## Group Theory and Linear Algebra

Last updated: 21 September 2014

## Lecture 12: Jordan Normal form

Let $A$ be an $n×n$ matrix. If $I=\left\{{i}_{1},\dots ,{i}_{m}\right\}$ is a subset of $\left\{1,2,\dots ,n\right\}$ and $J=\left\{{j}_{1},\dots ,{j}_{m}\right\}$ another subset of $\left\{1,2,\dots ,n\right\}$ the $\left(I,J\right)$ 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 $\text{det}\left({\left(t-A\right)}_{IJ}\right)$ is an ${m}^{\text{th}}$ order minor of $t-A\text{.}$

The gcd of the ${m}^{\text{th}}$ order minors of $t-A$ is $dm(t)=gcd { det(t-A)IJ | det(t-A)IJ is an mth order minor of t-A } .$ The similarity invariants of $A,$ or invariant factors of $t-A,$ are monic polynomials ${q}_{1}\left(t\right),\dots ,{q}_{n}\left(t\right)$ such that $dm(t)= q1(t) q2(t)⋯ qm(t),$ for $m=1,2,\dots ,n$ (i.e. ${q}_{m}\left(t\right)=\frac{{d}_{m}\left(t\right)}{{d}_{m-1}\left(t\right)}\text{).}$

The characteristic polynomial of $A$ is ${d}_{n}\left(t\right)\text{.}$

The minimal polynomial of $A$ is ${q}_{n}\left(t\right)\text{.}$

A Jordan block of type ${\left(t-\lambda \right)}^{e}$ is the $e×e$ matrix $( λ 1λ0 1λ ⋱ 0λ 1λ ) .$ Factor the ${q}_{1}\left(t\right),{q}_{2}\left(t\right),\dots ,{q}_{n}\left(t\right)\text{:}$ $q1(t) = (t-λ1)e11 (t-λ2)e12⋯ (t-λk)e1k q2(t) = (t-λ1)e21 (t-λ2)e22⋯ (t-λk)e2k ⋮ = qn(t) = (t-λ1)en1 (t-λ2)en2⋯ (t-λk)enk$

There exists an invertible matrix $P$ such that $PAP-1= ( J11 J12 ⋱ 0 J1k J21 J22 ⋱ ⋱ ⋱ ⋱ 0 ⋱ Jn,k-1 Jn,k )$ where ${J}_{m\ell }$ is a Jordan block of type ${\left(t-{\lambda }_{\ell }\right)}^{{e}_{m\ell }}\text{.}$

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

### Subsets of $\left\{1,2,3\right\}$

If $I$ and $J$ have $3$ elements then $I=\left\{1,2,3\right\}=J$ and ${\left(t-A\right)}_{IJ}=t-A$ and ${d}_{3}=\text{det}\left(t-A\right)\text{.}$ 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 ${\left(t-A\right)}_{IJ}$ is a single entry of $t-A\text{.}$ 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 ${d}_{1}\left(t\right)={q}_{1}t,$ ${d}_{2}\left(t\right)={q}_{1}\left(t\right){q}_{2}\left(t\right),$ ${d}_{3}\left(t\right)={q}_{1}\left(t\right){q}_{2}\left(t\right){q}_{3}\left(t\right)$ 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 ${q}_{1}\left(t\right),$ ${q}_{2}\left(t\right)$ and ${q}_{3}\left(t\right)$ 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 ${\alpha }_{1},{\alpha }_{2},{\alpha }_{3}$ 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.