Group Theory and Linear Algebra

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

Last updated: 21 September 2014

Lecture 13: Block decomposition

Let V be a vector space. Let B={b1,,bk} and C={c1,,c} be bases of V. Then k=.


Proof by contradiction.
Assume k<.
Let C={c2,,c}.
Then there exists bjB such that bjspan(C).


If all b1,,bkspan(C) then span(B)span(C) andVspan(C). Since c1V then c1span(C).
So c1=γ2c2++γc for some γ2,,γ𝔽.
So 0=-c1+γ2c2++γc.
This is a contradiction to the linear independence of C.
So there exists bj such that bjspan(C).

Reindex B so that b1span(C) and let C1={b1,c2,c3,,c}.

Claim C1 is a basis of V.

Proof of claim.

To show:
(a) C1 is linearly independent.
(b) span(C1)=V.
To show: If α1b1+α2c2++αc=0 then α1=0, α2=0, , α=0.
Assume α1b1+α2c2++αc=0.
Case 1 α0.
Then b1=-α2α1c2-α3α1c3--αα1a.
So b1span(C), which is a contradiction to the choice of b1.
So α1=0.
Case 2 α1=0.
Then α2c2++αc=0.
Since C is linearly independent α2=0, α3=0,, α=0.
To show: V=span{b1,c2,,c}.
To show: Vspan{b1,c2,,c}.
To show: span{c1,c2,,c}span{b1,c2,,c}.
To show: c1span{b1,c2,,c}.
Since C is a basis there exist α1,,α𝔽 with b1=α1c1+α2c2++αc.
Case 1 α1=0.
Then b1=α2c2++αc and b1span(C).
This is a contradiction to the choice of b1.
So α10.
Case 2 α20.
Then c1=α1-1(b1-α2c2--αc).
So c1span{b1,c2,,c}.
So C1={b1,c2,,c} is a basis of V.

Let C1={b1,c3,,c} and let bjB such that bjspan(C1).
Reindex B so that b2span(C1).
Let C2={b1,b2,c3,,c}.
Then, by a proof as for C1 above, C2 is a basis of V.
Continue this process to obtain Ck= {b1,b2,,bk,ck+1,ck+2,,c} which is a basis of V. Since B is a basis of V, then ck+1=α1b1 +α2b2++αk bk, for some α1,,αk𝔽.
So 0=α1b1++αkbk-ck+1.
This contradicts the linear independence of Ck.
So k.
A similar argument shows k.
So k=.

Let V be a vector space and let f:VV be a linear transformation. Let m(t) be the minimal polynomial of f. Assume that m(t)=p(t)q(t) with gcd(p,q)=1. Let k(t) and (t) be such that 1=p(t)k(t) +q(t)(t). Let U=p(f)k(f)V andW=q(f)(f) V. Then V=UW.

Let A= ( 20 12 30 13 ) . The minimal polynomial of A is m(t)= (t-2)2 (t-3)2 and p(t)=(t-2)2, q(t)=(t-3)2. p(t)=t2-4t+4and q(t)=t2-6t+9. t2-4t+4 = (t2-6t+9)+ (2t-5) t2-6t+9 = (2t-5) (12t-74)+14 since t2-4t+4= (t2-6t+9)+ (2t-5). 12t-74 2t-5 |t2-6t+9 | t2-52t |t2 -72t+9 |t2 -72t+354 0.5 |t2 -72t+ 14 So t2-6t+9= (12t-74) (2t-5)+14. So 1 = 4(t2-6t+9)- (2t-7)(2t-5) = 4(t2-6t+9)- (2t-7) ((t2-4t+4)-(t2-6t+9)) = (2t-3) (t2-6t+9)- (2t-7) (t2-4t+4) = (2t-3)(t-3)2 -(2t-7)(t-2)2. Then (2A-3) (A-3)2 = ( 10 21 30 23 ) ( -1 1-1 0 10 ) 2 = ( 1 21 3 23 ) ( 10 -21 00 00 ) = ( 10 01 00 00 ) and -(2A-7) (A-2)2 = ( 30 -23 10 -21 ) ( 00 10 10 11 ) 2 = ( 30 -23 10 -21 ) ( 00 00 10 21 ) = ( 00 00 10 01 ) .

Notes and References

These are a typed copy of Lecture 13 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on August 23, 2011.

page history