Lesson 13
Diagonalization of Symmetric Matrices
Recall. Given an n × n matrix A.
1. Matrix A is diagonalizable if and only if there are n linearly independent eigenvectors.
2. If A has n distinct eigenvalues, then A is diagonalizable.
Definition. A matrix A is symmetric if A⊤ = A.
Theorem. Let A be a symmetric matrix. Then all the eigenvalues of A are real numbers.
Recall.

−2
Consider the matrix A = −2
2
−2
1
−4

2
−4 (see previous topic)
1
• The matrix A is symmetric.
• characteristic polynomial: pA (λ) = (λ + 3)2 (λ − 6) ⇒ eigenvalues: λ = −3(with multiplicity 2), 6
   
 1 
 
−2 
 2
 2 
 1 
• Verify: EA (−3) =Span 1 ,  0  and EA (6) =Span −1 =Span −2






0
1
2
1




2 −2 1
2
5 4
1
• we can take P = 1 0 −2 ⇒ P −1 = −2 4 5
9
0 1
2
1 −2 2
Definition. Let B ∈ Mn×n (R). We say that B is an orthogonal matrix if B ⊤ = B −1 .
Theorem (Orthogonal Diagonalization Theorem). Let A be a symmetric matrix. Then there is an orthogonal
matrix Q such that
Q⊤ AQ = D,
where D is a diagonal matrix whose diagonal entries are the eigenvalues of A.
In the previous example

2
1
−2
9
1
5
4
−2

4 −2
5 −2
2
2
−2
1
−4

2
2
−4 1
1
0
−2
0
1
 
1
−3
−2 =  0
2
0
0
−3
0
where P is not an orthogonal matrix.
Recall.
A basis {e1 , e2 , . . . , en } of Rn is an orthonormal basis if
1. ∥ei ∥ = 1, for every i ∈ {1, 2, . . . , n}
(
0 i ̸= j
2. ei · ej =
1 i=j
Theorem (Characterization of Orthogonal Matrices). The following are equivalent:
1. The matrix Q is orthogonal.
2. The rows of Q form an orthonormal basis.
3. The columns of Q form an orthonormal basis.
Question: How do we find an orthogonal matrix Q such that D = Q−1 AQ = Q⊤ AQ?
Answer: Apply Gram-Schmidt orthogonalization process on the eigenspace basis.
1

0
0 ,
6
     
−2
1 
 2
Example 1. Let S = 1 ,  0  −2 . (Verify) By applying Gram-Schmidt orthogonalization process on S,


0
1
2
we have the orthonormal set

 
 
 
1 
−2
 1 2
1
1
√ 1 , √  4  , −2 .

 5
3
2
0 3 5 5
Therefore, we take
 2
√
 15
√
Q=
 5
0
We have
 2
√
5
 −2
 3 √5
1
3
√1
5
4
√
3 5
−2
3
0
−2
√
3 5
4
√
3 5
5
√
3 5

−2
5 
√
−2

3 5
2
2
3
 2
1
√
3
5

 −2
−2 
−1
⊤
√
⇒
Q
=
Q
=

3 
3 5
2
1
3
3

−2
1
−4

 √2
2  5
√1
−4 
 5
1
0
−2
√
3 5
4
√
3 5
5
√
3 5
√1
5
4
√
3 5
−2
3
0

5 
√

3 5
2
3
1

3
−3

−2 

0
=
3 
2
0
3

0
−3
0

0
0 ,
6
where Q is an orthogonal matrix.
Theorem. Let A be a symmetric matrix. Then the eigenvectors of A associated with different eigenvalues form an
orthogonal set.
1 2
Example 2. Consider the matrix A =
(symmetric):
2 −2
1−λ
2
pA (λ) = det(λI2 − A) = det(A − λI2 ) = det
= (λ + 3)(λ − 2)
2
−2 − λ
4 2
λ = −3 :
∼
2 1
−1 2
λ=2:
∼
2 −4
Exercises. Find an orthogonal matrix Q and a diagonal matrix D such that D = Q⊤ AQ.


0 10 10
0 , pA (x) = −x(x2 − 225)
1. A = 10 5
10 0 −5


11
2 −10
14
5 , pA (x) = −(x − 15)2 (x + 15)
2. A =  2
−10 5 −10


0 0 0 2
0 0 0 0
2

3. A = 
0 0 0 0, pA (x) = x (x − 2)(x + 2)
2 0 0 0
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