Download Homework 2. Solutions 1 a) Show that (x, y) = x1y1 + x2y2 + x3y3

Homework 2. Solutions
1
a)
b)
c)
d)
Show
Show
Show
Show
that
that
that
that
(x, y) = x1 y 1 + x2 y 2 + x3 y 3 defines a scalar product in R3 .
hx, yi = x1 y 1 + x2 y 2 does not define a scalar product in R3 .
(x, y) = x1 y 1 + x2 y 2 − x3 y 3 does not define a scalar product in R3 .
(x, y) = x1 y 1 + 3x2 y 2 + 5x3 y 3 defines a scalar product in R3 .
e† ) Find necessary and sufficient conditions for entries a, b, c of symmetrical matrix
the formula
µ
1
2
(x, y) = ( x , x )
a
b
b
c
¶µ
y1
y2
µ
a
b
b
c
¶
such that
¶
= ax1 y 1 + b(x1 y 2 + x2 y 1 ) + cx2 y 2
defines scalar product in R2 .
Recall that scalar product on a vector space V is a function B(x, y) = (x, y) on a pair of vectors which
takes real values and satisfies the the following conditions:
1) B(x, y) = B(y, x) (symmetricity condition)
2) B(λx + µy, z) = λB(x, z) + µB(y, z) (linearity condition (with respect to the first argument))
3) B(x, x) ≥ 0 , B(x, x) = 0 ⇔ x = 0 (positive-definiteness condition)
(The linearity condition with respect to the second argument follows from the conditions 2) and 1))
Check all these conditions for (x, y) = x1 y 1 + x2 y 2 + x3 y 3 :
1) (y, x) = y 1 x1 + y 2 x2 + y 3 x3 = x1 y 1 + x2 y 2 + x3 y 3 = (x, y). Hence it is symmetrical.
2) (λx + µy, z) = (λx1 + µy 1 )z 1 + (λx2 + µy 2 )z 2 + (λx3 + µy 3 )z 3 =
= λ(x1 z 1 + x2 z 2 + x3 z 3 ) + µ(y 1 z 1 + y 2 z 2 + y 3 z 3 ) = λ(x, y) + µ(y, z). Hence it is linear.
3) (x, x) = (x1 )2 + (x2 )2 + (x3 )2 ≥ 0. It is non-negative. If x = 0 then (x, x) = 0. If (x, x) =
1 2
(x ) + (x2 )2 + (x3 )2 = 0, then x1 = x2 = x3 = 0, i.e. x = 0. This we proved positive-definiteness.
All conditions are checked. Hence B(x, y) = x1 y 1 + x2 y 2 + x3 y 3 is indeed a scalar product in R3
Remark Note that x1 , x2 , x3 —are components of the vector, do not be confused with exponents!
Show that B(x, y) = x1 y 1 + x2 y 2 does not define scalar product in R3 .
To see that the formula (x, y) = x1 y 1 + x2 y 2 does not define scalar product check the condition 3) of
positive-definiteness: (x, x) = (x1 )2 + (x2 )2 may take zero values for x 6= 0. E.g. if x = (0, 0, −1) (x, x) = 0,
in spite of the fact that x 6= 0. The condition 3) of positive-definiteness is not satisfied. Hence it is not scalar
product.
Show that B(x, y) = x1 y 1 + x2 y 2 − x3 y 3 does not define scalar product in R3 .
To see that the formula (x, y) = x1 y 1 + x2 y 2 − x3 y 3 does not define scalar product check the condition
3): (x, x) = (x1 )2 + (x2 )2 − (x3 )2 may take negative values. E.g. if x = (0, 0, −1) (x, x) = −1 < 0. The
condition 3) of positive-definiteness is not satisfied. Hence it is not scalar product.
Now show that (x, y) = x1 y 1 + 3x2 y 2 + 5x3 y 3 is a scalar product in R3 .
We need to check all the conditions above for scalar product for (x, y) = x1 y 1 + 3x2 y 2 + 5x3 y 3 :
1) (y, x) = y 1 x1 + 3y 2 x2 + 5y 3 x3 = x1 y 1 + 3x2 y 2 + 5x3 y 3 = (x, y). Hence it is symmetrical.
2) (λx + µy, z) = (λx1 + µy 1 )z 1 + 3(λx2 + µy 2 )z 2 + 5(λx3 + µy 3 )z 3 =
= λ(x1 z 1 + 3x2 z 2 + 5x3 z 3 ) + µ(y 1 z 1 + 3y 2 z 2 + 5y 3 z 3 ) = λ(x, y) + µ(y, z). Hence it is linear.
3) (x, x) = (x1 )2 + 3(x2 )2 + 5(x3 )2 ≥ 0. It is non-negative. If x = 0 then obviously (x, x) = 0. If
(x, x) = (x1 )2 + 3(x2 )2 + 5(x3 )2 = 0, then x1 = x2 = x3 = 0. Hence it is positive-definite.
All conditions are checked. Hence (x, y) = x1 y 1 + 3x2 y 2 + 5x3 y 3 is indeed a scalar product in R3
1
e† )
The condition of linearity and symmetricity for the bilinear form
µ
¶µ 1 ¶
a b
y
B(x, y) = ( x1 , x2 )
= ax1 y 1 + b(x1 y 2 + x2 y 1 ) + cx2 y 2
b c
y2
are evidently obeyed.
The general answer on this question is: symmetric matrix is positive-definite if and only if all principal
minors are positive. For matrix under consideration it means that conditions a > 0 and ac − b2 > 0 are
necessary and sufficient conditions.
Give a proof for this special case.
Check the positive-definiteness condition.
For x = (1, 0) B(x, x) = a. Hence a > 0 is necessary condition. Now consider
B(x, x) = a(x1 )2 + 2bx1 x2 + c(x2 )2 =
(ax1 + bx2 )2 + (ac − b2 )(x2 )2
≥ 0 ⇔ ac − b2 ≥ 0
a
We see that B(x, x) > 0µfor all ¶
x 6= 0 iff a > 0 and (ac − b2 ) > 0.
T
α β
2 The matrix A =
obeys the conditions A A = I . Show that
γ δ
a) det A = ±1
b) if det A = 1 then there exists an angle ϕ : 0 ≤ ϕ < 2π such that A = Aϕ where
µ
¶
cos ϕ − sin ϕ
Aϕ =
(rotation matrix)
sin ϕ cos ϕ
µ
1
0
c) if¶det A = −1 then then there exists an angle ϕ : 0 ≤ ϕ < 2π such that A = Aϕ R, where R =
0
(a reflection matrix).
−1
The answers on this exercise see in lecture notes.
3 Show that for matrix Aϕ defined in the previous exercise the following relations are satisfied:
T
A−1
ϕ = Aϕ = A−ϕ ,
Aϕ+θ = Aϕ · Aθ .
µ
¶
cos ϕ − sin ϕ
We know (see lecture notes) that Aϕ =
. Then calculate inverse matrix A−1
ϕ . One
sin
ϕ
cos
ϕ
µ
¶
T
T
cos ϕ sin ϕ
can see that Aϕ = A−1
, because Aϕ Aϕ = I. On the other hand cos ϕ = cos(−ϕ) and
ϕ =
− sin ϕ cos ϕ
sin ϕ = − sin(−ϕ). Hence
µ
¶ µ
¶
T
cos ϕ sin ϕ
cos(−ϕ) − sin(−ϕ)
−1
Aϕ = Aϕ =
=
= A−ϕ .
− sin ϕ cos ϕ
sin(−ϕ) cos(−ϕ)
Now prove that Aϕ+θ = Aϕ · Aθ :
µ
¶µ
¶ µ
cos ϕ − sin ϕ
cos θ − sin θ
(cos ϕ cos θ − sin ϕ sin θ)
Aϕ · Aθ =
=
sin ϕ cos ϕ
sin θ cos θ
(cos ϕ sin θ + sin ϕ cos θ)
µ
cos(ϕ + θ) − sin(ϕ + θ)
sin(ϕ + θ) cos(ϕ + θ)
−(cos ϕ sin θ + sin ϕ cos θ)
(cos ϕ cos θ − sin ϕ sin θ)
¶
=
¶
= Aϕ+θ
4 Show that under the transformation (e01 , e02 ) = (e1 , e2 ) Aϕ an orthonormal basis transforms to an
orthonormal one.
2
How coordinates of vectors change if we rotate the orthonormal basis (e1 , e2 ) on the angle ϕ =
anticlockwise?
π
3
We have to check that scalar products (e01 , e01 ) = (e02 , e02 ) = 1 and (e01 , e02 ) = 0. Calculations show that
= (cos ϕe1 + sin ϕe2 , cos ϕe1 + sin ϕe2 ) = cos2 ϕ(e1 , e1 ) + 2 cos ϕ sin ϕ(e1 , e2 ) + sin2 ϕ(e2 , e2 ) = 1,
= (− sin ϕe1 +cos ϕe2 , − sin ϕe1 +cos ϕe2 ) = 1, (e01 , e02 ) = (cos ϕe1 +sin ϕe2 , − sin ϕe1 +cos ϕe2 ) = 0.
(e01 , e01 )
(e02 , e02 )
Now answer the second question.
If a = xex + ye2 =
x0 e0x
+
y 0 e02
Ã
and Aϕ = A π3 =
then we have:
1
√2
3
2
−
√
3
2
1
2
!
is the matrix of bases transformation
Ã
µ 0¶
µ ¶
µ 0¶
1
x
x
x
= (ex , ey )A π3
= (e0x , e0y )
a = (ex , ey )
= (ex , ey ) √23
0
0
y
y
y
2
Hence
µ ¶
µ 0¶ Ã 1
x
x
= A π3
= √23
y
y0
2
and
µ
x0
y0
¶
−1
= Tπ
3
−
√
3
2
1
2
!µ
µ ¶
µ ¶ Ã 1
x
x
2√
= A− π3
=
y
y
− 3
2
x0
y0
√
3
2
1
2
−
√
3
2
1
2
!µ
x0
y0
¶
¶
!µ
x0
y0
¶
5 Let {e1 , e2 , e3 } be an orthonormal basis of Euclidean space E3 . Consider the ordered set of vectors
which is expressed via basis {e1 , e2 , e3 } as in the exercise 6 of the Homework 1.
{e01 , e02 , e03 }
Write down explicitly transition matrix from the basis {e1 , e2 , e3 } to the ordered set of the vectors
What is the rank of this matrix? Is this matrix orthogonal?
{e01 , e02 , e03 }.
Find out is the ordered set of vectors {e01 , e02 , e03 } a basis in E3 . Is this basis an orthonormal basis of E3 ?
(you have to consider all cases a),b) c) and d)).
0
0
0
Case
a) The
 ordered set {e1 , e2 , e3 } = {e2 , e1 , e3 } is evidently orthonormal basis. Transition matrix

0 1 0
T =  1 0 0 , (e01 , e02 , e03 ) = (e1 , e2 , e3 )T . This matrix is non-degenerate, its rank is equal to 3 (det T =
0 0 1
1 6= 0). It is orthogonal because both bases are orthonormal.
is not a basis because vectors are linear
Case b) The ordered set {e01 , e02 , e03 } = {e1 , e1 +
 3e3 , e3 } 
1 1 0
dependent: e01 − e02 + 3te03 = 0. Transition matrix T =  0 0 0 , (e01 , e02 , e03 ) = (e1 , e2 , e3 )T . This matrix
0 3 1
is degenerate, its rank ≤ 2. One can see it noting that rows are linear dependent or noting that det T = 0.
Vectors {e01 , e02 , e03 } are linear dependent. On the other hand vectors {e01 , e02 } are linear independent. Hence
rank of the matrix T is equal to 2. This matrix is not orthogonal.
Case c) The ordered set {e01 , e02 , e03 } = {e1 − e2 , 3e1 − 3e2 , e3 } is not a basis because vectors are linear
dependent: 3e01 − e02 = 0.


1
3 0
One can see it also studying the transition matrix. Transition matrix T =  −1 −3 0 , (e01 , e02 , e03 ) =
0
0 1
(e1 , e2 , e3 )T . This matrix is degenerate, its rank ≤ 2, det T = 0. On the other hand second and third row of
this matrix are linear dependent. Hence rank of the matrix T is equal to 2. This matrix is not orthogonal.
Case d)
3
0
0
0
The
transitionmatrix from the basis {e1 , e2 , e3 } to the ordered triple {e1 , e2 , e3 } = {e2 , e1 , e1 +e2 +λe3 }
0 1 1
is T =  1 0 1 , (e01 , e02 , e03 ) = (e1 , e2 , e3 )T
0 0 λ
I-st case. λ 6= 0. The ordered set {e01 , e02 , e03 } is a basis because vectors are linear independent (see the
exercise 3), This basis is not orthogonal, because the length vector is not equal to 1 ((e03 , e03 ) = |e03 |2 = 2+λ2 ).
This matrix is not orthogonal, because the new basis is not orthonormal.
II-nd case λ = 0. The ordered set {e01 , e02 , e03 } is not a basis because vectors are linear independent:
0
e1 + e02 − e03 = 0. The transition matrix T has rank less or equal to 2, because vectors are linear dependent.
On the other hand vectors e01 , e02 are linear independent. Hence the rank of the matrix is equal to 2.
6† . Show that an arbitrary orthogonal transformation of two-dimensional Euclidean space can be considered as a composition of reflections.
If the determinant of orthogonal transformation is equal to −1 then
µ
T = T̃ϕ =
cos ϕ
sin ϕ
sin ϕ − cos ϕ
¶
One can see that this is reflection with respect to the axis which has an angle ϕ/2 with Ox axis.
If the determinant of orthogonal transformation is equal to 1 then
µ
T = Tϕ =
cos ϕ − sin ϕ
sin ϕ cos ϕ
¶
µ
=
cos ϕ
sin ϕ
sin ϕ
− cos ϕ
¶µ
1
0
0
−1
¶
µ
= T̃ϕ
1
0
0
−1
¶
i.e. rotation on the angle ϕ is a composition of two reflections.
7† Prove the Cauchy–Bunyakovsky–Schwarz inequality
(x, y)2 ≤ (x, x)(y, y) ,
where x, y are arbitrary two vectors and ( , ) is a scalar product in Euclidean space.
Hint: For any two given vectors x, y consider the quadratic polynomial At2 + 2Bt + C where A = (x, x),
B = (x, y), C = (y, y). Show that this polynomial has at most one real root and consider its discriminant.
Pn
Pn
Consider quadratic polynomial P (t) = i=1 (txi + y i )2 = At2 +2Bt +C, where A = i=1 (xi )2 = (x, x),
Pn
P
n
B = i=1 (xi y i ) = (x, y), C = i=1 (y i )2 = (y, y). We see that equation P (t) = 0 has at most one root (
and this is the case if only vector x is collinear to the vector y). This means that discriminant of this equation
is less or equal to zero. But discriminant of this equation is equal to 4B 2 − 4AC. Hence B 2 ≤ AC. It is just
CBS inequality. ((x, y)2 = (x, x)(y, y), i.e. discriminant is equal to zero ⇔ vectors x, y are colinear.
4