If all ai=0 then clearly W=V and e1,...,en suffice as a basis. Otherwise, let am≠0. For (x1,...,xn),(y1,...,yn)∈W we see(x1,...,xn)−r(y1,...,yn)=(x1−ry1,...,xn−ryn)as well as(x1−ry1)a1+...+(xn−ryn)an=(x1a1+...+xnan)−r(y1a1+...+ynan)=0so that W is a subspace. We claim ei−(ai/am)em for all i∈{1,...,m−1,m+1,...,n} is a basis for W. Foremost, all of these are seen to be vectors of W. Linear independence:b1(e1−(a1/am)em)+...+bm−1(em−1−(am−1/am)em)+bm+1(em+1−(am+1/am)em)+...+bn(en−(an/am)em)=0⇒b1e1+...+bm−1em−1+bm+1em+1+...+bnen−(∑i≠mbiai/am)em=0⇒b1,...bm−1,bm+1,...,bn=0This implies W is either of dimension n−1 or n. If it were the latter, then we would have W=V despite em∉W so that necessarily W is of dimension n−1 and now this set must be a basis. ◻
Sunday, July 7, 2013
Basis Calculation (11.1.1)
Dummit and Foote Abstract Algebra, section 11.1, exercise 1:
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Let V=Rn and let (a1,...,an)∈V be fixed. Let W⊆V be the set of vectors (x1,...,xn) such that x1a1+...+xnan=0. Prove W is a subspace and find a basis for W.
If all ai=0 then clearly W=V and e1,...,en suffice as a basis. Otherwise, let am≠0. For (x1,...,xn),(y1,...,yn)∈W we see(x1,...,xn)−r(y1,...,yn)=(x1−ry1,...,xn−ryn)as well as(x1−ry1)a1+...+(xn−ryn)an=(x1a1+...+xnan)−r(y1a1+...+ynan)=0so that W is a subspace. We claim ei−(ai/am)em for all i∈{1,...,m−1,m+1,...,n} is a basis for W. Foremost, all of these are seen to be vectors of W. Linear independence:b1(e1−(a1/am)em)+...+bm−1(em−1−(am−1/am)em)+bm+1(em+1−(am+1/am)em)+...+bn(en−(an/am)em)=0⇒b1e1+...+bm−1em−1+bm+1em+1+...+bnen−(∑i≠mbiai/am)em=0⇒b1,...bm−1,bm+1,...,bn=0This implies W is either of dimension n−1 or n. If it were the latter, then we would have W=V despite em∉W so that necessarily W is of dimension n−1 and now this set must be a basis. ◻
If all ai=0 then clearly W=V and e1,...,en suffice as a basis. Otherwise, let am≠0. For (x1,...,xn),(y1,...,yn)∈W we see(x1,...,xn)−r(y1,...,yn)=(x1−ry1,...,xn−ryn)as well as(x1−ry1)a1+...+(xn−ryn)an=(x1a1+...+xnan)−r(y1a1+...+ynan)=0so that W is a subspace. We claim ei−(ai/am)em for all i∈{1,...,m−1,m+1,...,n} is a basis for W. Foremost, all of these are seen to be vectors of W. Linear independence:b1(e1−(a1/am)em)+...+bm−1(em−1−(am−1/am)em)+bm+1(em+1−(am+1/am)em)+...+bn(en−(an/am)em)=0⇒b1e1+...+bm−1em−1+bm+1em+1+...+bnen−(∑i≠mbiai/am)em=0⇒b1,...bm−1,bm+1,...,bn=0This implies W is either of dimension n−1 or n. If it were the latter, then we would have W=V despite em∉W so that necessarily W is of dimension n−1 and now this set must be a basis. ◻
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