MATH 304

Linear Algebra
Lecture 31:
Eigenvalues and eigenvectors.
Characteristic equation.
Eigenvalues and eigenvectors

Definition. Let A be an n×n matrix. A number

λ ∈ R is called an eigenvalue of the matrix A if
Av = λv for a nonzero column vector v ∈ Rn .
The vector v is called an eigenvector of A
belonging to (or associated with) the eigenvalue λ.

Remarks. • Alternative notation:

eigenvalue = characteristic value,
eigenvector = characteristic vector.
• The zero vector is never considered an
eigenvector.
  
2 0
Example. A = .
0 3
      
2 0 1 2 1
= =2 ,
0 3 0 0 0
      
2 0 0 0 0
= =3 .
0 3 −2 −6 −2

Hence (1, 0) is an eigenvector of A belonging to the

eigenvalue 2, while (0, −2) is an eigenvector of A
belonging to the eigenvalue 3.
 
0 1
Example. A = .
1 0
         
0 1 1 1 0 1 1 −1
= , = .
1 0 1 1 1 0 −1 1
Hence (1, 1) is an eigenvector of A belonging to the
eigenvalue 1, while (1, −1) is an eigenvector of A
belonging to the eigenvalue −1.
Vectors v1 = (1, 1) and v2 = (1, −1) form a basis
for R2 . Consider a linear operator L : R2 → R2
given by L(x) = Ax. The matrix
 ofL with respect
1 0
to the basis v1 , v2 is B = .
0 −1
Let A be an n×n matrix. Consider a linear
operator L : Rn → Rn given by L(x) = Ax.
Let v1 , v2 , . . . , vn be a nonstandard basis for Rn
and B be the matrix of the operator L with respect
to this basis.
Theorem The matrix B is diagonal if and only if
vectors v1 , v2 , . . . , vn are eigenvectors of A.
If this is the case, then the diagonal entries of the
matrix B are the corresponding eigenvalues of A.
λ1 O
 
λ2
Avi = λi vi ⇐⇒ B = 
 
 ... 

O λn
Eigenspaces

Let A be an n×n matrix. Let v be an eigenvector

of A belonging to an eigenvalue λ.
Then Av = λv =⇒ Av = (λI )v =⇒ (A − λI )v = 0.
Hence v ∈ N(A − λI ), the nullspace of the matrix
A − λI .
Conversely, if x ∈ N(A − λI ) then Ax = λx.
Thus the eigenvectors of A belonging to the
eigenvalue λ are nonzero vectors from N(A − λI ).
Definition. If N(A − λI ) 6= {0} then it is called
the eigenspace of the matrix A corresponding to
the eigenvalue λ.
How to find eigenvalues and eigenvectors?
Theorem Given a square matrix A and a scalar λ,
the following statements are equivalent:
• λ is an eigenvalue of A,
• N(A − λI ) 6= {0},
• the matrix A − λI is singular,
• det(A − λI ) = 0.

Definition. det(A − λI ) = 0 is called the

characteristic equation of the matrix A.
Eigenvalues λ of A are roots of the characteristic
equation. Associated eigenvectors of A are nonzero
solutions of the equation (A − λI )x = 0.
  
a b
Example. A = .
c d
 
a −λ b 
det(A − λI ) =  
c d −λ
= (a − λ)(d − λ) − bc
= λ2 − (a + d)λ + (ad − bc).
  
a11 a12 a13
Example. A = a21 a22 a23 .
a31 a32 a33
 
 a11 − λ a 12 a 13

 
det(A − λI ) =  a21 a22 − λ a23 
 a31 a32 a33 − λ 
= −λ3 + c1 λ2 − c2 λ + c3 ,
where c1 = a11 + a22 + a33 (the trace of A),
     
 a11 a12   a11 a13   a22 a23 
c2 =  + + ,
a21 a22   a31 a33   a32 a33 
c3 = det A.
 Theorem. Let A = (aij ) be an n×n matrix.
Then det(A − λI ) is a polynomial of λ of degree n:
det(A − λI ) = (−1)n λn + c1 λn−1 + · · · + cn−1 λ + cn .
Furthermore, (−1)n−1 c1 = a11 + a22 + · · · + ann
and cn = det A.

Definition. The polynomial p(λ) = det(A − λI ) is

called the characteristic polynomial of the matrix A.

Corollary Any n×n matrix has at most n

eigenvalues.
  
2 1
Example. A = .
1 2  
2−λ 1 
Characteristic equation: 
 1
 = 0.
2−λ
(2 − λ)2 − 1 = 0 =⇒ λ1 = 1, λ2 = 3.
    
1 1 x 0
(A − I )x = 0 ⇐⇒ =
1 1 y 0
    
1 1 x 0
⇐⇒ = ⇐⇒ x + y = 0.
0 0 y 0
The general solution is (−t, t) = t(−1, 1), t ∈ R.
Thus v1 = (−1, 1) is an eigenvector associated
with the eigenvalue 1. The corresponding
eigenspace is the line spanned by v1 .
     
−1 1 x 0
(A − 3I )x = 0 ⇐⇒ =
1 −1 y 0
    
1 −1 x 0
⇐⇒ = ⇐⇒ x − y = 0.
0 0 y 0

The general solution is (t, t) = t(1, 1), t ∈ R.

Thus v2 = (1, 1) is an eigenvector associated with
the eigenvalue 3. The corresponding eigenspace is
the line spanned by v2 .
  
2 1
Summary. A = .
1 2
• The matrix A has two eigenvalues: 1 and 3.
• The eigenspace of A associated with the
eigenvalue 1 is the line t(−1, 1).
• The eigenspace of A associated with the
eigenvalue 3 is the line t(1, 1).
• Eigenvectors v1 = (−1, 1) and v2 = (1, 1) of
the matrix A form an orthogonal basis for R2 .
• Geometrically, the mapping x 7→ Ax is a stretch
by a factor of 3 away from the line x + y = 0 in
the orthogonal direction.