIYaGaaGOmaaqabaGccqGHsislcqaH7oaBaeaacaWGTbWaaS baaSqaaiaaikdacaaIZaaabeaaaOqaaiaad2gadaWgaaWcbaGaaG4m aiaaikdaaeqaaaGcbaGaamyBamaaBaaaleaacaaIZaGaaG4maaqaba GccqGHsislcqaH7oaBaaaacaGLOaGaayzkaaWaaWbaaSqabeaacqGH sislcaaIXaaaaOWaaeWaaeaafaqabeGacaaabaGaamyBamaaBaaale aacaaIYaGaaGOmaaqabaGccqGHsislcqaH7oaBaeaacaWGTbWaaSba aSqaaiaaikdacaaIZaaabeaaaOqaaiaad2gadaWgaaWcbaGaaG4mai aaikdaaeqaaaGcbaGaamyBamaaBaaaleaacaaIZaGaaG4maaqabaGc cqGHsislcqaH7oaBaaaacaGLOaGaayzkaaWaaeWaaeaafaqabeGaba aabaGaamODamaaBaaaleaacaaIYaaabeaaaOqaaiaadAhadaWgaaWc baGaaG4maaqabaaaaaGccaGLOaGaayzkaaGaeyypa0JaeyOeI0Ydbi aacckacaGGGcWdamaabmaabaqbaeqabiGaaaqaaiaad2gadaWgaaWc baGaaGOmaiaaikdaaeqaaOGaeyOeI0Iaeq4UdWgabaGaamyBamaaBa aaleaacaaIYaGaaG4maaqabaaakeaacaWGTbWaaSbaaSqaaiaaioda caaIYaaabeaaaOqaaiaad2gadaWgaaWcbaGaaG4maiaaiodaaeqaaO GaeyOeI0Iaeq4UdWgaaaGaayjkaiaawMcaamaaCaaaleqabaGaeyOe I0IaaGymaaaakmaabmaabaqbaeqabiqaaaqaaiaad2gadaWgaaWcba GaaGOmaiaaigdaaeqaaaGcbaGaamyBamaaBaaaleaacaaIZaGaaGym aaqabaaaaaGccaGLOaGaayzkaaaaaa@80D8@  

(1.3)

Since the matrix product on the left side of equation (1.3)is the unit matrix, we are left with the equation:

  ( v 2 v 3 )=−   ( m 22 −λ m 23 m 32 m 33 −λ ) −1 ( m 21 m 31 ) MathType@MTEF@5@5@+= feaahiart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaabaaaaaaaaape GaaiiOaiaacckapaWaaeWaaeaafaqabeGabaaabaGaamODamaaBaaa leaacaaIYaaabeaaaOqaaiaadAhadaWgaaWcbaGaaG4maaqabaaaaa GccaGLOaGaayzkaaGaeyypa0JaeyOeI0YdbiaacckacaGGGcWdamaa bmaabaqbaeqabiGaaaqaaiaad2gadaWgaaWcbaGaaGOmaiaaikdaae qaaOGaeyOeI0Iaeq4UdWgabaGaamyBamaaBaaaleaacaaIYaGaaG4m aaqabaaakeaacaWGTbWaaSbaaSqaaiaaiodacaaIYaaabeaaaOqaai aad2gadaWgaaWcbaGaaG4maiaaiodaaeqaaOGaeyOeI0Iaeq4UdWga aaGaayjkaiaawMcaamaaCaaaleqabaGaeyOeI0IaaGymaaaakmaabm aabaqbaeqabiqaaaqaaiaad2gadaWgaaWcbaGaaGOmaiaaigdaaeqa aaGcbaGaamyBamaaBaaaleaacaaIZaGaaGymaaqabaaaaaGccaGLOa Gaayzkaaaaaa@5C3B@  

(1.4)

which results in an eigenvector that agrees perfectly with our Power Method eigenvalue.

Again I want to emphasize that this method for eigenvector calculation works for any size square matrix.