04A realjord
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Section of a linear algebra text (apparently from Matthews' notes, folder 'matthews linear algebra'). It motivates a real analogue of the Jordan block, computes exp(tD) for D=aI+bJ, and defines K_n(a,b) with its exponential. It then derives the real Jordan form from complex conjugate eigenvalue decompositions, and works Example 4.7 with minimal polynomial (x^2+1)^2.
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4.10 The Real Jordan Form
4.10.1 Motivation
IfAis a realnnmatrix, the characteristic polynomial of Awill in general
have real roots and complex roots, the latter occurring in complex pairs.
In this section we show how to derive a canonical form BforAwhich has
real entries. It turns out that there is a simple formula for eBand this is
useful in solving _X=AX, as it allows one to directly express the complete
solution of the system of di erential equations in terms of real exponentials
and sines and cosines.
We rst introduce a real analogue of Jn(a+ib). It's the matrix Kn(a; b)2
M2n2n(R) de ned as follows:
LetD=a b
b a
=aI2+bJwhereJ2= I2(Jis a matrix version
ofi=p 1, whileDcorresponds to the complex number a+ib) then
eD=eaI2+bJ
=eaI2ebJ
=eaI2
I2+bJ
1!+(bJ)2
2!+
=ea
I2 b2
2!I2+b4
4!I2+
+b
1!J b3
3!J+
=ea[(cosb)I2+ (sinb)J]
=eacosbsinb
sinbcosb
:
Replacingaandbbytaandtb, wheret2R, gives
etD=eatcosbtsinbt
sinbtcosbt
:
DEFINITION 4.7
Letaandbbe real numbers and Kn(a; b)2M2n2n(R)be de ned by
Kn(a; b) =2
666664D0:::
I2D
0I2
...
D3
777775
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whereD=a b
b a
. Then it is easy to prove that
eKn(a;b)=2
66666664eD0:::
eD=1!eD
eD=2!eD=1!...
.........
eD=(n 1)! eD=1!eD3
77777775:
EXAMPLE 4.6
K2(0;1) =2
6640 1 0 0
1 0 0 0
1 0 0 1
0 1 1 03
775
and
etK2(0;1)=2
664costsint 0 0
sintcost 0 0
tcost tsint costsint
tsint tcost sintcost3
775:
4.10.2 Determining the real Jordan form
IfA= [aij] is a complex matrix, let A= [aij]. Then
1.
AB=AB;cA=cA c2C;AB=AB:
2. IfA2Mnn(R) anda0;:::;ar2C, then
a0In+arAr=a0In++arAr:
3. IfWis a subspace of Vn(C), then so is W=fwjw2Wg.
Moreover if W=hw1;:::;wri, then
W=hw1;:::;wri:
4. Ifw1;:::;wrare linearly independent vectors in Vn(C), then so are
w1;:::;wr. Hence ifw1;:::;wrform a basis for a subspace W, then
w1;:::;wrform a basis for W.
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5. LetAbe a realnnmatrix and c2C. Then
(a)
W=N((A cIn)h))W=N((A cIn)h):
(b)
W=W1Wr)W=W1Wr:
(c)
W=CTA;v)W=CTA;v:
(d)
W=rM
i=1CTA;vi)W=rM
i=1CTA;vi:
(e)
mTA;v= (x c)e)mTA;v= (x c)e:
LetA2Mnn(R). ThenmA2R[x] and so any complex roots will occur in
conjugate pairs.
Suppose that c1;:::;crare the distinct real eigenvalues and cr+1;:::;cr+s,
cr+1;:::; cr+sare the distinct non-real roots and
mA= (x c1)b1:::(x cr)br(x cr+1)br+1:::(x cr+s)br+s
(x cr+1)br+1:::(x cr+s)br+s:
For each complex eigenvalue ci; r+1ir+s, there exists a secondary
decomposition
N(A ciIn)bi=
iM
j=1CTA;vij; mTA;vij= (x ci)eij
Hence we have a corresponding secondary decomposition for the eigenvalue
ci:
N(A ciIn)bi=
iM
j=1CTA;vij; mTA;vij= (x ci)eij:
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For brevity, let c=ci,v=vij,e=eij. Let
P1=v;P2= (A cIn)P1; :::;Pe= (A cIn)Pe 1
and
P1=X1+iY1; P2=X2+iY2; :::; Pe=Xe+iYe;c=a+ib:
Then we have the following equations, posed in two di erent ways:
AP1=cP1+P2AX1=aX1 bY1+X2
AY1=bX1+aY1+Y2
......
APe=cPeAXe=aXe bYe
AYe=bXe+aYe:
In matrix terms we have
A[X1jY1jX2jY2jjXejYe] =
[X1jY1jX2jY2jjXejYe]2
6666666664a b
b a
1 0a b
0 1 b a
......
a b
0 b a3
7777777775:
The large \real jordan form" matrix is the 2 e2ematrixKe(a; b).
Note : Ife= 1, noI2block is present in this matrix.
The spaces CTA;vandCTA;vare independent and have bases P1;:::;Pe
and P1;:::; Pe, respectively.
Consequently the vectors
P1;:::;Pe;P1;:::; Pe
form a basis for CTA;v+CTA;v. It is then an easy exercise to deduce that the
real vectors X1; Y1;:::;Xe; Yeform a basis for theT{invariant subspace
W=CTA;v+CTA;v:
WritingT=TAfor brevity, the above right hand batch of equations tells
us that [TW]
=Ke(a; b). There will be ssuch real bases corresponding to
each of the complex eigenvalues cr+1:::;cr+s.
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Joining together these bases with the real elementary Jordan bases aris-
ing from any real eigenvalues c1;:::;crgives a basis forVn(C) such that
ifPis the non{singular real matrix formed by these basis vectors, then
P 1AP= [TA]
=JK;
where
J=rM
i=1
iM
j=1Jeij(ci); K =r+sM
i=r+1
iM
j=1Keij(ai; bi);
whereci=ai+ibiforr+ 1ir+s.
The matrix JKis said to be in real Jordan canonical form.
EXAMPLE 4.7
A=2
6641 1 0 0
2 0 1 0
2 0 0 1
2 1 1 13
775somA= (x2+ 1)2
= (x i)2(x+i)2:
Thus withp1=x i, we have the dot diagram
N2;p1
N1;p1=N(A iI4):
Thus we nd an elementary Jordan basis for N1;p1:
X11+iY11;(A iI4)(X11+iY11) =X12+iY12
yielding
AX11= Y11+X12
AY11=X11+Y12:(22)
Now we know
mTA;X11+iY11= (x i)2
)(A iI4)2(X11+iY11) = 0
)(A iI4)(X12+iY12) = 0
)AX12= Y12
AY12=X12:(23)
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Writing the four real equations (22) and (23) in matrix form, with
P= [X11jY11jX12jY12];
thenPis non-singular and
P 1AP=2
6640 1 0 0
1 0 0 0
1 0 0 1
0 1 1 03
775:
The numerical determination of Pis left as a tutorial problem.
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