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 12: Jordan Normal form

Let A be an n×n matrix. If I={i1,,im} is a subset of {1,2,,n} and J={j1,,jm} another subset of {1,2,,n} the (I,J) minor of t-A is det((t-A)IJ)= det ( (t-A)i1j1 (t-A)i1j2 (t-A)i1jm (t-A)imj1 (t-A)imj2 (t-A)imjm ) . If I and J have m elements det((t-A)IJ) is an mth order minor of t-A.

The gcd of the mth order minors of t-A is dm(t)=gcd { det(t-A)IJ| det(t-A)IJ is anmthorder minor oft-A } . The similarity invariants of A, or invariant factors of t-A, are monic polynomials q1(t),,qn(t) such that dm(t)= q1(t) q2(t) qm(t), for m=1,2,,n (i.e. qm(t)=dm(t)dm-1(t)).

The characteristic polynomial of A is dn(t).

The minimal polynomial of A is qn(t).

A Jordan block of type (t-λ)e is the e×e matrix ( λ 1λ0 1λ 0λ 1λ ) . Factor the q1(t),q2(t),,qn(t): q1(t) = (t-λ1)e11 (t-λ2)e12 (t-λk)e1k q2(t) = (t-λ1)e21 (t-λ2)e22 (t-λk)e2k = qn(t) = (t-λ1)en1 (t-λ2)en2 (t-λk)enk

There exists an invertible matrix P such that PAP-1= ( J11 J12 0 J1k J21 J22 0 Jn,k-1 Jn,k ) where Jm is a Jordan block of type (t-λ)em.

If A= ( 213 501 263 ) then t-A= ( t-213 5t1 26t-3 ) .

Subsets of {1,2,3}

If I and J have 3 elements then I={1,2,3}=J and (t-A)IJ=t-A and d3=det(t-A). So d3(t) = det(t-A)= (t-2)(t2-3t-6)- (5t-15-2)+3(30-2t) = t3-3t2-6t-2t2+ 6t+12-5t+17+90-6t = t3-5t2-11t+119. If I and J have 1 element then (t-A)IJ is a single entry of t-A. So d1(t)=gcd {t-2,1,3,5,t,1,2,6,t-3}=1. If I and J have 2 elements then Iis{1,2} or{1,3} or{2,3}and Jis{1,2} or{1,3} or{2,3}and d2(t) = gcd { det(t-215t), det(t-2351), det(1t31), det(t-2126), det(t-232t-3), det(136t-3), det(5t26), det(512t-3), det(t16t-3) } = gcd { t2-2t-5, t-10, 1-3t,, 5t-17, t2-3t-6 } =1. Wolfram alpha tells us that t3-5t2-11t+119= (t+4.23445) (t-(4.61723+2.60462i)) (t-(4.61723-2.60462i)). Since d1(t)=q1t, d2(t)=q1(t)q2(t), d3(t)=q1(t)q2(t)q3(t) and d1(t) = 1, d2(t) = 1, d3(t) = t3-5t2-11t+119, then q1(t) = 1, q2(t) = 1, q1(t) = t3-5t2-11t+119 and the factorization of q1(t), q2(t) and q3(t) are q1(t) = (t-α1)0 (t-α2)0 (t-α3)0, q2(t) = (t-α1)0 (t-α2)0 (t-α3)0, q3(t) = (t-α1)1 (t-α2)1 (t-α3)1 where α1=4.23445, α2 = 4.61723+i2.60462, α3 = 4.61723-i2.60462. The Jordan normal form theorem then says that there exists P such that PAP-1=P ( 213 501 263 ) P-1= ( α100 0α20 00a3 ) with α1,α2,α3 as above.

Notes and References

These are a typed copy of Lecture 12 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on August 19, 2011.

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