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MATH10212 • Linear Algebra • Brief lecture notes
Eigenvalues and Eigenvectors of n × n Matrices
Now that we have defined the determinant of an n × n matrix, we can continue our discussion of eigenvalues and eigenvectors in a general context.
Recall that a column-vector ~v ∈ Rn is an eigenvector of an n × n matrix A
for eigenvalue λ ∈ R if A~v = λ~v and ~v 6= ~0. Thus, ~v is a non-trivial solution
of the homogeneous linear system (A − λI)~x = λ~x, which is (A − λI)~x = ~0.
It has non-trivial solutions if and only if (A − λI) is non-invertible. By
Theorem 4.6, this is true if and only if det(A − λI) = 0. To summarize:
Theorem. The eigenvalues of a square matrix A are precisely the solutions λ of the equation det(A − λI) = 0.
When we expand det(A − λI), we get a polynomial in λ, called the characteristic polynomial of A. The equation det(A − λI) = 0 is called the characteristic equation of A. For example, if
·
¸
a b
A=
,
c d
its characteristic polynomial is
det(A − λI)
¯
¯ a−λ
b
= ¯¯
c
d−λ
¯
¯
¯
¯
= (a − λ)(d − λ) − bc
= λ2 − (a + d)λ + (ad − bc)
Let’s summarize the procedure we will follow (for now) to find the eigenvalues and eigenvectors (eigenspaces) of a matrix.
Let A be an n × n matrix.
1. Compute the characteristic polynomial det(A − λI) of A.
2. Find the eigenvalues of A by solving the characteristic equation det(A−
λI) = 0 for λ.
3. For each eigenvalue λ, find the null space of the matrix det(A − λI).
This is the eigenspace Eλ , the nonzero vectors of which are the eigenvectors of A corresponding to λ.
4. Find a basis for each eigenspace.


1 2 2
Example. For A = 2 1 2 the characteristic polynomial is
2 2 1
¯
¯
¯1 − λ
2
2 ¯¯
¯
1−λ
2 ¯¯ =
det(A − λI) = ¯¯ 2
¯ 2
2
1 − λ¯
MATH10212 • Linear Algebra • Brief lecture notes
45
(1 − λ)3 + 8 + 8 − 4(1 − λ) − 4(1 − λ) − 4(1 − λ) = · · · = −(λ − 5)(λ + 1)2 . (Of
course, we are lucky to spot factorization here; in general it is polynomial
of degree n for n × n matrix and roots may not be easy to find...). Thus,
roots=eigenvalues are 5 and −1.

 
 
2 2 2
x1
0
Let us find eigenspace E−1 : (A − (−1)I)~x = ~0; 2 2 2 x2  = 0;
x3  0
 2 2  2

 −s − t
x1 = −x2 − x3 , where x2 , x3 are free var.; E−1 =  s  | s, t ∈ R ;


t
   
−1 
 −1



1
a basis of E−1 :
, 0 .


1
0
   

−4 2
2
x1
0
Find the eigenspace E5 : (A − 5I)~x = ~0;  2 −4 2  x2  = 0; solve
2
2 −4 
0 
x
3 
 t

this system....: x1 = x2 = x3 , where x3 is a free var.; E5 = t | t ∈ R ;


t
 
 1 
1 .
a basis of E5 :


1
Definitions. The multiplicity of a given eigenvalue as a root of the characteristic polynomial det(A − λI) is called the algebraic multiplicity of λ.
The dimension of the eigenspace Eλ is called the geometric multiplicity of
λ.
In the above example, e.g., −1 was eigenvalue of algebraic multiplicity
2, and its geom. multiplicity was also 2. It can be shown that the geometric
multiplicity of λ is at most its algebraic multiplicity. And there are examples where geometric multiplicity is less than the algebraic multiplicity.
¸
·
3 2
Example. For A =
the characteristic polynomial is det(A − λI) =
0 3
¯·
¸¯
¯ 3−λ
2 ¯¯
¯
= (3 − λ)2 , so λ = 3 is a repeated root with multiplicity 2.
¯ 0
3−λ ¯
So λ = 3 is an eigenvalue of A of algebraic
·
¸ · multiplicity
¸
· ¸ 2. Let us find the
0
2
x
0
1
eigenspace E3 : (A − λI)~x = ~0 is
=
, so x2 = 0, while x1
0
0
x
0
2
½· ¸
¾
t
is a free var.; thus, E3 =
| t ∈ R has dimension 1, so the geometric
0
multiplicity of the eigenvalue λ = 3 is 1.
Theorem 4.15. The eigenvalues of a triangular matrix are the entries on
its main diagonal.
MATH10212 • Linear Algebra • Brief lecture notes
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Theorem 4.16. A square matrix A is invertible if and only if 0 is not an
eigenvalue of A.
Theorem 4.17.
Revisited
The Fundamental Theorem of Invertible Matrices,
Let A be an n × n matrix. The following statements are equivalent:
a. A is invertible.
b. A~x = ~b has a unique solution for every ~b in Rn .
c. A~x = ~0 has only the trivial solution.
d. The reduced row echelon form of A is In .
e. A is a product of elementary matrices.
f. rank (A) = n
g. nullity (A) = 0
h. The column vectors of A are linearly independent.
i. The column vectors of A span Rn .
j. The column vectors of A form a basis for Rn .
k. The row vectors of A are linearly independent.
l. The row vectors of A span Rn .
m. The row vectors of A form a basis for Rn .
n. det A 6= 0
o. 0 is not an eigenvalue of A.
Theorem 4.18. Let A be a square matrix with eigenvalue λ and corresponding eigenvector ~x.
a. For any positive integer n, λn is an eigenvalue of An with corresponding eigenvector ~x.
b. If A is invertible, then
eigenvector ~x.
1
λ
is an eigenvalue of A−1 with corresponding
c. For any integer n, λn is an eigenvalue of An with corresponding eigenvector ~x.
MATH10212 • Linear Algebra • Brief lecture notes
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Theorem 4.19. Suppose the n×n matrix A has eigenvectors ~v1 , ~v2 , . . . , ~vm
with corresponding eigenvalues λ1 , λ2 , . . . , λm . If ~x is a vector in Rn that
can be expressed as a linear combination of these eigenvectors – say,
~x = c1~v1 + c2~v2 + . . . + cm~vm
then for any integer k,
Ak ~x = c1 λk1 ~v1 + c2 λk2 ~v2 + . . . + cm λkm~vm
Theorem 4.20. Let A be an n × n matrix and let λ1 , λ2 , . . . , λm be distinct
eigenvalues of A with corresponding eigenvectors ~v1 , ~v2 , . . . , ~vm . Then
~v1 , ~v2 , . . . , ~vm are linearly independent.