Mathematical Methods for Physics and Engineering : A Comprehensive Guide

MATRICES AND VECTOR SPACES

8.14 Determination of eigenvalues and eigenvectors

The next step is to show how the eigenvalues and eigenvectors of a givenN×N

matrixAare found. To do this we refer to (8.68) and as in (8.69) rewrite it as

Ax−λIx=(A−λI)x= 0. (8.85)

The slight rearrangement used here is to writexasIx,whereIis the unit matrix

of orderN. The point of doing this is immediate since (8.85) now has the form

of a homogeneous set of simultaneous equations, the theory of which will be

developed in section 8.18. What will be proved there is that the equationBx= 0

only has a non-trivial solutionxif|B|= 0. Correspondingly, therefore, we must

have in the present case that

|A−λI|= 0 , (8.86)

if there are to be non-zero solutionsxto (8.85).

Equation (8.86) is known as thecharacteristic equationforAand its LHS as

thecharacteristicorsecular determinantofA. The equation is a polynomial of

degreeNin the quantityλ.TheNroots of this equationλi,i=1, 2 ,...,N, give

the eigenvalues ofA. Corresponding to eachλithere will be a column vectorxi,

which is theith eigenvector ofAand can be found by using (8.68).

It will be observed that when (8.86) is written out as a polynomial equation in

λ, the coefficient of−λN−^1 in the equation will be simplyA 11 +A 22 +···+ANN

relative to the coefficient ofλN. As discussed in section 8.8, the quantity

∑N
i=1Aii
is thetraceofAand, from the ordinary theory of polynomial equations, will be

equal to the sum of the roots of (8.86):

∑N

i=1

λi=TrA. (8.87)

This can be used as one check that a computation of the eigenvaluesλihas been

done correctly. Unless equation (8.87) is satisfied by a computed set of eigenvalues,

they have not been calculated correctly. However, that equation (8.87) is satisfied is

a necessary, but not sufficient, condition for a correct computation. An alternative

proof of (8.87) is given in section 8.16.

Find the eigenvalues and normalised eigenvectors of the real symmetric matrix

A=





11 3

11 − 3

3 − 3 − 3



.

Using (8.86),
∣
∣
∣∣
∣
∣

1 −λ 13

11 −λ − 3

3 − 3 − 3 −λ

∣

∣

∣∣

∣

∣

=0.