Linear Algebra Miscellaneous Proofs to Know

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1 Linear Algebra Miscellaneous Proofs to Know S. F. Ellermeyer Summer Semester 2010 Definition 1 An n n matrix, A, issaidtobeinvertible if there exists an n n matrix B such that AB BA I n (where I n is the n n identity matrix). Remark 2 We know that if A has an inverse, then that inverse is unique. Thus we denote the inverse of A by A 1. Definition 3 If a b A is a 2 2 matrix, then we define the determinant of A, denoted either by det (A) or A, tobe det (A) ad bc. Theorem 4 Suppose that is a 2 2 matrix. a b A 1. If det (A) 6 0,thenA is invertible and A 1 1 d b det (A) c a. 2. If det (A) 0,thenA is not invertible. 1

2 Proof. Suppose that det (A) 6 0and let 1 d b B. det (A) c a Then, by direct computation, we have µ a b 1 d b AB det (A) c a 1 a b d b det (A) c a 1 ad bc ab + ab ad bc cd cd bc + ad ad bc 0 ad bc ad bc 0 ad bc ad bc ad bc I 2. A similar computation shows that is is also true that BA I 2. Therefore A is invertible and A 1 B (as defined above). Now let us assume that det (A) 0.Then ad bc 0. (1) For the sake of obtaining a contradiction, let us now suppose that A is invertible. Then there is a 2 2 matrix, x1 x B 2, x 4 such that AB I 2.Thus a b x 3 x1 x 2 x 4 x By performing the above matrix multiplication, we see that it must then be true that ax 1 + bx 3 1 (2) ax 2 + bx 4 0 (3) cx 1 + dx 3 0 (4) cx 2 + dx 4 1. (5). 2

3 We will now consider two cases: Case 1: Suppose that a 0. Then, by equation (2), b 6 0. Consequently, by equation (1), c 0and, by equation (3), x 4 0. However, this means that equation (5) is not satisfied (because c 0and x 4 0). Thus it cannot bethecasethata 0. Case 2: Suppose that a 6 0. Then by performing the elementary operation c a E 1 + E 3 E 3, we obtain the system which can also be written as ax 1 + bx 3 1 µ bca + d x 3 c a ax 1 + bx 3 1 ad bc x 3 c a a. Since ad bc 0, then it must be the case that c 0. But then we must also have d 0by equation (1). However, this means that equation (1) is not satisfied,sowehaveonceagainarrivedat acontradiction. We conclude that if det (A) 0,thenA is not invertible. Theorem 5 Suppose that A and B are 2 2 matrices. Then det (AB) det(a)det(b). Proof. Wewillprovethisbycomputation.Let a b e f A and B g h. Then a b AB e f g h ae + bg af + bh ce + dg cf + dh 3

4 and thus det (AB) (ae + bg)(cf + dh) (af + bh)(ce + dg) acef + adeh + bcfg + bdgh acef adfg bceh bdgh adeh + bcfg adfg bceh. Also, det (A)det(B) (ad bc)(eh fg)adeh adfg bceh + bcfg. This shows that det (AB) det(a)det(b). Theorems and Proofs to Know For Exam 2 Definition 6 An indexed set of vectors {v 1, v 2,...,v n } in R m is said to be linearly independent if the vector equation x 1 v 1 + x 2 v x n v n 0 m has only the trivial solution (x 1 x 2 x n 0). If the above vector equation has non trivial solutions, then the set of vectors {v 1, v 2,...,v n } is said to be linearly dependent and any equation of the form c 1 v 1 + c 2 v c n v n 0 m with not all of the numbers c 1,c 2,...,c n equaltozeroiscalledalinear dependence relation for the set {v 1, v 2,...,v n }. Theorem 7 Suppose that {v 1, v 2,...,v n } is a set of two or more vectors in R m. This set of vectors is linearly dependent if and only if at least one of the vectors in this set is a linear combination of the other vectors in the set. Proof. Suppose that {v 1, v 2,...,v n } is a set of two or more vectors in R m and suppose that this set of vectors is linearly dependent. Then we have a linear dependence relation c 1 v 1 + c 2 v c n v n 0 m. 4

5 Notallofthenumbersc 1, c 2,..., c n are zero. In particular, there is some index j such that c j 60. This means that µ v j c µ 1 v c µ j 1 v j 1 + c µ j+1 v j c n v n c j c j c j c j showing that v j is a linear combination of the other vectors in the set. Conversely, suppose that there is some index j such that v j is a linear combination of the other vectors in the set. Then This means that v j c 1 v 1 + c 2 v c j 1 v j 1 + c j+1 v j c n v n c 1 v 2 + c 2 v c j 1 v j 1 +( 1) v j + c j+1 v j c n v n 0 m and hence that the set {v 1, v 2,...,v n } is linearly dependent (because c j 1 6 0). Theorem 8 Suppose that V is a vector space with addition operation and scalar multiplication operation. Then for any vector u V and any scalar k we have: 1. 0 u 0 2. k u ( 1) u 4. If k u 0, then either k 0or u 0. Proof. In proving this we will assume it to be known (to have already been proved) that the zero vector, 0, of V is unique and that the additive inverse of any vector in V is unique. Proof of Statement 1: and 0 u (0+0) u (simply because 00+0) (0 + 0) u 0 u +0 u (by the distributive property). 5

6 Thus 0 u 0 u +0 u. Since V is closed under scalar multiplication, we know that the vector 0 u is in V. Since all vectors in V have an additive inverse, then we know that (0 u) exists. Adding this vector to both sides of the above equation gives (0 u)+0 u (0 u)+(0 u +0 u). By using the associative property of addition, we obtain (0 u)+0 u ( (0 u)+0 u)+0 u and then by using the fact that (0 u) is the additive inverse of 0 u we obtain u. Finally, by using the fact that 0 is the additive identity of V,weobtain and this completes the proof. Proof of Statement 2: and 0 0 u k 0 k (0 + 0) (because ) k (0 + 0)k 0 + k 0 (by the distributive property). Thus k 0 k 0 + k 0. Since V is closed under scalar multiplication, we know that the vector k 0 is in V. Since all vectors in V have an additive inverse, then we know that (k 0) exists. Adding this vector to both sides of the above equation gives (k 0)+k 0 (k 0)+(k 0 + k 0). By using the associative property of addition, we obtain (k 0)+k 0 ( (k 0)+k 0)+k 0 6

7 and then by using the fact that (k 0) is the additive inverse of k 0 we obtain k 0. Finally, by using the fact that 0 is the additive identity of V,weobtain and this completes the proof. Proof of Statement 3: 0 k 0 u +( 1) u 1 u +( 1) u (because 1 u u) (1+( 1)) u (by the distributive property) 0 u (simply because 1+( 1) 0) 0 (by statement 1 of this theorem). Since u is the unique vector in V such that u +( u) 0 and since we have just shown that u +( 1) u 0, then it must be the case that u ( 1) u. Proof of Statement 4: Suppose that k u 0. Ifk 6 0, then we can multiply both sides of this equation by 1/k to obtain 1 k (k u) 1 k 0. Bytheassociativepropertyofmultiplicationweobtain µ 1 k k u 1 k 0. This gives 1 u 1 k 0. Since 1 u u, the left hand side of the above equation is equal to u. Also, by statement 2 of this theorem, the right hand side of the above equation is equal to 0. Thusu 0. In summary, we have proved that if k u 0 and k 6 0,thenitmust bethecasethatu 0. Thiscompletestheproof. 7

8 Theorem 9 Suppose that V is a vector space and suppose that W is a non empty subset of V. If W is closed under addition and closed under scalar multiplication, then W is a subspace of V. Proof. We are given that W is closed under both addition and scalar multiplication. To see that 0 W (where 0 is the zero vector of V ), we first note that since W 6, then there is at least one vector u W. Since W is closed under scalar multiplication, then 0 u W. By Statement 1 of the preceding theorem, 0 u 0. Thus0 W. Now let v be any vector in W and let us show that the additive inverse of v is also in W : First,wenotethatsinceW is closed under scalar multiplication, then ( 1) v W. However, by Statement 3 of the preceding theorem, we know that ( 1) v v. Thus v W. The remaining six vector space axioms, which are all algebraic properties, are automatically satisfied for all vectors in W and all scalars because they are satisfied for all vectors in V.Theproofofthetheoremisthuscomplete. 8