5 Eigenvalues and Eigenvectors
5.3
DIAGONALIZATION
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DIAGONALIZATION
7 2
Example 2: Let A = . Find a formula for
−4 1
given that A = PDP , where
−1
Ak,
1 1 5 0
P= and D =
−1 −2 0 3
Solution: The standard formula for the inverse of a
2 × 2 matrix yields
2 1
P =
−1
−1 −1
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DIAGONALIZATION
Then, by associativity of matrix multiplication,
A2 ( PDP
= −1
)( PDP −1 ) PD
= ( P −1 P ) DP −1 PDDP −1
I
1 1 5 2
0 2 1
PD
= P
2 −1
−1 −2 0 32 −1 −1
Again,
A3 ( PDP
= −1
) A2 ( PD
P −1 )=
P D 2 P −1 PDD
= P
2 −1
PD 3 P −1
I
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DIAGONALIZATION
In general, for k ≥ 1,
1 1 5 k
0 2 1
=A PD
= P k
−1 −2 0
k −1
k
3 −1 −1
2⋅5 − 3 k k
5 −3
k k
= k k
2 ⋅ 3 − 2 ⋅ 5 2⋅3 − 5
k k
A square matrix A is said to be diagonalizable if A is
similar to a diagonal matrix, that is, if A = PDP
−1
for some invertible matrix P and some diagonal,
matrix D.
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THE DIAGONALIZATION THEOREM
Theorem 5: An n × n matrix A is diagonalizable if
and only if A has n linearly independent eigenvectors.
In fact, A = PDP , with D a diagonal matrix, if
−1
and only if the columns of P and n linearly
independent eigenvectors of A. In this case, the
diagonal entries of D are eigenvalues of A that
correspond, respectively, to the eigenvectors in P.
In other words, A is diagonalizable if and only if
there are enough eigenvectors to form a basis of ℝ𝑛𝑛 .
We call such a basis an eigenvector basisof.ℝ𝑛𝑛
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THE DIAGONALIZATION THEOREM
Proof: First, observe that if P is any n × n matrix with
columns v1, …, vn, and if D is any diagonal matrix with
diagonal entries λ1, …, λn, then
AP A= [ v v v ] [ Av Av Av ] (1)
1 2 n 1 2 n
while
λ1 0 0
0 λ 0
PD
P= 2
[λ v
1 1
λ 2 v2 λ n vn ]
0 0 λ
n (2)
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THE DIAGONALIZATION THEOREM
Now suppose A is diagonalizable and A = PDP −1
. Then
right-multiplying this relation by P, we have
AP = PD.
In this case, equations (1) and (2) imply that
[ Av 1
Av 2 Av n ] = [ λ1v1 λ 2 v2 λ n vn ]
(3)
Equating columns, we find that
=Av1 λ=v , Av 2 λ 2 v 2 ,=
1 1
, Av n λ n v n (4)
Since P is invertible, its columns v1, …, vn must be
linearly independent.
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THE DIAGONALIZATION THEOREM
Also, since these columns are nonzero, the equations
in (4) show that λ1, …, λn are eigenvalues and v1, …,
vn are corresponding eigenvectors.
This argument proves the “only if ” parts of the first
and second statements, along with the third statement,
of the theorem.
Finally, given any n eigenvectors v1, …, vn, use them
to construct the columns of P and use corresponding
eigenvalues λ1, …, λn to construct D.
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THE DIAGONALIZATION THEOREM
By equations (1)–(3), AP = PD .
This is true without any condition on the
eigenvectors.
If, in fact, the eigenvectors are linearly independent,
then P is invertible (by the Invertible Matrix
Theorem), and AP = PD implies that A = PDP −1.
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DIAGONALIZING MATRICES
Example 3: Diagonalize the following matrix, if
possible. 1 3 3
3 −5 −3
A =−
3 3 1
That is, find an invertible matrix P and a diagonal
matrix D such that A = PDP . −1
Solution: There are four steps to implement the
description in Theorem 5.
Step 1. Find the eigenvalues of A.
Here, the characteristic equation turns out to involve a
cubic polynomial that can be factored:
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DIAGONALIZING MATRICES
0=det( A − λI ) =− λ − 3λ + 4
3 2
=−(λ − 1)(λ + 2) 2
The eigenvalues are λ = 1 and λ = −2 .
Step 2. Find three linearly independent eigenvectors
of A.
Three vectors are needed because A is a 3 × 3 matrix.
This is a critical step.
If it fails, then Theorem 5 says that A cannot be
diagonalized.
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DIAGONALIZING MATRICES
1
Basis for λ = 1: v1 = −1
1
−1 −1
Basis for λ =
−2 : v 2 =
1 and v3 = 0
0 1
You can check that {v1, v2, v3} is a linearly
independent set.
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DIAGONALIZING MATRICES
Step 3. Construct P from the vectors in step 2.
The order of the vectors is unimportant.
Using the order chosen in step 2, form
1 −1 −1
P= [v 1
v2 v3 ] = −1 1 0
1 0 1
Step 4. Construct D from the corresponding eigenvalues.
In this step, it is essential that the order of the eigenvalues
matches the order chosen for the columns of P.
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DIAGONALIZING MATRICES
Use the eigenvalue λ = −2 twice, once for each of the
eigenvectors corresponding to λ = −2 :
1 0 0
= D 0 −2 0
0 0 −2
To avoid computing P , simply verify that AD = PD.
−1
Compute
1 3 3 1 −1 −1 1 2 2
AP =−3 −5 −3 −1 1 0 =−1 −2 0
3 3 1 1 0 1 1 0 −2
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DIAGONALIZING MATRICES
1 −1 −1 1 0 0 1 2 2
PD =− 1 1 0 0 −2 0 =− 1 −2 0
1 0 1 0 0 −2 1 0 −2
Theorem 6: An n × n matrix with n distinct
eigenvalues is diagonalizable.
Proof: Let v1, …, vn be eigenvectors corresponding
to the n distinct eigenvalues of a matrix A.
Then {v1, …, vn} is linearly independent, by
Theorem 2 in Section 5.1.
Hence A is diagonalizable, by Theorem 5.
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MATRICES WHOSE EIGENVALUES ARE NOT
DISTINCT
It is not necessary for an n × n matrix to have n
distinct eigenvalues in order to be diagonalizable.
The 3 x 3 matrix in Example 3 is diagonalizable even
though it has only two distinct eigenvalues.
If an n × n matrix A has n distinct eigenvalues, with
corresponding eigenvectors v1, …, vn, and if
P = [ v1 vn2 ], then P is automatically invertible
because its columns are linearly independent, by
Theorem 2.
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MATRICES WHOSE EIGENVALUES ARE NOT
DISTINCT
When A is diagonalizable but has fewer than n
distinct eigenvalues, it is still possible to build P in
a way that makes P automatically invertible, as the
next theorem shows.
Theorem 7: Let A be an n × n matrix whose distinct
eigenvalues are λ1, …, λp.
a. For 1 ≤ k ≤ p , the dimension of the eigenspace
for λk is less than or equal to the multiplicity
of the eigenvalue λk.
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MATRICES WHOSE EIGENVALUES ARE NOT
DISTINCT
λk
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