Lecture 20
5 5 Complex Eigenvalues
As we saw the eigenvalues of an nxn matrix are found by finding the roots of
the characteristic polynomial We also know from algebra that many polynomials
don't have real roots e.g x't I o In this case the eigenvalues willbe complex
numbers
Example 1
a Find the eigenvalues of A If to
b Show that i and II are eigenvectors of A
solution
a We solve the characteristic equation
o det A XI
X I
l X
X2 t l
X t i Kil i where i l
Hence the eigenvalues of A are X i
Halil Ii oh I if
Thus il is an eigenvector of A corresponding to the eigenvalue i
at 3 Ii II it til it
Thus f il is of A an eigenvector corresponding to the eigenvalue i
, Remark
To explain why A doesn't have any real eigenvalues note that A is a
rotation matrix by 900 counterclockwise
K il ki 7
Hence there are no nonzero vectors which get mapped to a real mulitiple of
themselves and hence no real eigenvalues or eigenvectors in 1122
In Example 1 we did not need to find the eigenvectors ourselves they were
provided in the question Example 2 shows how we can find complex eigenvectors
Example 2
Let A I 575 Y Findtheeigenvalues and correspondingeigenvectors of A
Solution
The characteristic equation of A is
o
detfooo5
IS II x o.s x n x o.ts Ko.co
Xd 1.6X t 1
From the quadratic formula X L 1.6 Il 16 2 4 0.8 I 0.6 i
We calculate the eigenspace of 0.8 o bi Hence we must solve the
homogeneous system
A 0.8 o bi I e
Now A lo s o.si I f8575 f
b
fo8 o o.g.oo.fi
a'Is o si
Usually we would need to use standard row reduction techniques at this point
However this is quite difficult with complex numbers Fortunately there is a
trick we can use here
5 5 Complex Eigenvalues
As we saw the eigenvalues of an nxn matrix are found by finding the roots of
the characteristic polynomial We also know from algebra that many polynomials
don't have real roots e.g x't I o In this case the eigenvalues willbe complex
numbers
Example 1
a Find the eigenvalues of A If to
b Show that i and II are eigenvectors of A
solution
a We solve the characteristic equation
o det A XI
X I
l X
X2 t l
X t i Kil i where i l
Hence the eigenvalues of A are X i
Halil Ii oh I if
Thus il is an eigenvector of A corresponding to the eigenvalue i
at 3 Ii II it til it
Thus f il is of A an eigenvector corresponding to the eigenvalue i
, Remark
To explain why A doesn't have any real eigenvalues note that A is a
rotation matrix by 900 counterclockwise
K il ki 7
Hence there are no nonzero vectors which get mapped to a real mulitiple of
themselves and hence no real eigenvalues or eigenvectors in 1122
In Example 1 we did not need to find the eigenvectors ourselves they were
provided in the question Example 2 shows how we can find complex eigenvectors
Example 2
Let A I 575 Y Findtheeigenvalues and correspondingeigenvectors of A
Solution
The characteristic equation of A is
o
detfooo5
IS II x o.s x n x o.ts Ko.co
Xd 1.6X t 1
From the quadratic formula X L 1.6 Il 16 2 4 0.8 I 0.6 i
We calculate the eigenspace of 0.8 o bi Hence we must solve the
homogeneous system
A 0.8 o bi I e
Now A lo s o.si I f8575 f
b
fo8 o o.g.oo.fi
a'Is o si
Usually we would need to use standard row reduction techniques at this point
However this is quite difficult with complex numbers Fortunately there is a
trick we can use here
