The Rational Form

Abdullah Gumusoglu Burak Aldemir Busra Ozkır

Ihsan Demirel

July 1, 2014

Theorem for every linear operator T : V → V , there is an ordered basis βsuch that [T ]β is in rational form.

1 How to find rational form of linear trans-formation?

1. Determine the characteristic polynomialof T in form of f(x) = (−1)n(φ1(x)))n1(φ2(x))n2 ....(φk(x))nk .

2. Draw dot diagram for each φ.

• there are dim[ker(φ)]/d dots where d is degree of φ.

• there are r1 dots in first row, where r1 = 1d [(dim(V )− rank(φ(T ))]

• there are ri dots in in the ith row where 1d [rank(φ(T ))i−1−rank(φ(T )i)]

3. rational form of linear transformation ll be

R =

Z1,1 . . . . 0 0. Z1,2 . . . 0 0. . . . .. . Zi,j . .. . . . .0 0 . . . Zs−1 .0 0 . . . . Zk,l

where Zi,j is componion matrix of φai and a is number of dots in jth columnof φi’s dot diagram.

example let f(x)εP3(R) and T (f(x)) = f(0)x− f1′. let β = {1, x, x2, x3}

A = [T ]β =

0 −1 −2 −31 0 0 00 0 0 00 0 0 0

- characteristic polynomial is (t2 + 1)t2, φ1 = (t2 + 1) and φ2 = t.

1

- note that degree of φ1 is 2 and φ1(A) =

0 0 0 00 0 −2 −30 0 1 00 0 0 1

so dim(ker(φ1))=2

and dot diagram of φ1 has 2/2=1 dot.

- similarly degree of φ2=1 dim(ker(φ2))=2 so dot diagram of φ1 has 2/1=2 dots,first row has r1 dots where r1 = 1

d [(dim(V )− rank(φ(T ))] = 11 [(4− 2)] =

2(we found all dots so don’t need to check other rows.)

- dot diagram of phi1 dot diagram is

φ1 φ2• • •

- dot diagram of φ1 has only 1 column with 1 dot so Z1,1 is companion matrix

of φ11 = t2 + 1 that is

∣∣∣∣0 −11 0

∣∣∣∣- dot diagram of phi2 has only 2 column with 1 dot so Z2,1 = Z1,2 is companion

matrix of φ12 = t that is∣∣0∣∣

- so rational form of linear operation T is

Z1,1 . 0. Z2,1 .0 . Z2,2

=

0 −1 0 01 0 00 0 0 00 0 0 0

2 How to find ordered basis for rational form?

- assume that φ’s dot diagram has n column.

- for ith column, choose vi such that φ(vi)ai=0 where ai is number of dots in

column.let βvi = {vi, T (vi), ....., Tb−1(vi)} where b is degree of φ(vi)

ai .

- for (i+1)th column, choose vi+1 (not from span(βvi)) such that φ(vi+1)ai+1=0.letβvi+1

= {vi+1, T (vi+1), ....., T b−1(vi+1)} where b is degree of φai+1vi+1 .

- βφ =⋃ni=1 βi is basis for Kβ .

- for each φ, union of all βφ is ordered basis for rational form.

example consider previous example, we know that dot diagram of phi1 dotdiagram is

φ1 φ2• • •

- note that φ1(A)(1, 0, 0, 0) = 0 ,φ1 = t2 + 1 has degree 2, and its dot di-agram has 1 column with 1 dot so βφ1

= {(1, 0, 0, 0), A(1, 0, 0, 0)} ={(1, 0, 0, 0), (0, 1, 0, 0)}

- note that φ2(A)(0,−2, 1, 0) = 0 , φ2(A)(0,−3, 0, 1) = 0 , φ2 = t hass degree1 , and its dot diagram has 2 columns with 1 dot so β1 = {(0,−2, 1, 0)},β2 = {(0,−3, 0, 1)} and βφ2

= β1 ∪ β2 = {(0,−2, 1, 0), (0,−3, 0, 1)}

2

- ordered basis for rational form is β = βφ1∪βφ2

= {(1, 0, 0, 0), (0, 1, 0, 0), (0,−2, 1, 0), (0,−3, 0, 1)}

Q−1∗A∗Q =

1 0 0 00 1 2 30 0 1 00 0 0 1

*

0 −1 −2 −31 0 0 00 0 0 00 0 0 0

*

1 0 0 00 1 −2 −30 0 1 00 0 0 1

=

0 −1 0 01 0 0 00 0 0 00 0 0 0

3