5.3 Diagonalization
The goal here is to develop a useful
factorization A = PDP-1 where A is n x n and D
is a diagonal matrix (entries off the main
diagonal are all zeros). We can use the
factorization to compute Ak quickly for large k,
since Dk is easy to compute.
Example: Let
5 0
D
 . Compute D2, D3.
0
4


In general, what is Dk where k is a positive
integer?
2
5
0
5
0

5




2
D 




0 4 0 4  0
2

5
D3  D 2 D  
0
53 0 

3
0 4 
0
2
4 
0  5 0 

2 
4  0 4 
1
In general,
k

5
0
k
D 
k
0 4 
Example: Let
6  1
A

2
3


Find a formula for Ak given: A = PDP-1
 2  1
1 1
5 0 
1
P 
P
D



,
,

1
1
1
2
0
4






Start with symbols, and find a pattern.
Since A = PDP-1,
A2  PDP 1 PDP 1



 PDDP 1
 PD 2 P 1
A3  PD 2 P 1 PDP 1



 PD 2 DP 1
 PD3 P 1
2
So, Ak = PD k P-1.
We can write this explicitly:
6  1
2 3 


1 1 5k


1
2

 0
k
0   2  1

k 
4   1 1 
5k
4 k   2  1
 k

k 

1
1
5
2

4



 2  5k  4 k
 5k  4 k 
 k
k
k
k
2

5

2

4

5

2

4


This gives a nice compact formula which only
takes two matrix multiplications.
Definition: A square matrix is called
diagonalizable if it is similar to a diagonal
1
matrix, that is A  PDP where P is invertible
and D is a diagonal matrix.
3
When is a matrix diagonalizable?
Let’s look at the eigenvalues and
eigenvectors of the previous example.
Notice that:
6  1 1 5
1
2 3  1  5  51 ,

   

1
So 1 is an eigenvector corresponding to the
eigenvalue 5.
Also,
6  1 1 4
1 
 2 3   2   8   4  2  ,

   
 
1 
so  2  is an eigenvector corresponding to
λ = 4.
These vectors are the columns of the matrix
P, and the eigenvalues λ = 5, 4 are the entries
on the diagonal of D.
4
Now we write:
6  1 1 1 5 4
AP  





2
3
1
2
5
8


 

and,
5 4 1 1 5 0
5 8  1 2 0 4  PD .

 


So, AP = PD
6  1 1 1 1 1 5 0
2 3  1 2  1 2 0 4


 


Multiplying by P-1 on the right on both sides of
the equal sign gives: A = PD P-1
6  1 1 1 5 0 1 1
2 3   1 2 0 4 1 2

 



1
5
In general,
1 0
0 
2
Av1 v 2  v n   v1 v 2  v n 
 

0 0


A  v1 v 2  v n 


0
v1
0
0 
 

 n 
And if v1 v 2  v n  is invertible,
1 0  0 
0   0 
2



0
 

 n 
v 2  vn 
1
This brings us to …
The Diagonalization Theorem: An n x n matrix
A is diagonalizable iff A has n linearly
independent eigenvectors.
1
A

PDP
In fact,
with D a diagonal matrix iff
the columns of P are the n linearly
independent eigenvectors of A. In this case,
the diagonal entries of D are the eigenvalues
of A that correspond, respectively, to the
eigenvectors in P.
6
Example: Diagonalize the following matrix if
possible.
 2 0 0
A   1 2 1
 1 0 1
Step 1: Find the eigenvalues of A. Solve
det  A  I   0
0
0 
2  
det  1
2
1   0
  1
0
1   
2   2   1     0
2    1     0
2
  1,2
7
Step 2: Find three linearly independent
eigenvectors of A, if possible.
Solve  A  I x  0 for each value of λ.
For λ = 1:
0 
2  1 0
 A  1I    1 2  1 1 
  1
0 1  1
 1 0 0
  1 1 1
 1 0 0
The augmented matrix for
 1 0 0 0 1 0 0
 1 1 1 0 ~ 0 1 1

 
 1 0 0 0 0 0 0
This gives the solution
A  I x  0 is :
0
0
0
 0 
0
0
x   x3   x3  1  v1   1
 
 x3 
 1 
 1 
8
For λ = 2:
0
0 
2  2
 A  2I    1 2  2 1 
  1
0
1  2
0 0 0
  1 0 1 
 1 0  1
The augmented matrix for  A  I x  0 is:
 0 0 0 0 1 0 1 0
 1 0 1 0 ~ 0 0 0 0

 

 1 0  1 0 0 0 0 0
This gives the solution
 x3 
0
 1
x   x2   x2 1  x3  0 
 x3 
0
 1 
0 
 1
 v 2  1, v 3   0 
0
 1 
9
Step 3: Construct P from the vectors in step
2:
 0 0  1


P   1 1 0 
 1 0 1 
Step 4: Construct D from the corresponding
eigenvalues.
1 0 0
D  0 2 0
0 0 2
Step 5: Use your calculator to check by
verifying that AP = PD
 0 0  2
AP   1 2 0   PD
 1 0 2 
So, A is diagonalizable with
10
 1 0 1
1 0 0
 0 0  1
D  0 2 0 P   1 1 0  P -1   1 1 1

,
,
 1 0 0
0 0 2
 1 0 1 
11
Example:
Diagonalize the following matrix if possible.
2 4 6 
A  0 2 2
0 0 4
Since A is triangular, the eigenvalues are 2
and 4.
The basis for the eigenspace for λ = 2 is
1 
v1  0
0
and the basis for the e-space for λ = 4 is
5


v 2  1
1
Every eigenvector of A is a multiple of v1 or
v2, so A does not have three linearly
independent eigenvectors, so by Th. 5.5, A is
not diagonalizable.
12
Example:
 2 0 0
A  2 6 0
Why is
3 2 1 diagonalizable?
A has three distinct eigenvalues: λ = 2, 6, 1.
Since eigenvectors corresponding the distinct
eigenvalues are linearly independent, by Th.
5.2,  3 linearly independent eigenvectors, so
A is diagonalizable by Th 5.5.
This leads us to:
Theorem 6: An n x n matrix with distinct
eigenvalues is diagonalizable.
Note: The converse is not true. It is not
necessarily true that if a matrix is
diagonalizable, it has n distinct eigenvalues.
Example: Let
 2 0
 0 2
A
 24  12

0
0
0 0

0 0
2 0

0 2
13
λ = –2, 2, each with multiplicity 2.
Find the eigenvectors:
Solving (A – λI)x = 0
Gives the basis for the space of λ = –2
1
0 
0
1
v1    , v 2   
 6
 3
 
 
0
0 
and the basis for the space of λ = 2
0 
0 
0 
0 
v 3   , v 4   
1
0 
 
 
0 
1
These are clearly linearly independent, so
P  v1 v 2 v 3 v 4  is invertible; thus A is
1
diagonalizable, and A  PDP where
14
1
0
P
 6

0
0 0 0
1 0 0
3 1 0

0 0 1
Note the order of the columns.
and
 2 0
 0 2
D
0
0

0
0
0 0

0 0
2 0

0 2
and
1 0
0 1
-1
P 
6  3

0 0
0 0

0 0
1 0

0 1
15