1. The additive identity is unique. That is, if v + u = v, then u = 0. 2. The additive inverse of v is unique. That is, if v + u = 0, then u = -v. 3. Ov = 0 4. co - 0 5. If cv - 0, then c - 0 or v - 0. 6. -(-v) = v e the properties of vector addition and scalar multiplication from the following theorem. Properties of Vector Addition and Scalar Multiplication in R" Let u, v, and w be vectors in R", and let c and d be scalars. 1. u + v is a vector in R". 2. u + v- v + u 3. (u + u) + w - u + (u + w) 4. u + 0 -u 5. u + (-u) - 0 6. cu is a vector in R". 7. cu + v) = cu + cv 8. (c + d)u = cu + du 9. c(du) = (cơ)u 10. 1(u) = u Closure under addition Commutative property of addition Associative property of addition Additive identity property Additive inverse property Closure under scalar multiplication Distributive property Distributive property Associative property of multiplication Multiplicative identity property Step Justification c0 = c(0 + 0) --Select--- c0 = c0 + c0 ---Select-- D+ (-c0) = (c0 + c0) + (-c0) -Select-- 0 = (c0 + c0) + (-co) 0 = c0 + (c0 + (-co)) 0 = c0 + 0 ---Select--- --Select-- --Select-- 0 = c0 --Select--

1. The additive identity is unique. That is, if v + u = v, then u = 0. 2. The additive inverse of v is unique. That is, if v + u = 0, then u = -v. 3. Ov = 0 4. co - 0 5. If cv - 0, then c - 0 or v - 0. 6. -(-v) = v e the properties of vector addition and scalar multiplication from the following theorem. Properties of Vector Addition and Scalar Multiplication in R" Let u, v, and w be vectors in R", and let c and d be scalars. 1. u + v is a vector in R". 2. u + v- v + u 3. (u + u) + w - u + (u + w) 4. u + 0 -u 5. u + (-u) - 0 6. cu is a vector in R". 7. cu + v) = cu + cv 8. (c + d)u = cu + du 9. c(du) = (cơ)u 10. 1(u) = u Closure under addition Commutative property of addition Associative property of addition Additive identity property Additive inverse property Closure under scalar multiplication Distributive property Distributive property Associative property of multiplication Multiplicative identity property Step Justification c0 = c(0 + 0) --Select--- c0 = c0 + c0 ---Select-- D+ (-c0) = (c0 + c0) + (-c0) -Select-- 0 = (c0 + c0) + (-co) 0 = c0 + (c0 + (-co)) 0 = c0 + 0 ---Select--- --Select-- --Select-- 0 = c0 --Select--

Linear Algebra: A Modern Introduction

4th Edition

ISBN:9781285463247

Author:David Poole

Publisher:David Poole

Chapter4: Eigenvalues And Eigenvectors

Section4.2: Determinants

Problem 7AEXP

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