Course 111: Algebra, 2 March 2007
1. Apply the orthogonal diagonalisation process to


5 0 6


 0 11 6 
6 6 −2
The eigenvalues are determined from the characteristic equation:
(5−λ) ((11 − λ)(−2 − λ) − 36)+6(−6(11−λ)) = λ3 −14λ2 −49λ+686
By trial and error (enumerating the factors of 686 and finding one that
satisfies the polynomial in λ) λ = 7 is a factor.
Then by polynomial division you get
(λ − 7)(λ2 − 7λ − 98) = λ3 − 14λ2 − 49λ + 686
and
λ2 − 7λ − 98 = (λ − 14)(λ + 7)
The eigenvalues are −7, 7, 14.
The corresponding eigenvectors are solutions, x1 , x2 and x3 , to the
matrix-vector equations (A − 7I)x1 = 0, (A + 7I)x2 = 0 and (A −
14I)x3 = 0, where I is the identity matrix.
The solutions are
for λ = 7 x = (3, 3/2, 1)
for λ = −7 x = (−1/2, −1/3, 1)
for λ = 14 x = (2/3, 2, 1)
Then since the eigenvectors are all distinct the eigenvalues are orthogonal (check that this is true) and to make them orthonormal we just
need to divide each one by its norm.
||x1 || =
q
32
||x2 || =
q
(−1/2)2 + (−1/3)2 + 12 = 7/6,
||x3 || =
q
(2/3)2 + 22 + 12 = 7/3
+
(3/2)2
+
12
=
q
23/2,
Then A is orthogonalised by the matrix

q


3/ 23/2 −1/2/(7/6) 2/3/(7/3) 

q



 3/2/ 23/2 −1/3/(7/6)
=
2/(7/3)



q
1/ 23/2
1/(7/6)
1/(7/3)
and

√
3 2
√
23
√3
√46
√2
23
− 37
− 27
6
7
2
7
6
7
3
7





7 0 0

P T AP = D = 
 0 −7 0 
0 0 14
2. Suppose that A and B are both orthogonal n × n square matrices.
Prove that
• A−1 is an orthogonal matrix
Proof
Since A is orthogonal its column vectors form an orthogonal (in
fact they are orthonormal, but we only need orthogonal for this)
set of vectors. Now, by the definition of orthogonal matrices,
A−1 = AT . But this means that the rows of A−1 are the columns
of A and so are an orthogonal set of vectors and therefore (as
proved in the notes) A−1 is also an orthogonal matrix.
• AB is an orthogonal matrix
Hint: the norm preserving property can be useful here
Proof:
Consider ||ABx|| = ||A(Bx)||, where x ∈ Rn . A is an orthogonal
matrix so it preserves norms (proved in lectures). Therefore
||ABx|| = ||A(Bx)|| = ||Bx||
Now, B is also orthogonal and also preserves norms, so
||ABx|| = ||Bx|| = ||x||
Therefore the product AB preserves norms and therefore is itself
an orthogonal matrix.
• Either det(A) = 1 or det(A) = −1
Proof
Since A is orthogonal, AAT = I where I is the identity, n × n
matrix. Taking determinants on both sides
det(AAT ) = det(I) = 1
(1)
Then using properties of the determinant we rewrite the right
hand side of Eq.(1) as
det(A)det(AT ) = 1
det(A)det(A) = 1
[det(A)]2 = 1
and therefore det(A) = ±1.