Math 221: LINEAR ALGEBRA Chapter 6. Vector Spaces 6-2. Vector - PowerPoint PPT Presentation

Math 221: LINEAR ALGEBRA Chapter 6. Vector Spaces §6-2. Vector Spaces - Examples and Basic Properties Le Chen 1 Emory University, 2020 Fall (last updated on 10/09/2020) Creative Commons License (CC BY-NC-SA) 1 Slides are adapted from those by Karen Seyffarth from University of Calgary.

0 0 u v u v u u The proof of this theorem requires one to show that if the three properties listed above hold, then all the vector space axioms hold. Subspaces Definition Let V be a vector space and let U be a subset of V. Then U is a subspace of V if U is a vector space using the addition and scalar multiplication of V.

The proof of this theorem requires one to show that if the three properties listed above hold, then all the vector space axioms hold. Subspaces Definition Let V be a vector space and let U be a subset of V. Then U is a subspace of V if U is a vector space using the addition and scalar multiplication of V. Theorem (Subspace Test) Let V be a vector space and U ⊆ V. Then U is a subspace of V if and only if it satisfies the following three properties: 1. The zero vector of V is an element of U, i.e., 0 ∈ U where 0 is the zero vector of V. 2. U is closed under addition, i.e., if u , v ∈ U, then u + v ∈ U. 3. U is closed under scalar multiplication, i.e., if u ∈ U and k ∈ R , then k u ∈ U.

The proof of this theorem requires one to show that if the three properties listed above hold, then all the vector space axioms hold. Subspaces Definition Let V be a vector space and let U be a subset of V. Then U is a subspace of V if U is a vector space using the addition and scalar multiplication of V. Theorem (Subspace Test) Let V be a vector space and U ⊆ V. Then U is a subspace of V if and only if it satisfies the following three properties: 1. The zero vector of V is an element of U, i.e., 0 ∈ U where 0 is the zero vector of V. 2. U is closed under addition, i.e., if u , v ∈ U, then u + v ∈ U. 3. U is closed under scalar multiplication, i.e., if u ∈ U and k ∈ R , then k u ∈ U.

0 0 0 0 u u u u u 0 Important Note As a consequence of the proof, any subspace U of a vector space V has the same zero vector as V , and each u ∈ U has the same additive inverse in U as in V .

0 0 u u u u u 0 0 0 Important Note As a consequence of the proof, any subspace U of a vector space V has the same zero vector as V , and each u ∈ U has the same additive inverse in U as in V . Examples Let V be a vector space. 1. V is a subspace of V.

u u u u u 0 Important Note As a consequence of the proof, any subspace U of a vector space V has the same zero vector as V , and each u ∈ U has the same additive inverse in U as in V . Examples Let V be a vector space. 1. V is a subspace of V. 2. { 0 } is a subspace of V, where 0 denotes the zero vector of V. The proof uses the fact that in any vector space, a 0 = 0 for any a ∈ R .

Important Note As a consequence of the proof, any subspace U of a vector space V has the same zero vector as V , and each u ∈ U has the same additive inverse in U as in V . Examples Let V be a vector space. 1. V is a subspace of V. 2. { 0 } is a subspace of V, where 0 denotes the zero vector of V. The proof uses the fact that in any vector space, a 0 = 0 for any a ∈ R . 3. For any u ∈ V, R u = { k u | k ∈ R } is a subspace of V. The proof uses the fact that in any vector space, if u ∈ V, then 0 u = 0 .

0 0 0 0 0 0 0 0 M Problem Let A be a fixed (arbitrary) n × n real matrix, and define U = { X ∈ M nn | AX = XA } , i.e., U is the subset of matrices of M nn that commute with A. Prove that U is a subspace of M nn .

M Problem Let A be a fixed (arbitrary) n × n real matrix, and define U = { X ∈ M nn | AX = XA } , i.e., U is the subset of matrices of M nn that commute with A. Prove that U is a subspace of M nn . Solution ◮ Let 0 nn denote the n × n matrix of all zeros. Then A 0 nn = 0 nn and 0 nn A = 0 nn , so A 0 nn = 0 nn A. Thus 0 nn ∈ U.

M Problem Let A be a fixed (arbitrary) n × n real matrix, and define U = { X ∈ M nn | AX = XA } , i.e., U is the subset of matrices of M nn that commute with A. Prove that U is a subspace of M nn . Solution ◮ Let 0 nn denote the n × n matrix of all zeros. Then A 0 nn = 0 nn and 0 nn A = 0 nn , so A 0 nn = 0 nn A. Thus 0 nn ∈ U. ◮ Suppose X , Y ∈ U. Then AX = XA and AY = YA, implying that A ( X + Y ) = AX + AY = XA + YA = ( X + Y ) A , and thus X + Y ∈ U, so U is closed under addition.

Problem Let A be a fixed (arbitrary) n × n real matrix, and define U = { X ∈ M nn | AX = XA } , i.e., U is the subset of matrices of M nn that commute with A. Prove that U is a subspace of M nn . Solution ◮ Let 0 nn denote the n × n matrix of all zeros. Then A 0 nn = 0 nn and 0 nn A = 0 nn , so A 0 nn = 0 nn A. Thus 0 nn ∈ U. ◮ Suppose X , Y ∈ U. Then AX = XA and AY = YA, implying that A ( X + Y ) = AX + AY = XA + YA = ( X + Y ) A , and thus X + Y ∈ U, so U is closed under addition. ◮ Suppose X ∈ U and k ∈ R . Then AX = XA, implying that A ( kX ) = k ( AX ) = k ( XA ) = ( kX ) A ; thus kX ∈ U, so U is closed under scalar multiplication. By the subspace test, U is a subspace of M nn .

0 0 0 Problem Let t ∈ R , and let U = { p ∈ P | p ( t ) = 0 } , i.e., U is the subset of polynomials that have t as a root. Prove that U is a vector space.

Problem Let t ∈ R , and let U = { p ∈ P | p ( t ) = 0 } , i.e., U is the subset of polynomials that have t as a root. Prove that U is a vector space. Solution ◮ Let 0 denote the zero polynomial. Then 0 ( t ) = 0 , and thus 0 ∈ U.

Problem Let t ∈ R , and let U = { p ∈ P | p ( t ) = 0 } , i.e., U is the subset of polynomials that have t as a root. Prove that U is a vector space. Solution ◮ Let 0 denote the zero polynomial. Then 0 ( t ) = 0 , and thus 0 ∈ U. ◮ Let q , r ∈ U. Then q ( t ) = 0 , r ( t ) = 0 , and ( q + r )( t ) = q ( t ) + r ( t ) = 0 + 0 = 0 . Therefore, q + r ∈ U, so U is closed under addition.

Problem Let t ∈ R , and let U = { p ∈ P | p ( t ) = 0 } , i.e., U is the subset of polynomials that have t as a root. Prove that U is a vector space. Solution ◮ Let 0 denote the zero polynomial. Then 0 ( t ) = 0 , and thus 0 ∈ U. ◮ Let q , r ∈ U. Then q ( t ) = 0 , r ( t ) = 0 , and ( q + r )( t ) = q ( t ) + r ( t ) = 0 + 0 = 0 . Therefore, q + r ∈ U, so U is closed under addition. ◮ Let q ∈ U and k ∈ R . Then q ( t ) = 0 and ( kq )( t ) = k ( q ( t )) = k · 0 = 0 . Therefore, kq ∈ U, so U is closed under scalar multiplication. By the subspace test, U is a subspace of P , and thus is a vector space.

M M 0 Examples 1. It is routine to verify that P n is a subspace of P for all n ≥ 0 .

0 Examples 1. It is routine to verify that P n is a subspace of P for all n ≥ 0 . A ∈ M 22 | A 2 = A � � 2. U = is not a subspace of M 22 . To prove this, notice that I 2 , the two by two identity matrix, is in U, but 2 I 2 �∈ U since (2 I 2 ) 2 = 4 I 2 � = 2 I 2 , so U is not closed under scalar multiplication.

Examples 1. It is routine to verify that P n is a subspace of P for all n ≥ 0 . A ∈ M 22 | A 2 = A � � 2. U = is not a subspace of M 22 . To prove this, notice that I 2 , the two by two identity matrix, is in U, but 2 I 2 �∈ U since (2 I 2 ) 2 = 4 I 2 � = 2 I 2 , so U is not closed under scalar multiplication. 3. U = { p ∈ P 2 | p (1) = 1 } is not a subspace of P 2 since the zero polynomial is not in U: 0 (1) = 0 .

u u u u u u u u u u u u u u u u u u Linear Combinations and Spanning Sets Definitions Let V be a vector space and let { u 1 , u 2 , . . . , u n } be a subset of V. 1. A vector u ∈ V is called a linear combination of u 1 , u 2 , . . . , u n if there exist scalars a 1 , a 2 , . . . , a n ∈ R such that u = a 1 u 1 + a 2 u 2 + · · · + a n u n .

u u u u u u Linear Combinations and Spanning Sets Definitions Let V be a vector space and let { u 1 , u 2 , . . . , u n } be a subset of V. 1. A vector u ∈ V is called a linear combination of u 1 , u 2 , . . . , u n if there exist scalars a 1 , a 2 , . . . , a n ∈ R such that u = a 1 u 1 + a 2 u 2 + · · · + a n u n . 2. The set of all linear combinations of u 1 , u 2 , . . . , u n is called the span of u 1 , u 2 , . . . , u n , and is defined as span { u 1 , u 2 , . . . , u n } = { a 1 u 1 + a 2 u 2 + · · · + a n u n | a 1 , a 2 , . . . , a n ∈ R } .

Linear Combinations and Spanning Sets Definitions Let V be a vector space and let { u 1 , u 2 , . . . , u n } be a subset of V. 1. A vector u ∈ V is called a linear combination of u 1 , u 2 , . . . , u n if there exist scalars a 1 , a 2 , . . . , a n ∈ R such that u = a 1 u 1 + a 2 u 2 + · · · + a n u n . 2. The set of all linear combinations of u 1 , u 2 , . . . , u n is called the span of u 1 , u 2 , . . . , u n , and is defined as span { u 1 , u 2 , . . . , u n } = { a 1 u 1 + a 2 u 2 + · · · + a n u n | a 1 , a 2 , . . . , a n ∈ R } . 3. If U = span { u 1 , u 2 , . . . , u n } , then { u 1 , u 2 , . . . , u n } is called a spanning set of U.

as a linear combination of the other vectors. , impossible to express combination of the other vectors; otherwise, the system is inconsistent and it is as a linear If this system is consistent, then we have found a way to express . , and implying that Then such that Suppose that there exist Problem Is it possible to express x 2 + 1 as a linear combination of x + 1 , x 2 + x , x 2 + 2? and Equivalently, is x 2 + 1 ∈ span { x + 1 , x 2 + x , x 2 + 2 } ?