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

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

Naomi C

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Naomi C

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Diagonalisation

What is matrix diagonalisation?

  • A non-zero, square matrix is considered to be diagonal if all elements not along its leading diagonal are zero

  • A matrix bold italic P can be said to diagonalise matrix bold italic M, if bold italic D is a diagonal matrix where bold italic D equals bold italic P to the power of negative 1 end exponent bold italic M bold italic P

  • If matrix bold italic Mhas eigenvalues lambda subscript 1, lambda subscript 2 and eigenvectors bold italic x subscript 1, bold italic x subscript 2and is diagonisable by bold italic P, then

    • bold italic P equals left parenthesis bold italic x subscript 1 bold italic x subscript 2 right parenthesis, where the first column is the eigenvector bold italic x subscript 1and the second column is the eigenvector bold italic x subscript 2

    • 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

  • You will only need to be able to diagonalise 2 cross times 2 matrices

  • You will only need to consider 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

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 left parenthesis bold italic x subscript 1 bold italic x subscript 2 right parenthesis and bold italic P subscript 2 equals left parenthesis bold italic x subscript 2 bold italic x subscript 1 right parenthesis both diagonalise bold italic M.

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

Matrix Powers

One of the main applications of diagonalising a matrix is to make it easy to find powers of the matrix, which is useful when modelling transient situations such as the movement of populations between two towns.

How can the diagonalised matrix be used to find higher powers of the original matrix?

  • The equation to find the diagonalised matrix can be re-arranged for bold italic M:

 bold italic D equals bold italic P to the power of negative 1 end exponent bold italic M bold italic P rightwards double arrow bold italic M equals bold italic P bold italic D bold italic P to the power of negative 1 end exponent

  • Finding higher powers of a matrix when it is diagonalised 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

  • Therefore, we can easily find higher powers of the matrix  using the power formula for a matrix found in the formula booklet:

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

  • If you are asked to show this by hand, don’t forget to use your GDC to check your answer afterwards!

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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    Naomi C

    Author: Naomi C

    Expertise: Maths

    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.