Understanding Engineering Mathematics

=λ^2 − 7 λ+ 12

=(λ− 3 )(λ− 4 )= 0

So the eigenvalues areλ=3,λ=4.
To find corresponding eigenvectors we must solve the system of homogeneous equations
Au=λuwith each of the eigenvalues substituted. Forλ=3 the equations become

[A− 3 I]u=

[

3 − 3

2 − 2

][

u 1

u 2

]

= 0

(

u=

[

u 1

u 2

])

This gives two equations foru 1 ,u 2 which actually reduce to one, with solution

u 1 =u 2

In general then we get an eigenvector corresponding toλ=3 of the form

u=

[

u

u

]

=u

[

1

1

]

whereuis an arbitrary number, roughly analogous to the arbitrary constant in inte-
gration – we choose it to suit our purposes, usually to give a column vector with unit
magnitude (the magnitude of a column vector, like that of a vector (331

➤
), is the square
root of the sum of the squares of its components – if that puts you in mind of Pythagoras
again then so it should, the magnitude of a column vector is in a sense the ‘length’ of
a vector it might represent). In the above case choosinguto makeuof unit magnitude
gives

u=

1

√

2

[

1

1

]

Repeating the above arguments for the other eigenvalueλ=4 you should find a corre-
sponding eigenvector

v=v

[

3

2

]

wherevis again arbitrary. Takingvto be of unit magnitude gives

v=

1

√

13

[

3

2

]

Hopefully, you will agree that actually finding the eigenvalues and eigenvectors is relatively
routine – in this case simply expanding a determinant and solving a quadratic equation,
then solving a system of linear equations. In real engineering problems one might have
hundreds of components ordegrees of freedom, resulting in matrices with thousands of
entries, and things are then not so simple – we have no hope of expanding the corre-
sponding determinants or solving the resulting equations so easily. In such cases we have
to resort to completely different (numerical) methods for finding eigenvalues – but the
basic principles are still the same.