IBMathsDPApplications & Interpretation (AI)HLRevision Notes1. Number & AlgebraEigenvalues & EigenvectorsDiagonalisation & Powers of Matrices

Diagonalisation & Powers of Matrices (DP IB Applications & Interpretation (AI)): Revision Note

Naomi C

Written by: Naomi C

Reviewed by: Dan Finlay

Updated on

Diagonalisation

What is a diagonal matrix

  • A non-zero square matrix is considered to be diagonal if all elements are zero except the elements along its leading diagonal

    • e.g. open parentheses table row 1 0 0 row 0 3 0 row 0 0 cell negative 1 end cell end table close parentheses is diagonal but open parentheses table row 1 2 0 row 0 3 0 row 0 0 cell negative 1 end cell end table close parentheses is not

What is a diagonalisable matrix?

  • Matrix bold italic M is diagonalisable if there exists a matrix bold italic P such that

    • bold italic D equals bold italic P to the power of negative 1 end exponent bold italic M bold italic P is diagonal

  • This can also be written as bold italic M equals bold italic P bold italic D bold italic P to the power of negative 1 end exponent

Examiner Tips and Tricks

You will only need to be able to diagonalise matrices 2 cross times 2 matrices with real, distinct eigenvalues.

If there is only one eigenvalue, the matrix is either already diagonalised or cannot be diagonalised.

Diagonalisation of matrices with complex or imaginary eigenvalues is outside the scope of the course.

How can I diagonalise a matrix?

  • Consider the matrix bold italic M which has

    • real distinct eigenvalues lambda subscript 1 and lambda subscript 2 

    • with corresponding eigenvectors bold italic x subscript 1 equals open parentheses table row cell x subscript 1 end cell row cell y subscript 1 end cell end table close parentheses and bold italic x subscript 2 equals open parentheses table row cell x subscript 2 end cell row cell y subscript 2 end cell end table close parentheses

  • You can diagonalise matrix bold italic M using

    • bold italic P equals open parentheses table row cell x subscript 1 end cell cell x subscript 2 end cell row cell y subscript 1 end cell cell y subscript 2 end cell end table close parentheses

    • bold italic D equals open parentheses table row cell lambda subscript 1 end cell 0 row 0 cell lambda subscript 2 end cell end table close parentheses

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

    • the eigenvalues are 1 and 3

      • bold italic D equals open parentheses table row 1 0 row 0 3 end table close parentheses

    • corresponding eigenvectors are open parentheses table row 1 row cell negative 1 end cell end table close parentheses and open parentheses table row cell negative 3 end cell row 1 end table close parentheses

      • bold italic P equals open parentheses table row 1 cell negative 3 end cell row cell negative 1 end cell 1 end table close parentheses

      • bold italic P to the power of negative 1 end exponent equals negative 1 half open parentheses table row 1 3 row 1 1 end table close parentheses

    • open parentheses table row 1 0 row 0 3 end table close parentheses equals negative 1 half open parentheses table row 1 3 row 1 1 end table close parentheses open parentheses table row 4 3 row cell negative 1 end cell 0 end table close parentheses open parentheses table row 1 cell negative 3 end cell row cell negative 1 end cell 1 end table close parentheses

Examiner Tips and Tricks

Remember to use the formula booklet for the determinant and inverse of a matrix.

Worked Example

The matrix bold italic M equals open parentheses table row 5 4 row 3 1 end table close parentheses has the eigenvalues lambda subscript 1 equals 7 and lambda subscript 2 equals negative 1 with eigenvectors bold italic x subscript 1 equals open parentheses table row 2 row 1 end table close parentheses and bold italic x subscript 2 equals open parentheses table row 2 row cell negative 3 end cell end table close parentheses respectively.

Show that bold italic P subscript 1 equals open parentheses table row 2 2 row 1 cell negative 3 end cell end table close parentheses and bold italic P subscript 2 equals open parentheses table row 2 2 row cell negative 3 end cell 1 end table close parentheses both diagonalise bold italic M.

1-8-2-ib-ai-hl-applications-of-matrices-we-1-solution

Matrix powers

How can I find powers of a diagonalisable matrix?

  • Write the matrix in diagonalised form

    • bold italic M equals bold italic P bold italic D bold italic P to the power of negative 1 end exponent

  • Squaring this gives:

    • bold italic M squared equals bold italic P bold italic D bold italic P to the power of negative 1 end exponent bold italic P bold italic D bold italic P to the power of negative 1 end exponent

    • which simplifies to bold italic M squared equals bold italic P bold italic D squared bold italic P to the power of negative 1 end exponent

  • Powers can be found as the product of three matrices

    • bold italic M to the power of n equals bold italic P bold italic D to the power of n bold italic P to the power of negative 1 end exponent

Examiner Tips and Tricks

You are given this formula in the formula booklet.

  • Finding higher powers of a diagonal matrix is straight forward

    • open parentheses table row a 0 row 0 b end table close parentheses to the power of n equals open parentheses table row cell a to the power of n end cell 0 row 0 cell b to the power of n end cell end table close parentheses

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

    • open parentheses table row 4 3 row cell negative 1 end cell 0 end table close parentheses to the power of 4 equals negative 1 half open parentheses table row 1 cell negative 3 end cell row cell negative 1 end cell 1 end table close parentheses open parentheses table row 1 0 row 0 3 end table close parentheses to the power of 4 open parentheses table row 1 3 row 1 1 end table close parentheses

    • open parentheses table row 4 3 row cell negative 1 end cell 0 end table close parentheses to the power of 4 equals negative 1 half open parentheses table row 1 cell negative 3 end cell row cell negative 1 end cell 1 end table close parentheses open parentheses table row 1 0 row 0 81 end table close parentheses open parentheses table row 1 3 row 1 1 end table close parentheses

    • open parentheses table row 4 3 row cell negative 1 end cell 0 end table close parentheses to the power of 4 equals open parentheses table row 121 120 row cell negative 40 end cell cell negative 39 end cell end table close parentheses

Worked Example

The matrix bold italic M equals open parentheses table row 3 cell negative 2 end cell row cell negative 4 end cell 1 end table close parentheses has the eigenvalues lambda subscript 1 equals negative 1 and lambda subscript 2 equals 5 with eigenvectors bold italic x subscript 1 equals open parentheses table row 1 row 2 end table close parentheses and bold italic x subscript 2 equals open parentheses table row 1 row cell negative 1 end cell end table close parentheses respectively.

a) Show that bold italic M to the power of n can be expressed as 

bold italic M to the power of n equals negative 1 third open parentheses table row cell left parenthesis negative left parenthesis negative 1 right parenthesis to the power of n minus 2 left parenthesis 5 right parenthesis to the power of n right parenthesis end cell cell left parenthesis negative left parenthesis negative 1 right parenthesis to the power of n plus left parenthesis 5 right parenthesis to the power of n right parenthesis end cell row cell left parenthesis negative 2 left parenthesis negative 1 right parenthesis to the power of n plus 2 left parenthesis 5 right parenthesis to the power of n right parenthesis end cell cell left parenthesis negative 2 left parenthesis negative 1 right parenthesis to the power of n minus left parenthesis 5 right parenthesis to the power of n right parenthesis end cell end table close parentheses

1-8-2-ib-ai-hl-applications-of-matrices-we-2a-solution

b) Hence find bold italic M to the power of 5.

1-8-2-ib-ai-hl-applications-of-matrices-we-2b-solution

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    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.