 So we resume our study of dynamics in the phase plane and you recall that we were going to analyze the behavior of a two dimensional dynamical system near a critical point which we have taken to be the origin and the set of equations that we are interested in is a linear set of the form A x plus B y and y dot is C x plus D y where the constants A B C D are real constants and we also have assume that A D minus B C is equal to the determinant of the linear matrix L is not equal to 0. So this is the problem we would like to look at and understand the nature of the critical point at the origin based on the Eigen values of the matrix L and we pointed out that the general solution in this neighborhood could be written as linear combinations of the exponentials of the two Eigen values of the linear matrix L. L itself arose in a nonlinear system as the Jacobian matrix at the origin of the flow itself of the vector field F itself but right now we are focusing on the linear system per se as it stands and we have seen that the general solution as I said is a sum of exponentials. Now if you eliminate either x or y from the set of equations then this implies a second order differential equation for x which is of the form x double dot minus A plus D x dot plus A D minus B C equal to times x equal to 0 and similarly y double dot minus A plus D y dot plus A D minus B C y equal to 0. X and y of course would differ in general in their initial conditions as would x dot and y dot and you would get different combinations of e to the lambda 1 t and e to the lambda 2 t. If lambda 1 should happen to be equal to lambda 2 then the solutions are of the form linear combinations of e to the lambda t and t e to the lambda t and we have seen that too and now we are going to analyze the stability or otherwise of this set of equations depending on what the Eigen values of look like. Notice immediately that if you have a matrix L whose Eigen values matrix L A B C D whose Eigen values are lambda 1 and lambda 2 then we find the Eigen values by writing determinant lambda times I minus L equal to 0 which implies lambda squared minus A plus D lambda plus A D minus B C equal to 0. Do you notice anything interesting about these combinations this combination and that combination they are not arbitrary combinations of A B C and D they are very special combinations this is the trace of the matrix and that is the determinant of the matrix. So this immediately tells you that if I set t equal to trace of L and delta equal to the determinant of L this implies that the Eigen values are given by lambda squared minus lambda t plus determinant equal to 0. So the Eigen values of functions of the trace and the determinant of the matrix what does that imply lambda 1, 2 equal to t plus or minus square root of t squared minus 4 delta divided by 2 and recall that the solutions are exponentials in lambda 1 and lambda 2. So what does this tell you what does it tell you about these what does it tell you about these Eigen values the fact that it is dependent only on the trace and the determinant and not on the actual all 4 coefficient not on all 4 coefficients but just these 2 combinations that is going to play a role in what we are going to say next what are these combinations what is so special about these combinations what happens if I do a similarity transformation on the matrix the trace does not change and the determinant does not change which means that if I transform these variables from x and y to certain linear combinations of x and y by similarity transformation by coordinate transformation the Eigen values would not change and that has a physical implication as we will see shortly. So right now one is the sum of the Eigen values and the other is the determinant so it involves yes exactly the other is the product of the Eigen values because this quantity here the coefficient of the linear term is minus the sum of the Eigen values and the last coefficient is a product of the Eigen values these 2 quantities are independent of what sort of linear transformations you make on the matrix as we will see in a short while so the behavior of the Eigen values is going to be determined entirely by the behavior of T and delta therefore and we will take up this lesson more in more detail little later the entire behavior of the dynamical system near the critical point is going to be governed by just the combinations T and delta therefore we could regard these 2 as parameters and ask what kind of behavior you have in the T delta plane which we will do in a short while but first what are the various possibilities that you have for these Eigen values in terms of what kind of quantities can they possibly be and let us classify them so we now classify we start the classification first the first possibility of course is that lambda 1 and lambda 2 are both positive that is certainly a possibility so let us write lambda 1 greater than 0 lambda 2 greater than 0 the solutions as we have seen are exponentials in lambda 1 and lambda 2 e to the lambda 1 t and e to the lambda 2 t so as time goes on what happens to these exponentials if both are positive they increase they are growing exponentials and therefore the flow would be away from the critical point in all directions and you would say this is an unstable critical point let us look at a simple example and then we come back and formalize this in terms of the actual classification here so the example that we are going to look at and each time on this side of the board I am going to give an example x dot equal to x y dot equal to y what is the linear matrix L now in this case just the unit matrix and the two Eigen values are plus 1 and plus 1 each repeated Eigen value what would you say is the solution to this let us draw the phase diagram now here is x and here is y here is the critical point clearly unstable in this case because everything is going to flow outwards if I started with y equal to 0 at any point on the x axis I flow outwards if I start anywhere here I flow outwards anywhere here I flow outwards and anywhere here flows outwards if I start at any point here it is clear you are going to flow away in this fashion so the phase portrait in this case is exceedingly simple and it just the set of lines flowing out in all directions what would you say is the stability of this critical point unstable and this kind of critical point is called an unstable node what would have happened if the two Eigen values had been unequal in this case suppose this had been y dot equal to twice why what would the flow look like in this case the two Eigen values are 1 and 2 they both positive and the flow is again outwards but what sort of shape would the flow lines have it is evident from this that x goes like e to the power t while y goes like e to the power twice t so therefore y goes like x squared and what would the flow lines look like in this case if y you start off with y equal to 0 you are still going to be like that if you start off at x equal to 0 you are still going to do this but if you started off with finite values positive values say of both x and y the y is going to increase much more rapidly than the x like the square of x actually so it is actually going to flow off in this fashion or in this fashion here or here here therefore the phase portrait looks like this except for a flow along the y axis itself the rest of the flows have a common tangent which is the x axis this two is an unstable node it is just a different set of Eigen values of course if I change this to 3 y or any multiple of the exponent in x the coefficient in x you are going to have slightly different shapes of this parabola had the coefficient of x been larger than the coefficient of y the whole situation would have been reversed and you would have had parabolas flowing out such that the y axis is the common tangent but it is still an unstable node now what would you say would be the situation if you did not have decoupled equations of this kind remember that notice that in this case the coefficient of y is missing here and the coefficient of x is missing there so it is already a diagonal matrix but in general I do not have a diagonal matrix L has got both x and y on both sides what would you say would be the flow in that case can we deduce what the flow would be in geometrical shape based on the fact that in the decoupled case it looks like this in general after all you could regard ax plus b y and c x plus d y as linear transformations of the x and y coordinates by the matrix a b c d which is non singular and what does it do if I take two axes x and y and I go to linear combinations ax plus b y and c x plus d y what sort of coordinate axes do I get in general it could be rotated but rotation still leaves the angle between the x and y axis the two axes 90 degrees but in general that does not have to be the case I could go to oblique axes moreover the scale of x and the scale of y are also changed because these are factors which actually change they magnify or demagnify the most general linear transformation that you can induce which leaves the origin unchanged is precisely by a sort of linear equations of this kind if I put u equal to a x plus b y v c x plus d y then in general the u and v axes an order try tangles to each other the scale on one direction could have been demagnified or magnified and similarly for the other direction in pictures this means that I could start with a set of axes and if I have a little square of this kind and ask what happens to this after the transformation if it were a pure rotation it would just look like this still a square look like this if on the other hand it is not a rotation but a magnification in one direction and a contraction in the other direction it would perhaps look like this so I have contracted the x direction and expanded the y direction on top of these two transformations there is one more possibility and that is to change the perpendicular orthogonal axes into oblique axes in other words a shear in some direction so that would look like this a shear would take it to some shape of this kind and a general 2 by 2 transformation a general 2 by 2 matrix would take this little unit patch and in general push it in one direction squash it in some direction expand it in another direction rotate and shear and therefore the final product would actually be something like this perhaps so it has been rotated sheared and magnified or dilated in some direction or demagnified in the other direction how many parameters are there in this matrix in specifying L there are 4 real parameters here we must make sure that in this way of counting or decomposing a matrix a transformation a linear transformation into a rotation a dilation and a shear we stick to the same number of parameters and I put it to you that a general 2 by 2 linear transformation of this kind non-singular matrix induces a transformation which is a combination of a rotation a dilation or a magnification and a shear the number of parameters needed to specify a rotation is one just the rotation about from the reference axis x axis in the plane so there is one parameter here this has two parameters for the two axes that is 1 plus 2 is 3 and this gives the angle of shear that is one more that is 4 and that is exactly the same number there you can count this in several ways and this is not a unique decomposition but this is one of the ways of decomposing a general 2 by 2 matrix you can decompose it into a rotation a dilation and a shear if we accept that this is all it does then we can write down what the general flow would look like once you induce those transformations on this decouple case what would it look like it would simply look like some weak axis of this kind and perhaps so the moment I see a flow pattern of this kind I know it is an unstable node and it is just found from this simple case by a linear transformation of coordinates and therefore we assert that both eigenvalues positive corresponds to an unstable node notice there are two different kinds of nodes in pictures one of them is this generic case but the two roots are unequal and typically you would have a common tangent to all the trajectories except for an a single exceptional direction on the other hand the earlier case we looked at had a star pattern and a radial pattern in which you had straight lines emanating in all directions do not have to be straight lines but the fact is that all directions there are no exceptional directions at all all directions the flow moves off and there is no common tangent these are two subtle distinctions between further distinctions between kinds of nodes that one has but it is not really very relevant here and right now we are concerned with the fact that two positive eigenvalues implies an unstable node of the we will come back to this we will explain what sort of flows would have this extra this common tangent and specific instances in examples what would happen if both the eigenvalues were negative less than 0 lambda 2 less than 0 all the arrows would be reversed because you would have damped exponentials and this means that as t tends to infinity wherever you start you are going to flow into the critical point at the origin asymptotically therefore I would call this I would call this an asymptotically stable node the reason is I would like to be careful about the definition of stability so let us for the moment call it an asymptotically we will distinguish shortly between stability and asymptotic stability in the different concepts altogether this is asymptotic stability we are guaranteed that once a trajectory enters the neighborhood of this point the critical point then as t tends to infinity it inevitably falls into this critical point that is asymptotic stability the third possibility is if one of the eigenvalues is positive and the other eigenvalue is negative once again all we have to do to study this is to look at a decouple case a case where the matrix is already diagonal and then argue that the linear transformation of that picture would give us the true picture near a situation which corresponds to one positive eigenvalue and one negative eigenvalue so all we have to do is perhaps to do the following right x dot equal to minus x and y dot equal to plus 2 y in this instance x goes like e to the minus t flows into the fixed point where the critical point whereas y flows into 0 whereas y explodes x outwards like e to the 2 t so we know the solutions what does the flow look like well on the x axis since it flows in you have this picture and on the y axis since it flows out you have that picture if you start with some point here some value of x which is not 0 say some negative value then it is got to flow in towards 0 but the y has to flow outward and therefore the trajectories look like that and the continuity this is what they would look like would you call this a stable or an unstable critical point I call it unstable I call it unstable because except for very special initial conditions in other words y starts at 0 and remains at 0 if you are on the x axis to start with you flow in but wherever else you are on the plane you are actually going to go away to infinity in some direction or the other therefore this is unstable what is the shape of these trajectories it looks like hyperbolas but actually they are not hyperbolas unless of course the coefficients are equal in magnitude then of course you would have something like x y equal to constant and they would be rectangular hyperbolas if you had x dot is minus x and y dot is minus y then of course x y is constant and then they would look like rectangular hyperbolas but right now that is not happening the y variable is exploding much faster than the x variable is going to 0 but they look hyperbolic in shape roughly in shape this is an unstable critical point and it is called a saddle point so that was a saddle point again if I had a situation where I have a general L with two real eigenvalues one of which is positive and the other is negative the flow would look like a distortion of this picture by the same sort of coordinate transformation we looked at earlier and therefore in general perhaps this is what a saddle point flow would look like there would be two directions in which the flow goes in another one in which it goes out and the rest of the trajectories would follow curves of this kind and the whole plane in this case would be striated by these curves and that point itself is a trajectory by itself and it is unstable although I draw all these pictures with these lines going right through that point remember that point is a trajectory by itself but you can come arbitrarily close to it and that is the reason I draw these as continuous lines it is natural to call this the unstable direction and this the stable direction associated with this saddle point we will say much more about this as we go along so that is what a general flow near a saddle point looks like in this linear case these lines are actually straight lines as you can see but in a general case with the nonlinearity present the first thing that would happen is that these things although arbitrarily close very close to the critical point they would tend to be straight lines the actual flow lines would be curved in general so you would not have straight asymptotes as you have here in a nonlinear system that too will play a role so we have the third possibility namely it is a saddle point here remember we have ignored the case where one of the Eigen values can be 0 one or both Eigen values can be 0 because I said we are going to look at all those cases where the matrix is non singular therefore it cannot have 0 as an Eigen value we have to look at it separately we have to ask what happens when the Eigen value one or more of the Eigen values is 0 what are the possibilities exist have we exhausted everything complex Eigen values absolutely right we have complex Eigen values possible and the first thing that would happen is first thing we have to recognize is that since these coefficients are real the complex Eigen values would be a complex conjugate pair therefore all we have to look at is a situation in which for lambda 1, 2 is some lambda plus or minus i mu where lambda and mu are real numbers and lambda is greater than 0 the behavior of these trajectories the behavior of these trajectories is not really crucially dependent on whether mu is positive or negative it is a relevant because e to the i mu t is going to be just cosines and signs of mu t on the other hand you have a need to the lambda t and if lambda is positive it is going to flow outwards if lambda is negative it is going to flow inwards but there is going to be an oscillation in the sign of x and y because of e to the i mu t present there because that is going to read to cosines and signs which would change sign as t increases what then would the flow look like again I go to the face plane here is x and here is y if the flow has a positive value of mu it means as time goes on x and y would change sign but would flow away from the origin and they would oscillate with an increasing amplitude and go off so what sort of curve would you expect you would expect a spiral quite right but it would be an outward spinning spiral it would go off in this fashion you know that because e to the power so let me write it here e to the lambda 1 t for example would be e to the lambda t e to the i mu t which would be e to the lambda t times the cosine of mu t plus i sin and of course super positions of e to the lambda 1 t and e to the lambda 2 t would involve these cosines and signs and the coefficients would be adjusted so that the linear combination is real because we are looking at real variables but the fact is because of this cosines and signs although this increases as a function of time monotonically these functions would change sign they would oscillate between positive and negative values and all the while the amplitude would increase so in a picture like this if I started some value here of x a little later the value is larger in magnitude but opposite in sign and then once it comes back here it is again larger in magnitude but opposite in sign and the same thing is true for y so it is intuitively clear that the picture is actually that of an outward spiral outward moving spiral and this case therefore is called an unstable spiral point there are the names for the spiral point one of which is the word focus occasionally called a vertex as well but focus is much more common but I would like to stick to the word spiral point the phrase spiral point because this is evocative it tells you exactly what the trajectories look like once again if I have a linear transformation on the spiral what would what sort of distortion would the spiral undergo what would it look like that is not hard to see it could get squashed in one direction turned around and extended in the other direction may be therefore it to perhaps look this would be the general flow around a spiral point which is unstable and it is unstable because lambda is positive and exactly as in the case of the nodes we could assert that if lambda one two were of the form lambda plus or minus i mu with lambda less than zero the arrows would be reversed and the system would flow into the critical point as t tends to tends to infinity and this would lead to an asymptotically stable spiral point the arrows would just be reversed whether the spiral is wound in words clockwise or counter clockwise depends on the details of the problem it would actually depend on the direction in which the phase trajectories are reversed that would depend once again on the coefficients a b c d but this is what happens when you have a complex conjugate pair of points there is one more possibility what is that there are subclasses here between in these cases when I have for instance two equal roots both of which are positive both of which are negative or one of them is positive in the others negative and they equal in magnitude those are subclasses of what we looked at but what would have there is one more class which is distinct lambda is both the eigenvalues are pure imaginary they form a pure imaginary complex conjugate pair so lambda could be zero and yet determinant L is not zero because the eigenvalues are plus or minus I mu then that is the final case is right that down six lambda one comma two equal to plus or minus I mu when you use a real number the simplest example of this is in fact the simple harmonic oscillator you studied that over and over again but let us write that down in that if you recall corresponds to a problem in which x dot is p over m where p is the momentum of the oscillator and p dot which is the rate of change of momentum is equal to the force on the oscillator which is minus m omega squared x or minus kx where k is the spring constant what is the matrix look like this is a linear system in which L is equal to a zero here a one over m here a minus m omega squared here and a zero here this implies lambda one comma two equal to what are the two eigenvalues in this case plus or minus I omega and all as all of us know the solutions are linear combinations of e to the I omega t and e to the minus I omega t what are the phase trajectories look like in this case why I have p here they do not necessarily have to be circles the scales on the two axes may be different and their generalizations of circles so they are really ellipses of some kind the entire plane is striated by these laminated by these ellipses concentric ellipses would you call this critical point here would you call this stable or unstable or asymptotically stable what would you call it I certainly would not call it unstable because the trajectories do not disappear from its neighborhood they do not blow away from it on the other hand they do not fall into it either they just go round and round so we call this a stable critical point and this particular case is called a center so it is a stable center notice that I have said stable and not asymptotically stable so the time has come now to distinguish between stability and asymptotic stability and these are two different concepts all together one does not imply the other as we will see in a second I define a critical point to be stable if the following is true and this is doable rigorously but let me give an operational definition which is easy to understand stable here is the critical point and here is some neighborhood of this critical point then a stable critical point is one where a trajectory if it has once entered this neighborhood never leaves this neighborhood so something which has entered it may do all kinds of things inside here it may fall into it it may also keep going around or it may do this but it never leaves this neighborhood I call that a stable critical point if this is neighborhood is you once a phase trajectory enters the neighborhood you it never leaves it it does not have to be asymptotically stable this one you would say this is a situation I would say that this critical point these trajectories tend to this point and therefore we are going to have asymptotic stability but stability does not require that it suffices that a trajectory which once enters once it enters this neighborhood remains in this neighborhood forever and that is exactly what happens here if this is a neighborhood there exists a trajectory there exist trajectories which remain in this neighborhood forever the concept of stability depends on the neighborhood so like everything else you are going to have to perturb the system a little bit away from the critical point and ask what the trajectories do always that is how stability is defined in any case it is not a concept applicable to a single point but to a region yes you need to know if there are other critical points in there so the question you are asking really is how big can this neighborhood be in this case there is just one critical point so the entire plane there is a single critical point and no matter what neighborhood you give me there exist trajectories which having once entered the neighborhood never leave this neighborhood altogether if I have more than one critical point then the question that you raise comes up and we will see what happens when you have multiple critical points but right now just think in very elementary terms if I look at mechanical examples I would say that a stable equilibrium of a in a potential problem is if the potential has a minimum but the idea of a minimum is not a concept restricted to one point it says something about not only the function at that point and its slope but also about its curvature at that point so it is really a non-local concept in that sense talks about a neighborhood of this point and that is always going to be the case so I say that a critical point is a stable one if there exists a neighborhood of this critical point such that once a trajectory enters this point neighborhood it never leaves it as t tends to infinity what then is asymptotic stability well asymptotically stable again a critical point this point is asymptotically stable if a trajectory it starts in this neighborhood tends to this point as t tends to infinity you guaranteed that every trajectory that starts in this neighborhood tends to that point as t tends to infinity so every point trajectory starting in u tends the cp so something starts here it is guaranteed to fall into this asymptotically as t tends to infinity does asymptotic stability imply stability it is clear that stability does not imply asymptotic stability because it could keep going around does not have to fall in but does asymptotic stability imply stability not necessarily because you could start here you could in fact do a thing like this so it could be a spiral it goes out very far away but it is guaranteed to come back so it starts off does this goes around and eventually falls into this point here so it certainly does not have to be stable therefore stability and asymptotic stability they are not concepts they are independent concepts altogether you could have a trajectory that stable as well as asymptotically stable you could have a critical point which has the property but it is not necessary centers particularly are stable not asymptotically stable and spiral points stable asymptotically stable spiral points could have behavior like this they need not be stable but they would definitely be asymptotically stable the statement is that they are given a neighborhood there exist this sort of behavior we can make this much more rigorous but I am just trying to do this in heuristic terms right now till we have a few examples under our belt and then of course we could look at more tests for stability and so on so one of the things we are going to do is to have a kind of non-linear test for stability which does not depend on linearization as we done here and that is Lyapunov's direct method and we will talk about it later yeah did anyone have a question so as a center in particular is stable but not asymptotically stable this is an important point and we will see why in a short while this exhausts all possibilities of those cases where the determinant is not zero and now can we make some sense out of this I already mentioned that everything seems to depend on T versus del T and delta just these two combinations these two combinations which are unchanged under similarity transformations on the matrix and therefore it is clear that the nature of the critical point is kind of independent of these transformations it does not depend on the particular choice of coordinates X and Y linear combinations of this X and Y would still leave the nature of the critical point unchanged and that is very important to know because this is a very robust property that we are talking about okay now what can we say about this classification in general terms we can draw a little picture in parameter space so I draw T on this axis and the determinant delta on this axis it is obvious from here that the case delta equal to zero is on this line and that is the case I am going to ignore at the moment and the curve T squared equal to 4 delta would separate regions where the eigenvalues are complex from those where the eigenvalues are real and what is the curve T squared equal to 4 delta it is a parabola so on this curve T squared is 4 delta and outside T squared is greater than 4 delta whereas inside T squared is less than 4 and then the roots are complex what sort of behavior do you have here but delta is negative if delta is negative the square root of this number is certainly bigger than T in magnitude and therefore one eigenvalue is positive and the other eigenvalue is negative therefore below this axis you have entirely saddle points one positive eigenvalue another negative eigenvalue so this whole place is saddles saddle points and the flow looks like this whether you are on this side of the graph or on this side does not matter you just have saddle points and the whole thing is unstable everywhere in parameter space in this region you are guaranteed to have unstable saddle points what happens when you move up here if you are here say T squared is greater than 4 delta certainly but what happens now to these eigenvalues both are positive it is an unstable this place is unstable but what sort of unstable point what sort of unstable point you have two positive eigenvalues unstable nodes so you have unstable nodes therefore the picture would be perhaps something like this what would you have here you would have stable nodes asymptotically stable nodes so asymptotically stable nodes in this region and these are unstable nodes in that region what would you have here when T squared is less than 4 delta T is positive remember you are on the right hand side so T is positive and this becomes pure imaginary so the square root gives you plus or minus i times something or the other but then you have a positive real part therefore you have unstable spiral points so unstable spiral points flowing away and what would you have here you would have stable asymptotically stable spiral points so in this case the flow is inwards what you have on the vertical axis T is 0 you have centers so on this case in this case right here you have center all along this point you have centers so essentially you have this sort of behavior we now come to a very important concept this picture is sufficient to tell us why this is so why this happens to be so if I start somewhere here and I perturb these parameters a little bit I change ABCD a little bit I may wander to this place but what was an unstable node remains an unstable node similarly if I move around a little bit here nothing much happens if I move across this line I go from an unstable node to an unstable saddle point if I move across this line of course is a big change I go from something that is asymptotically stable to something that is unstable but then we have crossed the line of degenerate or higher order critical points where L becomes singular but with that exception everywhere else if I cross this line what was an unstable node becomes an unstable spiral point what was an asymptotically stable node becomes an asymptotically stable spiral point but if I am on this line and I perturb parameters a little bit if I move to this side it becomes unstable if I move to this side it becomes asymptotically stable so you can see that this line of points the line of centers is structurally unstable in the sense that small change of parameters can completely change the qualitative behavior of the critical point so we have our first statement which is somewhat general which is centers are structurally unstable when mathematicians say structurally unstable they mean that if you change parameters slightly then there is a qualitative change in the behavior of the system and this is exactly what happens to centers what sort of physical motion does that example of a center correspond to you have pure imaginary eigen values so what is special about this motion periodic motion periodic motion so this implies that periodic motion is structurally unstable small changes can get you completely away from periodic motion I add a little bit of friction here I had a tiny bit of friction here and this becomes this thing has another term m gamma x dot where gamma is a friction constant assuming the friction to be proportional to the instantaneous velocity with the retarding coefficient gamma and I use an m gamma there so that gamma has dimensions of time inverse and you immediately see that this thing here has minus m gamma here this point sorry m gamma times x dot so let us call this minus gamma p gamma here then of course the eigen values not equal to plus or minus i omega what happens to the eigen values what are the eigen values of this the minus gamma the gamma minus gamma plus or minus i omega right therefore the eigen values do not look like this at all what sort of critical point is the origin now asymptotically stable spiral provided gamma is positive provided the friction acts as a retarding force the friction does not act as a retarding force but pushes you along the same direction even further if there is positive feedback then of course it becomes unstable so this means that the phase trajectories in this case would reduce to something like this this is the case of ordinary friction gamma greater than 0 so introducing a small gamma pushes you from here to this region to asymptotically stable spirals if gamma had the wrong sign then of course you would become unstable and it would move into the other direction altogether so that is why periodic motion is so fragile it is not robust at all conditions have to be just so in order to have periodic motion and they are really the exception rather than the rule this has further implications in dynamical systems for example although I will say it in loose terms now we will look at it more rigorously later heart beats if they are too regular that is not very good so if you look at the heart beats of young infants with very robust hearts they are actually extremely irregular in some sense they are on something like a chaotic attractor which is very robust structurally stable small perturbations do not push you off this attractor on the other hand if the heartbeat gets extremely regular it becomes periodic you have to worry because this implies that a small change could cause you to have an explosive growth and either direction which is not very good so again this is a lesson of some generality centers are structurally unstable this is important to remember here now once we have this classification under our belt we really can generalize this and go on in very many directions and the first question I want to ask which will dispose off right away is that the Eigen values depended on the trace and the determinant in this case if you go to higher dimensions instead of 2 by 2 systems x y systems if I look at higher dimensional dynamical systems n dimensional systems then the linearized matrix would be n by n it would then have n Eigen values and once again we expect that the Eigen values are independent of similarity transformations they are stable against similarity their invariant under similarity transformations which would imply that the Eigen values should be writable in terms of quantities which are invariant under similarity transformations of the matrix in the 2 by 2 case we know that the trace does not change and the determinant does not change what happens in the 3 by 3 case you need 3 such combinations where are you going to get them from once again the trace of this matrix does not change in the determinant does not change but you need one more where is this going to come from and in the n by n case we need to have n invariant combinations where are these going to come from I need n of these quantities the determinant is just one of them I am willing to put that down but that does not give you n of them for the 2 by 2 case that was sufficient where are the others going to come from well we do know we do know that the coefficients in the secular equation would be the product the sum of all the Eigen values the products to at a time 3 at a time and so on and so forth so what invariant quantities are the functions of you need to put everything in terms of quantities which do not change under similarity transformations no not quite I mean in terms of L what are the quantities that would be unchanged what happens if I square the matrix L and then take the trace what would that be for the 2 by 2 case what is the square what is the trace of L squared well imagine you have diagonalized L you do not cannot always diagonalize the matrix but imagine you have diagonalized it what does L squared look like in the diagonal form exactly lambda 1 squared plus lambda 2 squared and what is the trace of that what is the trace of a diagonal matrix with elements lambda 1 squared and lambda 2 squared just the sum of these 2 so it is quite clear that if you give me lambda 1 plus lambda 2 and lambda 1 squared plus lambda 2 squared I could certainly find lambda 1 and lambda 2 because I straight away find lambda 1 lambda 2 by combining these 2 and then I could find lambda 1 and lambda 2 independently so what would be the generalization of that absolutely it would be trace of L squared trace of L cubed up to the trace of L to the top it is easy to check that AD minus BC in the 2 by 2 case can really be written as a combination of the trace of L squared and the trace of L the whole square so these are the invariant combinations you guaranteed that these combinations do not change under similarity transformations and that is what the Eigen values are functions of after all the statement is that if you give me lambda 1 plus lambda 2 up to lambda n lambda 1 squared plus lambda 2 squared up to lambda n squared and similarly lambda 1 to the power n some till lambda n to the power n then I can find all the lambdas in terms of these quantities they are invariant so this generalizes what we know for the 2 by 2 case just a side remark when can you diagonalize a matrix we can always find its Eigen values by writing determinant lambda I minus L equal to 0 and that is an algebraic equation of the nth degree and it has n roots in the complex plane what would be the case and what would be the condition for diagonalizing a matrix by a similarity transformation when can I take a matrix M and find the similarity transformation S such that this is a diagonal matrix and of course once it is diagonal its elements are just the Eigen values of the matrix and what conditions can you diagonalize a matrix by a similarity transformation this is not always possible you can always find the Eigen values that is a different problem from diagonalizing the matrix there are lots and lots of matrices which are not diagonalizable by similarity transformations a sufficient condition for you to diagonalize a matrix a sufficient condition is that the matrix M commute with its trans with its conjugate Hermitian conjugate if this is true by this dagger I mean the Hermitian conjugate of this matrix which is the transpose complex conjugated that is a sufficient condition for M to be diagonalizable by a similarity transformation such a matrix is called a normal matrix so we stop at this stage and take it up from here.