Group Theory and Linear Algebra

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last updated: 20 September 2014

Lecture 8: Bases and dimension

Let 𝔽 be a field. Let V be a vector space over 𝔽. Let S be a subset of V. A linear combination of elements of S is c1s2+ c2s2++ cksk, with k>0, c1,x2,,ck𝔽, s1,s2,,skS.

The span of S is span(S)= { c1s1++cksk |k>0, c1,,ck𝔽, s1,s2,,skS } .

The set S is linearly independent if S satisfies: If k>0, c1,,ck𝔽, s1,s2,,skS and c1s1+c2s2++cksk=0 then c1=0 and c2=0 and c3=0 and ... and ck=0.

A basis of V is a subset BV such that

(a) span(B)=V,
(b) B is linearly independent.

The dimension of V is Card(B), the number of elements in B, for a basis B of V.

[t]= { c0+c1t+c2t2+| ciand all but a finite number ofciequal 0 } is a vector space over : (3+4t+t2)+ (0+7t+1t2)= 3+11t+2t2 and 8.6(1+2t2)= 8.6+17.2t2. Let S={1,t,t2,t3,}. Then span(S)=[t] and S is linearly independent since

(a) If p[t] then p=p0+p1t+ p2t2++ pNtN with N>0 and 1,t,t2,,tNS and p0,p1,,pN,
(b) If p=p0+p1t++pNtN=0 then p0=0 and p1=0 and ... and pN=0. Since Card(S)=, the dimension of [t] is .

Let V be a vector space over 𝔽. Let f:VV be a linear transformation. Let B={b1,b2,} be a basis of V. The matrix of f with respect to B is Bf=(fij) given by f(bj)= f1jb1+ f2jb2+.

Let [t]3={c0+c1t+c2t2|c0,c1,c2}. Then B={1,t,t2} is a basis of [t]3.

Let f:[t]3[t]3 be given by f(p)=p(t-3). For example, f(3+2t-5t2) = 3+2(t-3)- 5(t-3)2 = 3+2t-6-5 (t2-6t+9) = 3+2t-6-5t2+ 30t-45 = -48+32t-5t2. To calculate the matrix of f with respect to B, f(1) = 1=1+0t+0t2, f(t) = t-3=-3+t+0t2, f(t2) = (t-3)2=t2 -6t+9=9-6t+t2 and Bf= ( 1-39 01-6 001 ) . Note that f(3+2t-5t2)= -48+32t-5t2 and ( 1-39 01-6 001 ) (32-5)= ( 3-6-45 2+30 -5 ) = ( -48 32 -5 ) .

Let f:VV be a linear transformation. The kernel, or nullspace, of f is kerf= {vV|f(v)=0} and the nullity of f is dim(kerf).

The image of f is imf= {f(v)|vV} and the rank of f is dim(imf).

Let f:[t][t] be given by f(p)=tp. For example, f(1+3t+2t2)=t+3t2+2t3. Then f(1) = t, f(t) = t2, f(t2) = t3, and the matrix of f with respect to the basis B={1,t,t3,} is Bf= ( 000 100 010 001 000 ) . Also, kerf = {0}, imf = span{t,t2,t3,}. So f is injective but not surjective.

Let d:[t][t] be given by d(p)=ddt·p. For example d(1+3t+2t2)=3+4t. Then d(1) = 0, d(t) = 1, d(t2) = 2t, d(t3) = 3t2, and the matrix of d with respect to the basis B={1,t,t2,t3,} is Bd= ( 01000 00200 00030 00004 ) . Also kerd = {c0|c0} =span{1}, imd = [t], since d(p0t+p1t22+p2t33+)= p0+p1t+p2t2 +p3t3+. So d is not injective, d is surjective, the nullity of d is 1, and the rank of d is .

Notes and References

These are a typed copy of Lecture 8 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on August 10, 2011.

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