04B rjalg
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Section 4.10.3 of a linear algebra text (Matthews, per the folder name), apparently with no mention of Phil's own work. It builds the 2n-by-2n real matrix Z from A and a complex eigenvalue a+ib, proves lemmas and a corollary linking null spaces of Z and p(A), and gives a procedure for choosing basis vectors. Example 4.8 works a 4x4 matrix with minimal polynomial (x^2+1)^2, producing P and the real Jordan form.
AI-written summary; may contain errors. This description is approximate.
Extracted text (machine-read; may contain errors)
4.10.3 A real algorithm for nding the real Jordan form
Referring to the last example, if we write Z=A I 4
I4A
, then
ZX11
Y11
=X12
Y12
;
Z Y11
X11
= Y12
X12
;
ZX12
Y12
=0
0
;
Z Y12
X12
=0
0
:
Then the vectorsX11
Y11
; Y11
X11
; ZX11
Y11
; Z Y11
X11
actually form an R{basis forN(Z). This leads to a method for nding the
real Jordan canonical form using real matrices. (I am indebted to Dr. B.D.
Jones for introducing me to the Zmatrix approach.)
More generally, we observe that a collection of equations of the form
AXij1=aiXij1 biYij1+Xij2
AYij1=biXij1+aiYij1+Yij2
...
AXijeij=aiXijeij biYijeij
AYijeij=biXijeij+aiYijeij
can be written concisely in real matrix form, giving rise to an elementary
Jordan basis corresponding to an elementary divisor xeijfor the following
real matrix: Let
Zi=A aiInbiIn
biInA aiIn
:
Then
ZiXij1
Yij1
=Xij2
Yij2
...
ZiXijeij
Yijeij
=0
0
:
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LEMMA 4.2
IfVis aC{vector space with basis v1;:::;vn, thenVis also anR{vector
space with basis
v1; iv1;:::;vn; ivn:
Hence
dimRV= 2 dimCV:
DEFINITION 4.8
LetA2Mnn(R)andc=a+ibbe a complex eigenvalue of Awith
b6= 0. LetZ2M2n2n(R)be de ned by
Z=A aInbIn
bInA aIn
= (A aIn)
In In
(bJ):
Also letp=x c.
LEMMA 4.3
Let :V2n(R)!Vn(C)be the mapping de ned by
X
Y
=X+iY; X; Y2Vn(R):
Then
(i)is anR{ isomorphism;
(ii) Y
X
=i(X+iY);
(iii)
ZhX
Y
=ph(A)(X+iY);
(iv)
Zh Y
X
=iph(A)(X+iY);
(v)mapsN(Zh)ontoN(ph(A);
COROLLARY 4.4
If
pe1 1(A)(X1+iY1);:::;pe
1(A)(X
+iY
)
form aC{basis forN(p(A)), then
Ze1 1X1
Y1
; Ze1 1 Y1
X1
;:::;Ze
1X
Y
; Ze
1 Y
X
form anR{basis forN(Z)and conversely.
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Remark: Consequently the dot diagram for the eigenvalue 0 for the matrix
Zhas the same height as that for the eigenvalue cofA, with each row
expanded to twice the length.
To nd suitable vectors X1; Y1;:::;X
; Y
, we employ the usual algo-
rithm for nding the Jordan blocks corresponding to the eigenvalue 0 of the
matrixZ, with the extra proviso that we always ensure that the basis for
Nh;xis chosen to have the form
Zh 1X1
Y1
; Zh 1 Y1
X1
;:::;Zh 1Xr
Yr
; Zh 1 Yr
Xr
;
wherer= ( nullityZh nullityZh 1)=2.
This can be ensured by extending a spanning family for N(Zh):
X1
Y1
;:::;"
X(Zh)
Y(Zh)#
to the form
X1
Y1
; Y1
X1
;:::;"
X(Zh)
Y(Zh)#
;"
Y(Zh)
X(Zh)#
:
EXAMPLE 4.8
A=2
6641 1 0 0
2 0 1 0
2 0 0 1
2 1 1 13
7752M44(R)hasmA= (x2+1)2. Find a real
non{singular matrix Psuch thatP 1APis in real Jordan form.
Solution:
Z=2
666666666641 1 0 0 1 0 0 0
2 0 1 0 0 1 0 0
2 0 0 1 0 0 1 0
2 1 1 1 0 0 0 1
1 0 0 0 1 1 0 0
0 1 0 0 2 0 1 0
0 0 1 0 2 0 0 1
0 0 0 1 2 1 1 13
77777777775
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basis forN(Z2) :2
666666666641 1 1 =2 1=2
2 1=2 0 1=2
2 1=2 1 3 =2
2 3=2 2 3=2
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 13
77777777775
blown{up basis for N(Z2) :
2
666666666641 1 1 0 1 =2 0 1 =2 0
2 0 1=2 1 0 0 1=2 0
2 0 1=2 0 1 1 3=2 0
2 0 3=2 0 2 0 3=2 1
1 1 0 1 0 1 =2 0 1 =2
0 2 1 1=2 0 0 0 1=2
0 2 0 1 =2 1 1 0 3 =2
0 2 0 3=2 0 2 1 3=23
77777777775
!left{to{right basis for N(Z2) :2
666666666641 1 1 0
2 0 1=2 1
2 0 1=2 0
2 0 3=2 0
1 1 0 1
0 2 1 1=2
0 2 0 1 =2
0 2 0 3=23
77777777775
We then derive a spanning family for N2;x:
Zbasis matrix =2
666666666640 0 1=2 0
0 0 1=2 1=2
0 0 1=2 1=2
0 0 1=2 1=2
0 0 0 1 =2
0 0 1=2 1=2
0 0 1=2 1=2
0 0 1=2 1=23
77777777775!basis forN2;x:
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2
666666666641=2 0
1=2 1=2
1=2 1=2
1=2 1=2
0 1=2
1=2 1=2
1=2 1=2
1=2 1=23
77777777775
Consequently we read o that Zh
X11
Y11i
=h
X12
Y12i
is a basis for N2;x=N1;x=
N(Z). where
P= [X11jY11jX12jY12] =2
6641 0 1=2 0
1=2 1 1=2 1=2
1=2 0 1=2 1=2
3=2 0 1=2 1=23
775:
Then
P 1AP=2
6640 1 0 0
1 0 0 0
1 0 0 1
0 1 1 03
775;
which is in real Jordan form.
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