IBMathsDPApplications & Interpretation (AI)HLRevision Notes1. Number & AlgebraEigenvalues & EigenvectorsThe Characteristic Polynomial, Eigenvalues & Eigenvectors

The Characteristic Polynomial, Eigenvalues & Eigenvectors (DP IB Applications & Interpretation (AI)): Revision Note

Naomi C

Written by: Naomi C

Reviewed by: Dan Finlay

Updated on

Characteristic polynomials

What is the characteristic polynomial of a matrix?

  • The characteristic polynomial of an n cross times n matrix bold italic A is:

p left parenthesis lambda right parenthesis equals det space left parenthesis lambda bold italic I minus bold italic A right parenthesis

Examiner Tips and Tricks

In this course you will only be expected to find the characteristic equation for a 2 cross times 2 matrix and this will always be a quadratic.

How do I find the characteristic polynomial?

  • STEP 1
    Write lambda bold italic I minus bold italic A

    • Remember that the identity matrix must be of the same order as bold italic A

    • e.g. Error converting from MathML to accessible text.

  • STEP 2
    Find the determinant of lambda bold italic I minus bold italic A using the formula given to you in the formula booklet

    • det space bold italic A equals vertical line bold italic A vertical line equals a d minus b c

    • e.g. det space open parentheses table row cell lambda minus 4 end cell cell negative 3 end cell row 1 lambda end table close parentheses equals open parentheses straight lambda minus 4 close parentheses open parentheses straight lambda close parentheses minus open parentheses negative 3 close parentheses open parentheses 1 close parentheses

  • STEP 3
    Simplify the polynomial 

    • e.g. the characteristic polynomial of open parentheses table row 4 3 row cell negative 1 end cell 0 end table close parentheses is lambda squared minus 4 lambda plus 3

Examiner Tips and Tricks

You need to remember the characteristic equation as it is not given in the formula booklet.

Worked Example

Find the characteristic polynomial of the following matrix

bold italic A equals open parentheses table row 5 4 row 3 1 end table close parentheses.

1-8-1-ib-ai-hl-eigenvalues--eigenvectors-we-1-solution

Eigenvalues & eigenvectors

What are eigenvalues and eigenvectors of a matrix?

  • An eigenvector of a matrix is a non-zero vector that gives a scalar multiple when multiplied by the matrix

  • The corresponding eigenvalue is the value of the scalar multiple

  • If bold italic A bold italic x equals lambda bold italic xwhen bold italic x is a non-zero vector and lambda a constant

    • bold italic x is an eigenvector of the matrix bold italic A

    • lambda is the corresponding eigenvalue of the matrix bold italic A

  • If bold italic x is an eigenvector then any scalar multiple is also an eigenvector with the same eigenvalue

    • For each eigenvalue there are an infinite number of corresponding eigenvectors

  • For example, open parentheses table row 4 3 row cell negative 1 end cell 0 end table close parentheses open parentheses table row cell negative 3 end cell row 1 end table close parentheses equals open parentheses table row cell negative 9 end cell row 3 end table close parentheses equals 3 open parentheses table row cell negative 3 end cell row 1 end table close parentheses

    • open parentheses table row cell negative 3 end cell row 1 end table close parentheses is an eigenvector with eigenvalue 3

How do you find the eigenvalues of a matrix?

  • The eigenvalues of matrix bold italic A are found by solving the characteristic polynomial of the matrix

    • This is because bold italic A bold italic x equals lambda bold italic x can be written as open parentheses lambda bold italic I minus bold italic A close parentheses bold italic x equals bold 0

  • For this course, as the characteristic polynomial will always be a quadratic, the polynomial will always generate one of the following:

    • two real and distinct eigenvalues,

    • one real repeated eigenvalue or

    • complex eigenvalues

  • For example, the characteristic polynomial of open parentheses table row 4 3 row cell negative 1 end cell 0 end table close parentheses is lambda squared minus 4 lambda plus 3

    • lambda squared minus 4 lambda plus 3 equals 0 rightwards double arrow lambda equals 1 comma space 3

    • 1 and 3 are eigenvalues

How do I find an eigenvector of a matrix for a given eigenvalue?

Two distinct eigenvalues

  • STEP 1
    Write bold italic x equals open parentheses table row x row y end table close parentheses

  • STEP 2
    Substitute the eigenvalue into the equation left parenthesis lambda bold italic I minus bold italic A right parenthesis bold italic x equals bold 0 and form two equations in terms of x and y

    • The two equations will be scalar multiples of each other

      • e.g. Error converting from MathML to accessible text.

      • For lambda equals 1, Error converting from MathML to accessible text. gives negative 3 x minus 3 y equals 0 and x plus y equals 0

      • Both give y equals negative x

  • STEP 3
    Set one of the variables equal to a non-zero value

    • It is easiest to use x equals 1

      • e.g. if x equals 1 then y equals negative 1

      • open parentheses table row 1 row cell negative 1 end cell end table close parentheses is an eigenvector corresponding to the eigenvalue 1

Examiner Tips and Tricks

  • You can do a quick check on your calculated eigenvalues as the values along the leading diagonal of the matrix you are analysing should sum to the total of the eigenvalues for the matrix

Worked Example

Find the eigenvalues and associated eigenvectors for the following matrices.

a) bold italic A equals open parentheses table row 5 4 row 3 1 end table close parentheses  .

1-8-1-ib-ai-hl-eigenvalues--eigenvectors-we-2ai-solution
1-8-1-ib-ai-hl-eigenvalues--eigenvectors-we-2aii-solution

b) bold italic B equals open parentheses table row 1 cell negative 5 end cell row 2 3 end table close parentheses .

1-8-1-ib-ai-hl-eigenvalues--eigenvectors-we-2bi-solution
1-8-1-ib-ai-hl-eigenvalues--eigenvectors-we-2bii-solution

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

    Test yourselfFlashcards
    Previous:Solving Systems of Linear Equations with MatricesNext:Diagonalisation & Powers of Matrices
    Author
    Reviewer
    Naomi C

    Author: Naomi C

    Expertise: Maths Content Creator

    Naomi graduated from Durham University in 2007 with a Masters degree in Civil Engineering. She has taught Mathematics in the UK, Malaysia and Switzerland covering GCSE, IGCSE, A-Level and IB. She particularly enjoys applying Mathematics to real life and endeavours to bring creativity to the content she creates.

    Dan Finlay

    Reviewer: Dan Finlay

    Expertise: Maths Subject Lead

    Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.