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Lecture 26
Andrei Antonenko
April 11, 2003
1 Orthogonality
Let V be a Euclidean space, and let v and u be 2 vectors in this space. Then we can define the angle between these 2 vectors.
Definition 1.1. The angle θ between two vectors v and u from the vector space V can be defined by the following formula:
cos θ = 〈u, v〉 ‖u‖‖v‖ Actually, we should check that this definition is correct — we have an expression for cosine, and it should belong to the interval [− 1 , 1]! It is easy to check. From Cauchy-Bunyakovsky- Schwartz inequality we have |〈u, v〉| ≤ ‖u‖‖v‖,
and so we will have − 1 ≤ (^) ‖〈uu, v‖‖v〉‖ ≤ 1
Let’s note that this is a general definition which works in many different spaces, and not only in R^2 and R^3. For example, by this formula we can define the angle between two functions from C[a, b] or between two polynomials.
Example 1.2. Let u = (1, 2 , 3) and let v = (− 1 , 2 , −2). Then 〈u, v〉 = −1 + 4 − 6 = − 4 , ‖u‖ =
14 , and ‖v‖ =
1 + 4 + 4 = 3. So, the angle θ between these two vectors can be defined by the following formula:
cos θ = 〈u, v〉 ‖u‖‖v‖
= −^4
Another important concept is a normalization of the vector.
Definition 1.3. If v is a vector from the vector space V then the vector
v ‖v‖
is called a normalization of v.
The main property of normalization is that it’s norm is equal to 1. So, we take a vector which is proportional to v with the “length” 1. Now we’ll give the very important definition — main definition of this lecture.
Definition 1.4. Two vectors u and v are called orthogonal if
〈u, v〉 = 0.
Let we have a vector v from the Euclidean space V. Let’s consider all vectors u which are orthogonal to v, i.e. the set of vectors u such that 〈v, u〉 = 0:
v⊥^ = {u ∈ V |〈u, v〉 = 0} (read “v-perp”)
This set is called the orthogonal complement to the vector v. Now let S be a set of vectors. Then we can define S⊥^ as the following set:
S⊥^ = {u ∈ V |〈u, v〉 = 0 for all v ∈ S}
This set S⊥^ is called the orthogonal complement to the set S. Let’s note that S⊥^ is a vector space. First, 0 is orthogonal to any other vector, since 〈 0 , v〉 = 0 for all v. So, 0 ∈ S⊥. Now, let u 1 , u 2 ∈ S⊥, such that 〈u 1 , v〉 = 0 for all v and 〈u 2 , v〉 = 0 for all v. So it follows that 〈u 1 + u 2 , v〉 = 〈u 1 , v〉 + 〈u 2 , v〉 = 0, and thus u 1 + u 2 belongs to S⊥. Moreover, it is obvious to check that if u belongs to S⊥^ then ku belongs to it. Thus, we proved that S⊥^ is a vector space. Geometrically speaking, let’s consider the case of 3-dimensional space. Let v be a vector from the 3-dimensional space, then v⊥^ is the plane which is perpendicular to this vector. Now let’s consider more difficult case, when S consists of 2 vectors v and u. In this case the line L will be orthogonal to both vectors. It is illustrated on the picture below.
°
°
° °
°
° °
°
°
6
u v L
Our goal is to describe u⊥^ or S⊥^ somehow, for example give a basis of it. Actually, for the simple spaces, like Rn^ the basis of the orthogonal complement can be found as a basis in the solution space of the corresponding homogeneous system. We’ll show it on the example.
Example 1.5. Let v = (1, − 3 , 4). Let’s find the basis of the orthogonal complement to u, i.e. the basis of u⊥. Vector u = (x, y, z) is orthogonal to v if 〈v, u〉 = x − 3 y + 4z = 0. So, we have an equation: x − 3 y + 4z = 0.
Using linearity of the scalar product we get:
a 1 〈v 1 , v 1 〉 + a 2 〈v 2 , v 1 〉 + · · · + an〈vn, v 1 〉 = 0.
All terms except the first one are equal to 0, so we get
a 1 〈v 1 , v 1 〉 = 0.
Since 〈v 1 , v 1 〉 6 = 0 then a 1 = 0. In the same way we can prove that a 2 , a 3 ,... , an = 0. So, vectors are linearly independent.
Another theorem gives us a mathematically exact formulation of the fact known from the school geometry.
Theorem 1.8 (Pythagoras theorem). If u and v are orthogonal vectors, then
‖u + v‖^2 = ‖u‖^2 + ‖v‖^2.
Proof. We have:
‖u + v‖^2 = 〈u + v, u + v〉 = 〈u, u〉 + 〈u, v〉 + 〈v, u〉 + 〈v, v〉 = 〈u, u〉 + 2〈u, v〉 + 〈v, v〉.
Since v and u are orthogonal, 〈u, v〉 = 0, and so ‖u + v‖^2 = 〈u, u〉 + 〈v, v〉 = ‖u‖^2 + ‖v‖^2.
In 2-dimensional space this theorem is exactly the Pythagoras theorem from the school geometry. The following picture illustrates it:
? -
ZZ u (^) vZu+vZ~
From this theorem it follows that if we have n orthogonal vectors in the n-dimensional vector space, then they form a basis. Moreover, this basis is nice, and we can find coordinates in this basis very easily. We’ll demonstrate it in the next example.
Example 1.9. Let v 1 = (1, 2 , 1), v 2 = (2, 1 , −4) and v 3 = (3, − 2 , 1). We can check that these vectors are orthogonal, i.e. 〈v 1 , v 2 〉 = 0, 〈v 1 , v 3 〉 = 0, and 〈v 2 , v 3 〉 = 0. So, they are linearly independent by theorem (1.7), and since there are 3 vectors, they form a basis in R^3 , i.e. each vector can be represented as a linear combination of them. Now let u = (7, 1 , 9). We want to represent u as a linear combination of v 1 , v 2 , and v 3 , i.e. find coefficients a 1 , a 2 , and a 3 such that
u = a 1 v 1 + a 2 v 2 + a 3 v 3 , (1)
i.e. (7, 1 , 9) = a 1 (1, 2 , 1) + a 2 (2, 1 , −4) + a 3 (3, − 2 , 1). The familiar way to do it is to set up a linear system and solve it for unknowns a 1 , a 2 and a 3. But there is an easier way to do it. Let’s multiply the equality (1) by v 1 , v 2 , and then by v 3. We will have:
〈u, v 1 〉 = a 1 〈v 1 , v 1 〉 + a 2 〈v 2 , v 1 〉 + 〈v 3 , v 1 〉 = a 1 〈v 1 , v 1 〉,
since 〈v 2 , v 1 〉 = 〈v 3 , v 1 〉 = 0. So,
a 1 = 〈u, v^1 〉 〈v 1 , v 1 〉
=^18
Now,
〈u, v 2 〉 = a 1 〈v 1 , v 2 〉 + a 2 〈v 2 , v 2 〉 + 〈v 3 , v 2 〉 = a 2 〈v 2 , v 2 〉,
since 〈v 1 , v 2 〉 = 〈v 3 , v 2 〉 = 0. So,
a 2 =
〈u, v 2 〉 〈v 2 , v 2 〉 =
4 + 1 + 16 =^
21 =^ −^1.
Now,
〈u, v 3 〉 = a 1 〈v 1 , v 3 〉 + a 2 〈v 2 , v 3 〉 + 〈v 3 , v 3 〉 = a 3 〈v 3 , v 3 〉,
since 〈v 1 , v 3 〉 = 〈v 2 , v 3 〉 = 0. So,
a 3 = (^) 〈〈vu, v^3 〉 3 , v 3 〉^
=^21 9 + 4 + 1^ −^ 2 + 9 =^2814 = 2.
So, we got that u = 3v 1 − v 2 + 2v 3.
