Jordan Canonical Form II - Computation

Converting from the Rational Canonical Form to the Jordan Canonical Form

To compute the Jordan canonical form for an nΓ—n matrix A, we first compute the invariant factors a1(x),…,am(x). For each invariant factor a(x), we factor a(x) completely into linear factors:

a(x)=(xβˆ’Ξ»1)Ξ±1β‹―(xβˆ’Ξ»s)Ξ±s,

where the Ξ»i are the distinct roots. For this factor, the corresponding elementary divisors are (xβˆ’Ξ»1)Ξ±1,…,(xβˆ’Ξ»s)Ξ±s. We can then write down the corresponding Jordan blocks J1,…,Js. Doing this for every invariant factor allows us to write down the Jordan canonical form J for A.

How about if we want to find a change-of-basis matrix Q such that Qβˆ’1AQ=J? We can compute that, as well. First note that if v∈V is the F[x]-module generator for the summand corresponding to F[x]/(a(x)), then the elements

a(x)(xβˆ’Ξ»1)Ξ±1β‹…v,…,a(x)(xβˆ’Ξ»s)Ξ±sβ‹…v

are F[x]-module generators for the summands corresponding to the quotients

F[x]/(xβˆ’Ξ»1)Ξ±1,…,F[x]/(xβˆ’Ξ»s)Ξ±s

(To see why this is true, first note that for each i the polynomials (xβˆ’Ξ»i)Ξ±i and a(x)(xβˆ’Ξ»i)Ξ±i are relatively prime; the former is the annihilator of the summand F[x]/(xβˆ’Ξ»i)Ξ±i, while the latter is the annihilator of the complement of that summand in the invariant summand F[x]/(a(x)). As such, multiplication by a(x)(xβˆ’Ξ»i)Ξ±i provides a surjective morphism F[x]/(a(x))β†’F[x]/(xβˆ’Ξ»i)Ξ±i.)

For each one of these summands, if w=a(x)(xβˆ’Ξ»)Ξ±β‹…v is the F[x]-module generator for the summand corresponding to F[x]/(xβˆ’Ξ»)Ξ±, then an F-vector space basis for that summand is

(Aβˆ’Ξ»In)Ξ±βˆ’1w,…,(Aβˆ’Ξ»In)w,w.

Note that the basis vectors have been listed in this order so that the corresponding matrix for T (on that summand) is the Jordan block, matching the way in which we chose the basis {(xβ€•βˆ’Ξ»)Ξ±βˆ’1,…,xβ€•βˆ’Ξ»,1} for F[x]/(xβˆ’Ξ»)Ξ±.

We can use this to algorithmically compute a change-of-basis matrix Q. We first compute an auxiliary matrix Pβ€² just as we did for the rational canonical form. The nonzero columns of Pβ€² correspond to vectors that generate the invariant factor summands as F[x]-modules. We can thus follow the recipe describe above, where v takes the role of a nonzero column of Pβ€². It's easiest to follow this through specific examples.

Examples

Example 1

Let A be the 3Γ—3 matrix

[2βˆ’21403βˆ’7002]

We have seen that the invariant factors of A are a1(x)=xβˆ’2 and a2(x)=x2βˆ’5x+6=(xβˆ’2)(xβˆ’3). The elementary divisors of A are therefore xβˆ’2, xβˆ’2, xβˆ’3, and so the Jordan canonical form is

J=[200020003]

For a change-of-basis matrix P, first recall an auxiliary matrix Pβ€² for this matrix was

Pβ€²=[0βˆ’7βˆ’1071010]

For the first nonzero column v1 of Pβ€² (which is a Q[x]-module generator for the summand corresponding to the first invariant factor a1(x)=xβˆ’2), we compute

w1=a1(x)(xβˆ’2)β‹…v1=v1=[βˆ’771].

Since the summand Q[x]/(xβˆ’2) is 1-dimensional, the single vector w1 is our Q-basis for that summand.

For the second nonzero column v2 of Pβ€² (which is a Q[x]-module generator for the summand corresponding to the second invariant factor a2(x)=(xβˆ’2)(xβˆ’3)), we compute the projections of v2 into the two summands Q[x]/(xβˆ’2) and Q[x]/(xβˆ’3), respectively:

w2=a2(x)(xβˆ’2)β‹…v2=(Aβˆ’3I)v2=(Aβˆ’3I)[βˆ’110]=[βˆ’100]w3=a2(x)(xβˆ’3)β‹…v2=(Aβˆ’2I)v2=(Aβˆ’2I)[βˆ’110]=[βˆ’210]

Again, since each of those summands is 1-dimensional, these are our Q-bases for those two summands.

A change-of-basis matrix is then

Q=[βˆ’7βˆ’1βˆ’2701100]

One can verify Qβˆ’1AQ=J.

Example 2

Consider the 4Γ—4 matrix

A=[12βˆ’442βˆ’14βˆ’8101βˆ’201βˆ’23]

We have seen that the invariant factors of this matrix are a1(x)=(xβˆ’1)2, a2(x)=(xβˆ’1)2. The elementary divisors of A are therefore (xβˆ’1)2, (xβˆ’1)2, and so the Jordan canonical form is

J=[1100010000110001]

An auxiliary matrix for A was

Pβ€²=[0010000100000000]

The first nonzero column v1 of Pβ€² is a Q[x]-module generator for the summand Q[x]/(xβˆ’1)2. There is nothing to project here, since this summand does not decompose further. (In other words, the invariant factor a1(x)=(xβˆ’1)2 is also an elementary divisor). That summand is 2-dimensional over Q, so we compute its Q-basis:

w1=(Aβˆ’I)1v1=[0210]w2=v1=[2βˆ’101]

The second nonzero column v2 of Pβ€² is a Q[x]-module generator for the second summand Q[x]/(xβˆ’1)2. By the identical reasoning used for the previous summand, a Q-basis for this summand is then

w3=(Aβˆ’I)1v2=[2βˆ’201]w4=v2=[0100]

Thus, a change-of-basis matrix Q is

Q=[012020βˆ’2110000010]