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2.2 Finding the eigenvalues and eigenvectors of a two multiplication two matrix

For a given square matrix, normal cap a, it is possible to solve the equation

normal cap a times bold v equals lamda times bold v
Equation label:(2)

where bold v are column vectors known as eigenvectors and lamda is a scalar called an eigenvalue.

A procedure to find the eigenvectors and eigenvalues of a two multiplication two square matrix normal cap a equals matrix row 1column 1 ab row 2column 1 cd is

  • Solve the quadratic equation lamda squared minus left parenthesis a plus d right parenthesis times lamda plus left parenthesis a times d minus b times c right parenthesis equals zero to find the two values of lamda which are the required eigenvalues.

  • For each eigenvalue found, write down the eigenvector equations

    left parenthesis a minus lamda right parenthesis times x plus b times y equals zero
    c times x plus left parenthesis d minus lamda right parenthesis times y equals zero

  • This pair of equations usually reduces to a single equation that is readily solved for x and y. The eigenvector is given by bold v equals vector element 1 x element 2 y with x and y replaced by their solved values.

  • It is often useful to normalise bold v by writing it as a unit vector. It this case, the unit vector is given by

    bold v sub u equals one divided by Square root of x squared plus y squared times vector element 1 x element 2 y full stop

Exercise 5

Find the eigenvalues and eigenvectors of the following matrix:

normal cap lamda equals matrix row 1column 1 41 row 2column 1 23

Answer

Following the prescription described above: a equals four, b equals one, c equals two and d equals three. So we first need to solve the quadratic equation

lamda squared minus left parenthesis four plus three right parenthesis times lamda plus left parenthesis left parenthesis four multiplication three right parenthesis minus left parenthesis one multiplication two right parenthesis right parenthesis equals zero

which is simply

lamda squared minus seven times lamda plus 10 equals zero

This can be written as

left parenthesis lamda minus two right parenthesis times left parenthesis lamda minus five right parenthesis equals zero

So it has solutions lamda sub one equals two and lamda sub two equals five. These are the two eigenvalues.

We now write the two eigenvector equations:

multiline equation row 1 left parenthesis four minus lamda right parenthesis times x plus one times y equals zero row 2 two times x plus left parenthesis three minus lamda right parenthesis times y equals zero

For eigenvalue lamda sub one equals two these reduce to

multiline equation row 1 two times x plus y equals zero row 2 two times x plus y equals zero

Both equations imply that y equals negative two times x, so x equals one and y equals negative two and the first eigenvector is

bold v sub one equals vector element 1 one element 2 negative two

For eigenvalue lamda sub two equals five these reduce to

multiline equation row 1 negative x plus y equals zero row 2 two times x minus two times y equals zero

Both equations imply that y equals x, so x equals one and y equals one and the second eigenvector is

bold v sub two equals vector element 1 one element 2 one
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