The atatched draf can be perhaps useful

I have some extra comment after thinking ot it. the above response of mine needs some extension. It proves stability for the parameter  epsilon in the interval ( 1-delta 1, 1] but not ensures it  for epsilon in ( 0, 1 -delta1].

On the other hand, byy writing :

A+epsilon*K=A+K-delta*K                 , ( delta = 1-epsilon)

and since A+K is non singular( since all its eigenvalues stable), we can also write:

A+epsilon*K=( A+k)*( I-delta*inverse[(A+K)]*K)

so that A+epsilon* K  is non singular , and stable , if

delta=1-epsilon< 1/norm[inverse (A+K)*K] ( from Banach perturbation lemma)

, eqiuivalently  ,  if epsilon in ( max (0, 1-1/norm [inverse (A+K)*K]) , 1  ]

Dear Manuel De la Sen - Nice calculations! I am reminded of the calculations in the book - Introduction to Matrix Computations by G.W. Stewart.

My text book - Numerical Linear Algebra (Reprint to be released in Aug end) - also deals with such problems in the chapter on "Vector and Matrix Norms".

As you have given detailed calculations, I don't need to add anything!

With best wishes

Sundar

http://phindia.com/bookdetails/numerical_linear_algebra_by-sundarapandian_v_-isbn-978-81-203-3436-6

I owe many thanks to Prof. De la Sen for your quick and kind help!

But, I'm afraid the proof in the attached doc has flaws.

The eigenvalue is indeed continuous, but it is not linear.

Let us see a simple continuous nonlinear function: sin(x)

sin(0)=0, sin(pi*5/6)=0.5, but sin(pi/2)=1>0.5

Dear Prof. Vaidyanathan, thank you very much for your help! Let me pay my sincere respects to your academic reputation!

Dear Ning: Prof. De La Sen's proof is correct! The eigenvalues of a matrix are the roots of the characteristic polynomial, which is given by a determinant function. We all know that determinant function is continuous. So, his arguments are valid. "Linearity" is not in the picture at all!

WIth best wishes

Sundar

Ning:

Suppose that

A = diag(0, 0, ..., 0)   [n X n zero matrix] - This is critically stable.

Take

K = diag(-1, -1, ..., -1) = - I.

Then A + K = -I, which is stable (or Hurwitz).

Now, what is A + eps K? A + eps K = -eps I and it is very clear that this matrix is stable for all values of eps > 0, and in particular for 0 < eps < 1.

Likewise, visualize this for any diagonal critically stable matrix. Work out a proof yourself. Then try general cases. Please use MATLAB - calculate eigenvalues for the perturbed matrix. Do numerical simulations, get some intuition and then work out a mathematical proof.

Professor has given a very nice proof for different cases. Study both.

With best wishes

Sundar

I tend to agree to the second proof by Prof. De la Sen. But the result is conservative to exclude very very small ε. Indeed, the actual problem I am seeking to solve is to prove that A can be stabilized by arbitrarily small ε, (to avoid any possible actuator saturation). This problem seems not that simple as it looks.

I myself have a proof.

Suppose that there is a common positive definite matrix P which can construct Lyapunov function for both dynamic systems x'=Ax and x'=(A+K)x.

Then

ATP+PA=-Q1

(A+K)TP+P(A+K)=-Q2

where Q1 is positive semi-definite and Q2 is  positive definite.

So KTP+PK=Q1-Q2, which is negative definite.

So (A+εK)TP+P(A+εK)=-Q1+ε(Q1-Q2) must be negative definite!

The only limitation of my proof is that a common P is required...Actually I think this limitation can be removed.

It seems that  something like thios can be also useful to get conditions:

If

1) A marginally stable

2)A+K stability matrix

3) additional condition:

either     K(transpose)*A+A*K

or     K(transp)*K 

is positive definite ( brief notation" >0")

then

B(epsilpon)=A+epsilon*K

is a stability matrix for all epsilon in (0,1).

Define

C(epsilon)=(A+epsilon*K)(transpose)*(A+epsilon*K)

                    =A(transpose)*A+(epsilon**2)K(transp)*K+ epsilon*(K(transp)*A+K*A)>0

for epsilon in (0,1) since

A(trans)*A>=0( symmetric, semidef. positive)

C(1)>0(symmetric,  positive definite) : otherwise, A+K would be singular, then it cannot be a stability matrix.

C(epsilon)>0 for epsilon >0 since A(transp)*A>=0, (epsilon**2)*K(transp)*K>0, epsilon*(K(trans)*A+A*K)>=0 ( the sum of a positive definite matrix with two  semidefinite ones)

eigenvalues of C(epsilon) are real non negative ( since symmetric positive at least semidef pos) and are strictly increasing as epsilon increases since

C(epsilon)/d(/epsilon)=2*epsilon*K(transp.)*K +K(transp)+A*K>0 for epsilon nonnegative

so its eigenvalues are real , nonnegative and grow with epsilon.

It can be also considered to relax

(K(transp)*A+K*A) pos. definite to semidefinite if, in addition,  K is  nonsingular since then

K(transp)*K is positive definite and C(epsilon)>0 for epsilon >0.