Linear Algebra

Section I. Definition of Vector Space 81There are five conditions in item (1). For (1), closure of addition, note thatfor any v 1 , v 2 , w 1 , w 2 ∈ R the result of the sum( ) ( ) ( )v1 w1 v1 + w+ =1v 2 w 2 v 2 + w 2is a column array with two real entries, and so is in R 2 . For (2), that additionof vectors commutes, take all entries to be real numbers and compute( ) ( ) ( ) ( ) ( ) ( )v1 w1 v1 + w+ =1 w1 + v=1 w1 v1= +v 2 w 2 v 2 + w 2 w 2 + v 2 w 2 v 2(the second equality follows from the fact that the components of the vectors arereal numbers, and the addition of real numbers is commutative). Condition (3),associativity of vector addition, is similar.((v1v 2)+(w1w 2)) +(u1) ( )(v1 + w=1 ) + u 1u 2 (v 2 + w 2 ) + u 2( )v1 + (w=1 + u 1 )v 2 + (w 2 + u 2 )( ) ( )v1 w1= + ( +v 2 w 2(u1u 2))For the fourth condition we must produce a zero element — the vector of zeroesis it.( ) ( ( )v1 0 v1+ =v 2 0)v 2For (5), to produce an additive inverse, note that for any v 1 , v 2 ∈ R we have( ) ( ) (−v1 v1 0+ =−v 2 v 2 0)so the first vector is the desired additive inverse of the second.The checks for the five conditions having to do with scalar multiplication arejust as routine. For (6), closure under scalar multiplication, where r, v 1 , v 2 ∈ R,( ) ( )v1 rv1r · =v 2 rv 2is a column array with two real entries, and so is in R 2 . Next, this checks (7).( ) ( ) ( ) ( ) ( )v1 (r + s)v1 rv1 + sv(r + s) · ==1 v1 v1= r · + s ·v 2 (r + s)v 2 rv 2 + sv 2 v 2 v 2For (8), that scalar multiplication distributes from the left over vector addition,we have this.( ) ( ) ( ) ( ) ( ) ( )v1 w1 r(v1 + wr · ( + ) =1 ) rv1 + rw=1 v1 w1= r · + r ·v 2 w 2 r(v 2 + w 2 ) rv 2 + rw 2 v 2 w 2