Diagonalizable matrices
• If matrix A has n real and distinct eigenvalues λi, then it also has n real eigenvectors vi which are linearly independent:
A vi = λivi, i ∈ {1, 2, 3, . . . , n}.
In this case the matrix A can be diagonalized using a transformation matrix T having the eigenvectors vi as its columns:
T = v1 v2 v3 . . . vn
• The transformed matrix A = T−1AT is diagonal. The coefficients on the diagonal of matrix A are equal to the eigenvalues λi of matrix A:
A = T−1AT =
λ1
λ2 λ3
. ..
λn
• The order of the eigenvalues λi on the diagonal of matrix A if equal to the order of the the corresponding eigenvectors vi within matrix T.
• A matrix A can be diagonalized if and only if it is possible to find a number of linear independent eigenvalues vi equal to the dimension n of matrix A.
• The matrices which are not diagonalizable can be transformed in the Jordan canonical form. This canonical form is the form more likely diagonal.
• A matrix A can be diagonalized also if its eigenvalues λi are complex conjugate. In this case the eigenvectors vi are complex conjugate and the transformation T is complex.
Jordan canonical form
• Let λi, for i = 1, . . . , h, be the “distinct” eigenvalues of matrix A and let ri
be the corresponding molteplicity degree within the characteristic polynomial:
∆A(λ) = (λ − λ1)r1(λ− λ2)r2 . . .(λ− λh)rh
• A matrix A transformed in the Jordan canonical form A has the following block diagonal form:
A = T−1AT =
J1 0 . . . 0 0 J2 . . . 0 ... ... ... ...
0 0 . . . Jh
• To each distinct eigenvalue λi of matrix A corresponds Jordan block Ji of dimension equal to the algebraic molteplicity ri of eigenvalue λi, that is the molteplicity ri of eigenvalue λi within the characteristic polynomial ∆A(λ).
• Each Jordan block Ji has itself the structure of a block diagonal matrix:
Ji =
Ji,1 0 . . . 0 0 Ji,2 . . . 0 ... ... ... ...
0 0 . . . Ji,mi
dim Ji = ri
i = 1, . . . , h
• On the diagonal of the Jordan block Ji are present mi Jordan miniblocks Ji, j, where mi is the geometric molteplicity of the eigenvaalue λi, that is the number of linear independent eigenvectors vi,j associated to the eigenvalue λi.
• The structure of all the Jordan miniblocks Ji, j is the following:
Ji,j =
λi 1 0 . . . 0 0 0 λi 1 . . . 0 0 0 0 λi . . . 0 0 ... ... ... ... ... ...
0 0 0 . . . λi 1 0 0 0 . . . 0 λi
dim Ji,j = νi,j
j = 1, . . . , mi
• The following relations hold:
n =
h
X
i=1
ri, ri =
mi
X
j=1
νi,j.
• The matrix A is diagonalizable if and only if the dimension νi,j of all the Jordan miniblocks Ji, j is unitary: νi,j = 1.
• In this case all the Jordan miniblocks Ji, j = λi are equal to the corresponding eigenvalue λi and all the Jordan blocks Ji are diagonal and characterized by the same eigenvalue λi:
Ji, j = λi, Ji =
λi 0 . . . 0 0 λi . . . 0 ... ... ... ...
0 0 . . . λi
,
dim Ji = ri
i = 1, . . . , h
• A matrix A is diagonalizable if and only if the algebraic molteplicity ri of each eigenvalue λi is equal to the geometric molteplicity mi, that is if the number mi of linearly independent eigenvectors vi,j associated to each eigenvalue λi is equal to the algebraic molteplicity ri of the eigenvalue λi
within the characteristic polynomial ∆A(λ).
• Example. Matrix in Jordan canonical form:
A=
−1 1 0 −1
−3
−3 1 0 −3
=
J1,1
J2,1 J2,2
=J1 J2
,
∆A(λ) = (λ + 1)2(λ + 3)3 n = 5, h = 2 λ1 = −1, λ2 = −3
r1 = 2, r2 = 3 m1 = 1, m2 = 2 The matrix A has two distinct eigenvalues λ1 = −1 and λ2 = −3. The eigenvalue λ1 has algebraic molteplicity r1 = 2 and geometric molteplicity m1 = 1. The eigenvalue λ2 has algebraic molteplicity r2 = 3 and geometric molteplicity m2 = 2. The Jordan block J1 has dimension r1 = 2 and it is composed by only one miniblock J1,1. The second Jordan block J2 has dimension r2 = 3 and it is composed by m2 = 2 miniblocks J2,1 and J2,2.
• The mi linear independent eigenvectors vi,j associated to the eigenvalue λi
can be determined solving the following autonomous linear system:
(λiI− A)vi,j = 0, j = 1, . . . , mi.
• The number mi of linearly independent eigenvectors vi,j is equal to the number of miniblocks Ji,j present within the Jordan block Ji.
• In the case mi < ri, the number of linearly independent eigenvectors vi,j
is not sufficient for diagonalizing the matrix. In this case, the chains vi,j(k) of generalized eigenvectors for i = 1, 2, . . . , h, j = 1, 2, . . . , mi and k = 1, 2, . . . , νi,j must be determined.
• The generalized eigenvectors v(k)i,j can be determined solving the following linear equations:
(A − λiI)vi,j(2) = vi,j(1) = vi,j
(A − λiI)vi,j(3) = vi,j(2)
... ...
(A − λiI)v(νi,ji,j) = vi,j(νi,j−1)
The first eigenvector v(1)i,j = vi,j is known. Solving the first equation one obtains the generalized eigenvector v(2)i,j. Substituting vi,j(2) and solving the next equation one obtains the generalized eigenvector v(3)i,j, . . ., and so on.
• The particular “almost diagonal” form of the Jordan miniblocks Ji,j is ob- tained inserting the chains of generalized eigenvectors v(k)i,j as columns of the transformation matrix T:
T = h
. . . vi,j(1) vi,j(2) . . . v(νi,ji,j) . . . i
• If the geometric molteplicity mi is equal to the algebraic molteplicity ri, that is if mi = ri, the chain of generalized eigenvectors v(k)i,j has unitary length and it is formed by only one eigenvector vi,j.
• Numeric example in Matlab:
--- clc; echo on
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Matrix to be transformed in Jordan canonical form
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
A =sym([...
73/129, -18/43, 1/129, 383/129, -1213/258;
2610/989, -2419/989, -687/989, 1759/989, -6889/1978;
2885/989, 655/989, -4039/989, 2310/989, -3905/989;
2273/989, -2079/989, 1565/989, -326/989, -5915/1978;
4456/2967, -3052/989, 9190/2967, 4778/2967, -13964/2967]);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% The command "jordan(A)" transforms matrix A in Jordan canonical form
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
[V,AJ]=jordan(A);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
AJ %%% Jordan canonical form of marix A
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
AJ =
[ -3, 1, 0, 0, 0]
[ 0, -3, 0, 0, 0]
[ 0, 0, -1, 1, 0]
[ 0, 0, 0, -1, 0]
[ 0, 0, 0, 0, -3]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
V %%% Matrix of generalized eigenvectors of matrix A
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
V =
[ -522/989, -242/989, 4760/2967, 1231/989, 6450201/4106690]
[ -406/989, -3334/2967, 2380/2967, 3334/2967, -142348/293335]
[ -290/989, -3275/2967, 2975/2967, 3275/2967, -903187/1642676]
[ -348/989, -418/989, 1785/989, 418/989, -186498/2053345]
[ -580/989, -718/2967, 4760/2967, 718/2967, 966065/821338]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Verification: the columns of V are the generalized eigenvectors of A
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
(A-(-3)*eye(5))*V(:,1)==0 % --> Vero (A-(-3)*eye(5))*V(:,2)==V(:,1) % --> Vero (A-(-1)*eye(5))*V(:,3)==0 % --> Vero (A-(-1)*eye(5))*V(:,4)==V(:,3) % --> Vero (A-(-3)*eye(5))*V(:,5)==0 % --> Vero
---
• The free evolution of a discrete linear system is the following:
x(k) = Akx0 = (TAT−1)kx0 = TAkT−1x0
= T
J1 0 . . . 0 0 J2 . . . 0 ... ... ... ...
0 0 . . . Jh
k
T−1x0 = T
Jk1 0 . . . 0 0 Jk2 . . . 0 ... ... ... ...
0 0 . . . Jkh
T−1x0
• The free evolution of a time-continuous linear system is the following:
x(t) = eAtx0 = e(TAT−1)tx0 = TeAtT−1x0
= T e
J1 0 . . . 0 0 J2 . . . 0
.. .
.. .
.. .
.. . 0 0 . . . Jh
t
T−1x0 = T
eJ1t 0 . . . 0 0 eJ2t . . . 0 ... ... ... ...
0 0 . . . eJht
T−1x0
• The power Ak and the exponential eAt of matrix A can be determined if it is known how to compute the power Jki and the exponential eJit of the generic Jordan miniblock Ji of dimension ν:
J =
λ 1 0 . . . 0 0 0 λ 1 . . . 0 0 0 0 λ . . . 0 0 ... ... ... ... ... ...
0 0 0 . . . λ 1 0 0 0 . . . 0 λ
= λI + N
• Matrix J can be expressed as a sum of the diagonal matrix λI and the nilpotent matrix N which has non zero and unitary elements only on the first over-diagonal. In the case ν = 5, it is:
N =
0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0
• The powers of matrix N have the following structure:
N2 =
0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0
N3 =
0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
. . .
that is, the matrix Nk has non zero elements only on the k-th over-diagonal.
• The matrix N is nilpotent of order ν if it satisfies the following relation:
Nν = 0 dove ν = dim N.
• The k-th power of matrix J can be expressed in the following way:
Jk = (λI + N)k = λkI + k 1
λk−1N+ k 2
λk−2N2 + . . . + Nk
• Knowing that Nk are zero for k ≥ ν, it follows that:
Jk = (λI + N)k
= λkI + k 1
λk−1N+ k 2
λk−2N2 + . . . +
k ν−1
λk−ν+1Nν−1
=
λk kλk−1 k(k−1)2 λk−2 . . .
k ν−2
λk−ν+2
k ν−1
λk−ν+1
0 λk kλk−1 . . . ...
k ν−2
λk−ν+2
0 0 λk . . . ... ...
... ... ... ... ... ...
0 0 0 . . . λk kλk−1
0 0 0 . . . 0 λk
• The symbol k h
has been used to denote the following binomial coefficient:
k h
= k(k − 1) . . . (k − h + 1) h!
• The functions f(λ, k) present within matrix Jk are called the modes of the discrete system associated to the eigenvalue λ.
• In the continuous-time case the exponential eJt can be computed as follows:
eJt = e(λI+N)t = eλIteNt = eλtIeNt
= eλt
∞
X
n=0
tn
n!Nn = eλt
ν−1
X
n=0
tn n!Nn
= eλth
I+ tN + t22N2 + . . . + (ν−1)!tν−1 Nν−1i
=
eλt teλt t22eλt t3!3eλt . . . (ν−3)!tν−3 eλt (ν−2)!tν−2 eλt (ν−1)!tν−1 eλt 0 eλt teλt t22eλt . . . ... (ν−3)!tν−3 eλt tν−2
(ν−2)!eλt 0 0 eλt teλt . . . ... ... (ν−3)!tν−3 eλt
0 0 0 eλt . . . ... ... ...
... ... ... ... ... ... ... ...
0 0 0 0 . . . eλt teλt t22eλt
0 0 0 0 . . . 0 eλt teλt
0 0 0 0 . . . 0 0 eλt
• The time functions f(λ, t) which are present within matrix eJt are called the modes of the continuous-time system associated too the eigenvalue λ.
• The Jordan canonical form can be obtained also in the case of complex conjugate eigenvalues λi. Numerical example in Matlab:
---
A =[ 1 6 -3 -7; --> AJ=[-5.15+4.77i 0 0 0;
-5 -8 2 7; 0 -5.15-4.77i 0 0;
-20 -16 -15 -17; 0 0 -2.85+3.86i 0;
10 6 8 6] 0 0 0 -2.85-3.86i]
[T,AJ]=eig(A) % Transform matrix A in Jordan canonical form AJ T=[-0.49-0.17i -0.49+0.17i 0.67 0.67;
0.54 0.54 -0.60+0.25i -0.60-0.25i;
0.30-0.42i 0.30+0.42i -0.31-0.05i -0.31+0.05i;
-0.22+0.36i -0.21-0.36i -0.01-0.13i -0.01+0.13i]
---
• In the case of complex conjugate eigenvalues λi also the matrices T and AJ are “complex” and therefore “not easily” to be used. In this case, the Jordan real form can be used.
• Jordan real form. Let us consider a matrix A of dimension 6 charac- terized by two complex conjugate eigenvalues λ1,2 with algebraic molteplicity r = 3 and geometric molteplicity m = 1:
λ1,2 = σ ± jω, ∆A(λ) = (λ − λ1)3(λ − λ2)3 = [(λ − σ)2 + ω2]3 Let v1, v2 and v3 be the chain of complex generalized eigenvectors associated to the complex eigenvalue v1. Applying to A the following “complex” state space transformation:
x = T x, T = v1 v2 v3 v∗1 v∗2 v3∗ one obtains a “complex” matrix A in Jordan canonical form:
A = T−1AT =
λ1 1 0 0 0 0 0 λ1 1 0 0 0 0 0 λ1 0 0 0 0 0 0 λ∗1 1 0 0 0 0 0 λ∗1 1 0 0 0 0 0 λ∗1
Two “complex” Jordan miniblocks of dimension 3 are obtained.
• Let viR and viI be the real and the imaginary parts of the generalized eigenvector vi. Using the following state space transformation:
x = ˜Tx,˜ T˜ = v1R v1I v2R v2I v3R v3I the matrix A can be transformed in the “Jordan real form”:
A˜ = ˜T−1A ˜T =
σ ω 1 0 0 0
−ω σ 0 1 0 0
0 0 σ ω 1 0
0 0 −ω σ 0 1
0 0 0 0 σ ω
0 0 0 0 −ω σ
(1)
• In this case the free evolution of linear systems can be expressed as linear combinations of only real terms:
x(k) = Akx0 = ˜T ˜AkT˜−1x0, x(t) = eAtx0 = ˜TeAt˜ T˜−1x0.
• Let |λ| and θ denote the module and the phase of the complex number λ = σ + jω:
|λ| = √
σ2 + ω2 θ = arctan ω
σ
( σ = |λ| cos θ ω = |λ| sin θ
|λ| λ θ
σ ω
• The Jordan real miniblock J can be expressed in the following way:
J =
σ ω
−ω σ
= |λ|
cos θ sin θ
− sin θ cos θ
.
• The following relations hold:
eJt = eσt
cos ωt sin ωt
− sin ωt cos ωt
, Jk = |λ|k cos kθ sin kθ
− sin kθ cos kθ
.
• Example in Matlab:
--- syms sig w t J=[sig w; -w sig] EJt=simplify(expm(J*t))
J = EJt =
[ sig, w] [ exp(sig*t)*cos(t*w), exp(sig*t)*sin(t*w)]
[ -w, sig] [ -exp(sig*t)*sin(t*w), exp(sig*t)*cos(t*w)]
---
• The k-th power ˜Ak of the real Jordan matrix ˜A in (1) is the following:
A˜k=
|λ|k cos kθ sin kθ
− sin kθ cos kθ
k|λ|k−1 cos(k−1)θ sin(k−1)θ
− sin(k−1)θ cos(k−1)θ
k(k−1)
2 |λ|k−2 cos(k−2)θ sin(k−2)θ
− sin(k−2)θ cos(k−2)θ
0 |λ|k cos kθ sin kθ
− sin kθ cos kθ
k|λ|k−1 cos(k−1)θ sin(k−1)θ
− sin(k−1)θ cos(k−1)θ
0 0 |λ|k cos kθ sin kθ
− sin kθ cos kθ
• The exponential eAt˜ of the real Jordan matrix ˜A in (1) is the following:
eAt˜ =
eσtcos ωt eσtsin ωt teσtcos ωt teσtsin ωt t22eσtcos ωt t22eσtsin ωt
−eσtsin ωt eσtcos ωt −teσtsin ωt teσtcos ωt −t22eσtsin ωt t22eσtcos ωt 0 0 eσtcos ωt eσtsin ωt teσtcos ωt teσtsin ωt 0 0 −eσtsin ωt eσtcos ωt −teσtsin ωt teσtcos ωt
0 0 0 0 eσtcos ωt eσtsin ωt
0 0 0 0 −eσtsin ωt eσtcos ωt
Example. Given the following dynamic system:
˙x(t) =
1 0 1 2 1 1 1 −1 2
x(t), x0 =
1 1 1
compute the free evolution x(t) of the system starting from the initial condition x(0) = x0. The “formal solution” of this problem is:
x(t) = eAtx0 = TeAtT−1x0
where T is the transformation matrix which “diagonalize” the matrix A. Matrix T can be determined computing the eigenvalues and the eigenvectors of matrix A:
det(sI − A) =
s− 1 0 −1
−2 s − 1 −1
−1 1 s− 2
= s(s2 − 4s + 5) = s[(s − 2)2 + 1)]
The eigenvalues are s1,2 = 2 ± j e s3 = 0. The complex eigenvector v1 corresponding to the eigenvalue s1 = 2 + j is the following:
(s1I−A)v1= o ⇔
1 + j 0 −1
−2 1 + j −1
−1 1 j
v1= o ⇒ v1=
1 2 − j 1 + j
= v1R+ jv1I
The real eigenvector v3 corresponding to te eigenvalue s3 = 0 is:
(s3I− A)v3 = −Av3 = o ↔ −
1 0 1
2 1 1
1 −1 2
v3 = o → v3 =
1
−1
−1
The following transformation matrix T:
T = v1R v1I v3 =
1 0 1
2 −1 −1 1 1 −1
, T−1 = 1
5
2 1 1
1 −2 −3 3 1 −1
transforms matrix A in the “real” Jordan canonical form:
A = T−1AT =
2 1 0
−1 2 0 0 0 0
Therefore, the free evolution x(t) of the system starting from initial condition x0 is the following:
x(t) = T
e2tcos t e2tsin t 0
−e2tsin t e2tcos t 0
0 0 1
T−1x0.
