Linear Algebra

82 Chapter Two. Vector SpacesThe ninth( ) ( )v1 (rs)v1(rs) · = =v 2 (rs)v 2( ) ( )r(sv1 )v1= r · (s · )r(sv 2 )v 2and tenth conditions are also straightforward.( ) ( )v1 1v11 · = =v 2 1v 2In a similar way, each R n is a vector space with the usual operations ofvector addition and scalar multiplication. (In R 1 , we usually do not write themembers as column vectors, i.e., we usually do not write ‘(π)’. Instead we justwrite ‘π’.)(v1v 2)1.4 Example This subset of R 3 that is a plane through the origin⎛ ⎞P = { ⎝ x y⎠ ∣ x + y + z = 0}zis a vector space if ‘+’ and ‘·’ are interpreted in this way.⎛⎝ x ⎞ ⎛1y 1⎠ + ⎝ x ⎞ ⎛2y 2⎠ = ⎝ x ⎞ ⎛1 + x 2y 1 + y 2⎠ r · ⎝ x ⎞ ⎛y⎠ = ⎝ rx⎞ry⎠z 1 z 2 z 1 + z 2 z rzThe addition and scalar multiplication operations here are just the ones of R 3 ,reused on its subset P . We say that P inherits these operations from R 3 . Thisexample of an addition in P⎛⎝ 1 ⎞ ⎛1 ⎠ + ⎝ −1⎞ ⎛0 ⎠ = ⎝ 0 ⎞1 ⎠−2 1 −1illustrates that P is closed under addition. We’ve added two vectors from P —that is, with the property that the sum of their three entries is zero — and theresult is a vector also in P . Of course, this example of closure is not a proof ofclosure. To prove that P is closed under addition, take two elements of P⎛⎝ x ⎞1y 1⎠z 1⎛⎝ x ⎞2y 2⎠z 2(membership in P means that x 1 + y 1 + z 1 = 0 and x 2 + y 2 + z 2 = 0), andobserve that their sum ⎛⎝ x ⎞1 + x 2y 1 + y 2⎠z 1 + z 2