1. No category

Orthogonal Transformations
1. Suppose A is an n × m matrix and B is an m × p matrix. Decide whether each of the following
expressions makes sense; if so, what size is the matrix?
(a) (AB)T
Solution. This makes sense; the product AB is an n × p matrix, so the transpose (AB)T is a
p × n matrix.
(b) AT B T
Solution. This doesn’t make sense: AT is an m × n matrix and B T is a p × m matrix, so the
product AT B T is not defined (unless n happens to equal p).
(c) B T AT
Solution. This makes sense; B T is a p × m matrix and AT is an m × n matrix, so B T AT is a
p × n matrix. In fact, B T AT is equal to (AB)T .
2. (a) If M is a matrix whose columns are orthonormal, what can you say about M T M ?
Solution. Let ~u1 , . . . , ~um be the columns of M ; we know they are orthonormal. Then,
 T
 T

~u1
~u1 ~u1 · · · ~uT1 ~um

 
.. 
..
M T M =  ...  ~u1 · · · ~um =  ...
.
. 
T
T
T
~um
~um ~u1 · · · ~um ~um
That is, M T M is the m × m matrix whose (i, j) entry is ~uTi ~uj = ~ui · ~uj . But since ~u1 , . . . , ~um are
orthonormal,
(
0 if i 6= j
~ui · ~uj =
1 if i = j
So, the diagonal entries of M T M are 1, and all other entries are 0; that is, M T M is the
identity matrix Im .
(b) If M is a square matrix whose columns are orthonormal, is M invertible? If so, what is its
inverse?
A square matrix with orthonormal columns is, rather confusingly, called an orthogonal matrix.
There is no special term for a square matrix with orthogonal columns.
Solution. If M is an n × n matrix with orthonormal columns, then its columns must form a basis
of Rn (n linearly independent vectors must form a basis of Rn , and we have seen in a previous
class that orthonormal vectors are linearly independent, so M must be invertible.
The fact that M T M = In then shows that the inverse of M is M T .
3. If M and N are orthogonal matrices, which of the following matrices are orthogonal as well?
(a) M −1
Solution. Yes, it is orthogonal. Here are two arguments. If M is orthogonal, then it preserves
length, and so does M −1 , which means that M −1 is orthogonal. One could also check that
(M −1 )T M −1 = (M M T )−1 = I −1 = I
1
(b) M T
Solution. Note that since M T M = I and M is square, then M T = M −1 . So by previous item it
is orthogonal.
This implies that if M is orthogonal, not only the columns of M form an orthonormal basis, but
also the rows of M form another orthonormal basis.
(c) M N
Solution. If M and N preserve lengths, the composition M N also does, and hence is orthonormal. Alternatively, one could check that
(M N )T M N = (N T M T )M N = N T (M T M )N = N T N = I
(d) M + I
Solution. Not always! For example, N = I is orthonormal, but N + I = 2I is not.
4. Which of the following are orthogonal transformations?
(a) Reflection about a line in R2 .
Solution. Yes, preserves lenghts.
(b) Dilation.
Solution. No, does not preserve lengths (unless dilation factor is ±1).
(c) Rotation.
Solution. Yes, rotations preserve lengths.
(d) Projection onto a plane.
Solution. No, it is not even invertible!
(e) Shear.
Solution. Does not preserve lenghts (unless the shear factor is zero, that is, it is the identity
transformation).
5. Let M be an orthogonal 2 × 2 matrix.
(a) Show that M is a reflection about a line or a rotation.
a c
Solution. The matrix has to be of the form M =
with the constraints that
b d
a
The only solutions are
b
a2 + b2 = c+ d2 = 1 and ac + bd = 0
−b
a b
or
which represent a rotation or a reflection respectively.
a
b −a
2
Geometrically, M e1 and M e2 are a pair of orthogonal vectors of length one. If their order did not
flip, then the rotation that takes e1 to M e1 also takes e2 to M e2 . Hence M is this rotation.
Otherwise, if their order flipped, let L be the line bissecting the angle formed by e1 and M e1 .
Then it bissects the angle for e2 and M e2 as well. So the reflection about L takes e1 to M e1 and
e2 to M e2 . Hence M is this reflection!
(b) If A, B : R2 → R2 are reflections, transformation is the composition AB?
Solution. The composition is still orthogonal. Hence it is either a rotation or a reflection. As
reflections flip the orientation, applying two refletions reverts to the origianl orientation. So AB
preserves orientation and therefore is not a reflection. Hence it is a rotation!
Bonus question: what is the angle of the rotation? Think in terms of the angle between the
reflection lines for A and B.
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