Lecture 11 Orthogonal transformations
A linear transformation A of a Euclidean space is called orthogonal if it preserves the scalar product of any two vectors x and y of the space, i.e. (Ax, Ay) = (x, y).
The length of a vector x for an orthogonal transformation is not changed, i.e. |Ax| = |x|.
Thus,
The last equality implies that an orthogonal transformation A does not change the angle between any two vectors x and y.
An orthogonal transformation transfers any orthonormal basis in orthonormal. Conversely, if a linear transformation transfers an orthonormal basis in orthonormal then it is orthogonal.
Example.
The scalar product of elements 
and
in
the linear hull L
= L(sin x, cos x)
has been introduced by the formula: 
Prove that a linear transformation D (differentiation) acting in L is orthogonal.
Solution: Find Df1 and Df2:
By
definition of scalar product 
Then
Thus, 
,
i.e. D
is an orthogonal transformation in L.
Quadratic forms
A
quadratic
form
of real variables
is a polynomial of the second degree according to these variables
which does not contain a free term and terms of the first degree.
If
is a quadratic form of variables
and
is a real number then
If
n
= 2 then
If
n
= 3 then
Further we give all necessary formulations and definitions for a quadratic form of three variables.
A
matrix
with
for all
is calledmatrix
of quadratic form
and the corresponding determinant – determinant
of this quadratic
form.
Since
A
is a symmetric matrix then the roots
and
of the characteristic equation
are real numbers.
Let
be
normalized eigenvectors corresponding to characteristic numbers
in an orthonormal basis
Then the vectors
form an orthonormal basis.
The
matrix
is the transition matrix from the basis
to
the basis
Formulas of transformation of coordinates at transition to the new orthonormal basis have the following form:
Transforming
by these formulas the quadratic form
we
obtain the following quadratic form:
.
It does not contain terms with
We
say that a quadratic form
has
been reduced to a canonical
type
by an orthogonal transformation B.
Here
arguments were performed in assumption that characteristic numbers
are
different. At solving exercises we show how it should act if there
are identical numbers among characteristic ones.
Example.
Reduce the quadratic form
to
a canonical type.
Solution:
Here
Compose
the characteristic equation:
or
Solving
this equation we obtain the characteristic numbers
Find
eigenvectors. If
we obtain the following system of equations:
.
Thus,
Assuming
we
have
,
i.e. an eigenvector corresponding to the characteristic number
is
If
we obtain the following system of equations:
.
And
consequently we obtain an eigenvector
corresponding
to
In
order to normalize the vectors u
and v
we should take
Thus,
we have found normalized eigenvectors:
The
transition matrix from the orthonormal basis
to
the orthonormal basis
is
the following:
Then formulas of transformation of coordinates are the following:
Thus,
This
result can be obtained at once since
Example.
Reduce the quadratic form
to
a canonical type.
Solution:
Here
Compose
the characteristic equation:
Solving
this equation we obtain the characteristic numbers
Find
eigenvectors. If
we obtain the following system of equations:
.
It
is reduced to one equation
The
solution of this system can be written as:
Thus,
we obtain the family of eigenvectors
depending
on two parameters a
and b.
If
we obtain the following system of equations:
.
Solving
for example the last two equations we have
or
Thus,
we obtain the one-parameter family of eigenvectors
Let’s
select two orthogonal vectors from the family of eigenvectors
.
Assuming for example a
=
0, b
= 1, we obtain an eigenvector
Choose parameters a
and
b
in order to hold the equality (u,
u1)
= 0. Then we obtain the equation
i.e. 4a
+ 5b
= 0. Now we can take a
= 5, b
= – 4. Thus, we find another vector of the considered family:
Thus,
we obtain three pairwise orthogonal vectors:
The eigenvectors u1
and u2
correspond to the characteristic number
= 7, and the eigenvector v
corresponds to the characteristic number
= – 2 (for c
=
1).
Normalizing
these vectors we obtain a new orthonormal basis and the transition
matrix to a new basis is the following:
Applying the formulas of transformation of coordinates
to
our quadratic form we obtain
Law of inertia of quadratic forms: Both the number of positive canonical coefficients and the number of negative canonical coefficients are constant and don’t depend on way of reducing a quadratic form to a canonical type.