For linear time invariant (LTI) systems, eigenvalues are the same as poles of the open-loop system. Let the system be described by

dx(t)/dt = A x(t) + B u(t) and y(t)=C x(t)

where x(t) is the state vector, u(t) the input vector and y(t) is output vector and the matrices A,B and C of appropriate sizes are constants. Without trying full transient solution using state transition matrices, for stability analysis of the LTI system, we simply assume that the initial conditions x(0)=0 and take laplace transform of both sides of the equations.

s x(s) = A x(s) + B u(s) and y(s)= C x(s) = C Inv(s I - A) B u(s)

G(s) = y(s)/u(s) = C Adj(s I - A) B/ det (s I - A)

where G(s) is the transfer function matrix, s is the complex variable and I is an identity matrix. The poles of G(s) are roots of det (sI - A) =0, i.e., a, b, .... and are easily related to roots of det (lamda * I -A) =0 or the eigenvalues.

The output y(t), when the input u(t)=delta(t) is an impulse or u(s)=1, is simply obtained from

y(t) = Inv Laplace G(s)

which can be obtained after partial fractions of the form G(s) = A/(s-a) + B/(s-b) + ..

The solution is y(t)=A*exp(a*t)+B*exp (b*t) + ... and it converges to zero if real parts of roots a, b, ... are negative. It is the same result that we expect from eigenvalue analysis.

Physically speaking, the eigenvalues identify the modal frequencies and the decay rates (effective damping) of decoupled/orthogonal modal oscillators. The whole unforced system response is a combination of all modal responses in a particular way (called modal participation) which depends on the initial conditions.

An interesting note about modal damping: It is not physical damping. Consider a unstable oscillating system needs to be stabilized and thus, it is mounted on a flexible damped foundation. If the foundation does not provide sufficient damping then the whole new system remains unstable, i.e., new mode has still negative modal damping. If the foundation provides good deal of damping then the foundation would not move (similar to being jammed) and the whole system would still remain unstable. The foundation damping needs to be tuned in such a way that it dissipates maximum energy exactly at the modal frequency of the unstable mode.

Such physical intuition comes from practice/experience. Similar problems related to damper optimization are found in many fields: floating mass/root damper in turbine blades where floating mass and its shape need optimization to improve friction damping, squeeze-film dampers in rotor systems where anchoring spring needs optimization to improve damping, etc. No book can explain these things. When you work on various industrial problems, you will learn by yourself.

Hello, I will try to answer. 

I will start with a summary.

Summary: To find the stability properties, you need to solve the system. You can solve it by the choice of convenient coordinates, in which the matrix of the system is diagonal and, hence, the entries of the matrix are eigenvalues. Therefore, it's the eigenvalues who determine the stability of the solution.

Details:

The reason is that eigenvalues tell you everything important about the matrix. What do I mean?

So, consider a general dynamical system, 

d x/ dt = f,

where x and f are vectors, t is a time. Suppose that x0 is a critical point, so that

f(x0) = 0 and the derivatives are zero. That is, critical point is an equilibrium configuration, which does not evolve in time.

The idea behind the stability is: if I choose initial condition which is close to x0, but not exactly x0, how will the system behave? Will the phase trajectory stay in the neighborhood of x0, or will it escape?

If you are close to x0, as you assume, you can Taylor expand the (vector) function f as 

f(x0+h) = f(x0) + h J + o(h^2),

where J is the Jacobi matrix of derivatives of f. And f(x0)=0, because it is a critical point.

Hence, in the small vicinity of x0, you approximate the original non-linear system by a linear one, which is represented by the Jacobi matrix.

If you want to decide whether the system is stable or unstable, you essentially need to solve the linear system, say, in two dimension,

x' = a x + b y

y' = c x + d y,

where a,b,c,d are the elements of the Jacobi matrix. In matrix form,

d x/dt =  J . x,

where x = (x,y).

So far, there is no mystery: if you want to know, how the system behaves, you need to solve the equation. For linear systems, this can be always done.

But instead of solving the system directly, we use a trick.

Your variables x = (x, y) can be understood as coordinates in the x-y plane. 

Notice that we have system of two linear equations, so there is no principal obstacle in solving it. The only complication is that x and y are mixed. We don't have one equation for x and another one for y. Instead, we have two coupled equations, each containing both variables. 

The trick: we can rotate our coordinate system in such a way, that these equations decouple. That is, we can find new coordinates (u, v), such that the system becomes

d u / dt = A u,

d v/dt = B v.

These equations can be solved easily:

u = u0 exp(A t),

v = v0 exp(B t).

Now, if the exponents A, B are real negative, the system is stable, if they are both positive, the system is unstable, otherwise it is a saddle point. If they are complex, you have stable or unstable focus, or a center.

You see: in the new coordinates we found the solution without any effort, while solving the original system can be painful, especially for many variables.

How is this related to the eigenvalues? Well, the coefficients A and B are exactly eigenvalues. Why?

Take our simple system in terms of u, v variables. In the matrix form, we can define

u = (u, v),

K = ( A, 0 

         0, B ) (matrix),

then our system is

d u/d t = K.u.

Obviously, A and B are eigenvalues of K. But they are also the eigenvalues of the original Jacobi matrix J. There is a procedure called diagonalization of the matrix, which is the similarity transformation

J ----> K = R.J.R^(-1),

such that the resulting matrix K is diagonal. You can easily check that K has the same eigenvalues as J. So, the solution of our original system (in terms of x, y), can be solved this way:

- first, find such matrix R which brings J into diagonal form

K = ( A, 0

         0, B )

Finding such a matrix is analogous to rotating your coordinate system.

- Transform the coordinates

(u, v ) = R.(x,y)

- in the now coordinates, the system is

 d u / dt = A u,

d v/dt = B v.

- Now you can solve it and eventually transform with matrix R^{-1} back to the original coordinates.

The point is when you are "rotating" your coordinates, i.e. diagonalizing the matrix J, you are essentially finding the eigenvalues of matrix J. So, if you don't need the entire solution, and it suffices to check the stability, it is enough to find the eigenvalues and their signs (or real and imaginary parts) and you are done. 

For linear time invariant (LTI) systems, eigenvalues are the same as poles of the open-loop system. Let the system be described by

dx(t)/dt = A x(t) + B u(t) and y(t)=C x(t)

where x(t) is the state vector, u(t) the input vector and y(t) is output vector and the matrices A,B and C of appropriate sizes are constants. Without trying full transient solution using state transition matrices, for stability analysis of the LTI system, we simply assume that the initial conditions x(0)=0 and take laplace transform of both sides of the equations.

s x(s) = A x(s) + B u(s) and y(s)= C x(s) = C Inv(s I - A) B u(s)

G(s) = y(s)/u(s) = C Adj(s I - A) B/ det (s I - A)

where G(s) is the transfer function matrix, s is the complex variable and I is an identity matrix. The poles of G(s) are roots of det (sI - A) =0, i.e., a, b, .... and are easily related to roots of det (lamda * I -A) =0 or the eigenvalues.

The output y(t), when the input u(t)=delta(t) is an impulse or u(s)=1, is simply obtained from

y(t) = Inv Laplace G(s)

which can be obtained after partial fractions of the form G(s) = A/(s-a) + B/(s-b) + ..

The solution is y(t)=A*exp(a*t)+B*exp (b*t) + ... and it converges to zero if real parts of roots a, b, ... are negative. It is the same result that we expect from eigenvalue analysis.

Physically speaking, the eigenvalues identify the modal frequencies and the decay rates (effective damping) of decoupled/orthogonal modal oscillators. The whole unforced system response is a combination of all modal responses in a particular way (called modal participation) which depends on the initial conditions.

An interesting note about modal damping: It is not physical damping. Consider a unstable oscillating system needs to be stabilized and thus, it is mounted on a flexible damped foundation. If the foundation does not provide sufficient damping then the whole new system remains unstable, i.e., new mode has still negative modal damping. If the foundation provides good deal of damping then the foundation would not move (similar to being jammed) and the whole system would still remain unstable. The foundation damping needs to be tuned in such a way that it dissipates maximum energy exactly at the modal frequency of the unstable mode.

Such physical intuition comes from practice/experience. Similar problems related to damper optimization are found in many fields: floating mass/root damper in turbine blades where floating mass and its shape need optimization to improve friction damping, squeeze-film dampers in rotor systems where anchoring spring needs optimization to improve damping, etc. No book can explain these things. When you work on various industrial problems, you will learn by yourself.

Let me try to explain quickly a reason, why the eigenvalues could be important.

You have a system

dx/dt = Ax

x, a vector, A a matrix. You want to know something *basic* about the system: not how x1 evolves, or whether x1>x2 at some time. You want to know whether the system is stable.

That cannot depend on how you present your data. In particular, if you make a change of variables

y = Tx,

with T a linear transformation, you can rewrite the system in terms of y, and the answer should be the same.

The transformation can be rewritten as

x = T-1y

Substituting in the original equation gives you

T-1(dy/dt) = AT-1y

or in other words

dy/dt = TAT-1y

So stability should be the same, whether you take the matrix to be A or TAT-1, whatever the invertible matrix T you consider.

Since it does not matter whether you take A or TAT-1, you choose T so that TAT-1becomes a lot simpler than A. But this is a math problem, and its solution is given by eigenvalue theory: choosing T appropriately, you can usually get A into diagonal form, if not, then into Jordan form.

Dear Arun Sir,

The following paragraph of yours holds special significance as it relates eigenvalues to the physical environment.

"Physically speaking, the eigenvalues identify the modal frequencies and the decay rates (effective damping) of decoupled/orthogonal modal oscillators. The whole unforced system response is a combination of all modal responses in a particular way (called modal participation) which depends on the initial conditions. "

Is there any resource (journal/book) that will take me deeper into this aspect of eigen values/eigen vectors ??

There are good deal of books on the subject (See Chapter 5, Theory of Vibration with Applications by W.T. Thomson or Thomson and Dahleh). But books do not have animations. A good way to understand is to look at the interactive animations and examples given in the following link.

http://lpsa.swarthmore.edu/MtrxVibe/Anims/VibrationAnimations.html

I hope that this would be helpful.