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Lecture 8
Andrei Antonenko
February 19, 2003
1 Matrix equations and the inverse
1.1 Discussion of the algorithm - Part 2
Last time we proved the following result:
Lemma 1.1. 1. If the square matrix A is invertible, then its RREF is the identity matrix.
- If we can reduce the matrix A by elementary row operations to the identity matrix, i.e. if its RREF is identity matrix, then this algorithm gives us A−^1 B in the right half of the augmented matrix.
Actually, the opposite assertion of this lemma is also true.
Lemma 1.2. Let RREF of a square matrix A be the identity matrix. Then A is invertible.
Proof. Let’s consider the process of reducing A to it’s RREF. It can be done by elementary row operations with matrices E 1 , E 2 ,... , Es, i.e. (EsEs− 1 · · · E 1 )A = I. From this equality we see that the product of matrices EsEs− 1 · · · E 1 satisfy the definition of the inverse for A.
So, from these 2 lemmas we get the interesting result, which is the main result about invertible matrices so far:
Theorem 1.3. The matrix A is invertible if and only if its RREF is the identity matrix.
2 Vector spaces
In this lecture we will introduce a new algebraic structure which is one of the most important structure in linear algebra. This would be a set with 2 operations — addition of its elements and multiplication of numbers by its elements.
Definition 2.1. Let k be any field. We didn’t study fields so far, so those who are not familiar with them can just treat the letter k as another notation for R. A set V is called vector space if there defined an operation of addition of elements of V such that ∀v, w ∈ V v + w ∈ V , and an operation of multiplication of elements of k by elements of V (often called scalar multiplication) such that ∀k ∈ k ∀v ∈ V kv ∈ V , and the following axioms are satisfied: Axioms of addition:
(A1) ∀v, u ∈ V v + u = u + v
(A2) ∀v, u, w ∈ V v + (u + w) = (v + u) + w
(A3) ∃ 0 ∈ V such that v + 0 = v
(A4) ∀v ∈ V ∃(−v) ∈ V such that v + (−v) = 0
Axioms of multiplication:
(M1) ∀a ∈ k ∀u, v ∈ V a(u + v) = au + av
(M2) ∀a, b ∈ k ∀v ∈ V (a + b)v = av + bv
(M3) ∀a, b ∈ k ∀v ∈ V a(bv) = (ab)v
(M4) ∀u ∈ V 1 u = u
Elements of the vector space are called vectors.
Now we’ll give a number of examples of a vector spaces.
Example 2.2 (Space Rn). Let V be a set of n-tuples of elements of R. We can define operations as follows:
Addition: (a 1 , a 2 ,... , an) + (b 1 , b 2 ,... , bn) = (a 1 + b 1 , a 2 + b 2 ,... , an + bn)
Scalar multiplication: k(a 1 , a 2 ,... , an) = (ka 1 , ka 2 ,... , kan).
The zero vector is 0 = (0, 0 ,... , 0) and the negative vector is −(a 1 , a 2 ,... , an) = (−a 1 , −a 2 ,... , −an).
Example 2.3 (Space P (t)). Let V be a set of all polynomials of the form
p(t) = a 0 + a 1 t + a 2 t^2 + · · · + asts, s ∈ N.
We can define operations as follows:
Addition: Usual addition of polynomials.
Scalar multiplication: Multiplication of a polynomial by a number.
- If u + w = v + w then u = v.
- ∀k ∈ k k 0 = 0.
Proof. k 0 = k( 0 + 0 ) = k 0 + k 0 , and so by the first property 0 = l 0.
Proof. 0 u = (0 + 0)u = 0u + 0u, and so by the first property 0 = 0u.
- If k 6 = 0 and ku = 0 then u = 0.
Proof. u = 1u = (k−^1 k)u = k−^1 (ku) = k−^10 = 0.
- ∀k ∈ k and u ∈ V (−k)u = k(−u).
Proof. 0 = k 0 = k(u + (−u)) = ku + k(−u), and 0 = 0u = (k + (−k))u = ku + (−k)u. So, k(−u) = (−k)u.
3 Subspaces
Definition 3.1. Let V be a vector space. The subset W ⊂ V is called a subspace of V if W itself is a vector space.
To check that W is a subspace we need to check the following properties:
- 0 ∈ W
- ∀v, w ∈ W v + w ∈ W
- ∀k ∈ k ∀u ∈ W ku ∈ W
Example 3.2. Consider a vector space R^2. Then its subset W = {(0, y)|y ∈ R} — set of pairs for which the first element equals to 0, is a subspace. We can prove it. First of all, (0, 0) ∈ W , since first element of it is 0. Moreover, let u = (0, a) ∈ W , v = (0, b) ∈ W. Then their sum u + v = (0, a + b) ∈ W since it has zero on the first place. Now let’s multiply any vector u = (0, a) ∈ W by any number k: ku = (0, ka), and it belongs to W , since it has 0 on the first place. So, this is a subspace.
Example 3.3. Consider a vector space R^2. Then its subset W = {(1, y)|y ∈ R} — set of pairs for which the first element equals to 1, is NOT a subspace. Here the first property is not satisfied — (0, 0) doesn’t belong to W. Other properties are not satisfied as well: (1, a) ∈ W , (1, b) ∈ W , but their sum (2, a + b) 6 ∈ W , since it has 2 on the first place.
Example 3.4. Consider a vector space R^2. Then its subset W = {(x, y)|x, y ∈ R, x = y} — set of pairs for which the first element is equal to the second element (geometrically, it is a line on the plane), is a subspace.
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Let’s check it. First of all, if a = (a, a) ∈ W , and b = (b, b) ∈ W then a+b = (a+b, a+b) ∈ W. Than, (0, 0) ∈ W. Moreover, for each k ∈ R we have k(a, a) = (ka, ka) ∈ W. So, this is a subspace.
One can prove that any line on the plane R^2 which goes through the origin is a subspace. Moreover, any plane in the space R^3 which contains the origin (0, 0 , 0) is a subspace.
Example 3.5. Consider a vector space R^2. Then its subset W = {(x, x^2 )|x ∈ R} — set of pairs for which the second element is equal to the square of the first element is NOT a subspace. Let’s prove it. First of all, if (0, 0) = (0, 02 ) ∈ W. Now let’s consider 2 elements of this set — (1, 1) ∈ W and (2, 4) ∈ W. Their sum (3, 5) doesn’t belong to W , since 5 6 = 3^2. So, we showed that there are two elements sum of which doesn’t belong to the set. So, this is not a vector space.
