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 51 ,
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
Av1 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
0
