Orthogonal # ‚ # Matrices
Suppose T is an 8 ‚ 8 orthogonal matrix. Since T is square and T " œ T X , we have
" œ detÐT T " Ñ œ detÐT T X Ñ œ Ðdet T ÑÐdetT X Ñ œ Ðdet T Ñ# , so det T œ „ "Þ
Theorem Suppose T œ 
+
-
,
is orthogonal. If
.
ñ if det T œ ", then the mapping B Ä T B is a rotation
ñ if det T œ  ",
then the mapping B Ä T B is a reflection across
a line P through the origin ÐP is one of the eigenspaces of T Ñ
Proof Since T is orthogonal, the columns are orthonormal vectors in ‘# , so
cos )
+
,
+
+#  - # œ " œ ,#  . # . Therefore  and   are on the unit circle. Write   œ 
,
sin ) 
.
+
,
+
where ) is the angle between the positive B-axis and  . Because  and   are
.
,
orthogonal,   must be in one of the two positions U or Uw (see picture); therefore
.
cos Ð)  1# Ñ
cos Ð)  $#1 Ñ
,
,
œ
or
œ
.
 .   sin Ð)  1 Ñ 
. 
sin Ð)  $#1 Ñ 
#
So T œ 
cos )
sin )
cos .
where either . œ ) 
sin . 
i) If . œ )  1# , then
cos )
so T œ 
sin )
cos .
sin .
1
#
or . œ ) 
$1
# Þ
œ cos ) cos 1#  sin ) sin 1#
œ sin ) cos 1#  cos ) sin 1#
 sin )
ß which is a rotation matrix.
cos ) 
œ  sin )
œ
cos )ß
ii) If . œ ) 
$1
# ,
det
the same trig identities give T œ 
cos )
sin )
sin )
Þ In this case,
 cos ) 
cos )  sin )
œ -#  ", so T has eigenvalues -" œ " and
 cos )  - 
sin )
-# œ  ". Pick corresponding eigenvectors ," and ,# .
Then ," † ,#
œ
T ," † T ,#
(because T is orthogonal)
œ
Ð"Ñ ," † Ð  "Ñ,#
œ  ," † ,# ß so ," † ,# œ !Þ
Since ," and ,# are orthogonal, they are linearly independent. (Another reason: they are
linearly independent because they are eigenvectors corresponding to different
eigenvectors.)
Therefore Ö," ß ,# × is an eigenvector basis for ‘# . So any B in ‘# can be written as
B œ -" ,"  -# ,# ,
so
T B œ -" Ð"Ñ,"  -# Ð  "Ñ,# œ -" ,"  -# ,# Þ
This formula shows that the mapping B È T B is a reflection across the line
(eigenspace) P œ SpanÖ," × Ðsee the figure below).