 So, the next result that we are going to see is going to be a little more non trivial it is going to require a little bit more involvement. So, do you think a positive matrix can be nilpotent? So, the question is can a positive matrix be nilpotent? Think in terms of what we have done until previous lecture until Jordan canonical form not specifically today. What is the trace of a matrix equal to? What do we know the trace of a matrix to be equal to? Some of the Eigen values. What do we know about the Eigen values of a nilpotent matrix? They are all zeros. So, what must the trace of a nilpotent matrix be? Zero. What is the trace of a positive matrix? Positive non-zero. So, there you have it right you do not have to write down everything. So, the answer is an emphatic rule of course not. There are several ways of arguing this, but of course, one straight forward way to argue is that a positive matrix can never be nilpotent right. So, your positive matrix definitely has at least one non-zero Eigen value. Therefore, the spectral radius of a is not equal to 0 if a is positive. The only way it could have been 0 is if all its Eigen values were 0 or if it were nilpotent, but it cannot be nilpotent like we have argued. So, in general a positive matrix must have a non-zero spectral radius yeah by definition spectral radius is positive. So, hereafter the moment we are given a can we not just convert it to a upon r a. So, that we call this object as a hat and we are assured that the spectral radius of a hat is unity. So, in general whatever we talk about positive matrices now with respect to the spectral radius and all we can just reduce it to the question of a normalized positive matrix correspondingly yeah you agree. We need not worry about the specific spectral radius of a given positive matrix whatever is its spectral radius just normalize it by that. So, you get a new matrix this is scaling all the written I mean division of course, you cannot divide a matrix by a matrix. So, this is a division by a scalar which is non-zero guaranteed because it is a positive matrix. So, therefore, this matrix we know that such that the spectral radius of a hat is definitely unity. So, it is just a bit of normalization that we do all right. Now, suppose see what are the claims I made I made a claim that you have exactly one Eigen value on the spectral circle that is that is on equaling the spectral radius. You cannot have any other Eigen value anywhere on the complex plane which is just have the picture in mind you have this circle here. So, now it is boiled down to the unit circle because this has unity. So, this basically this one and none of these other points that you can potentially check out for Eigen values will ever yield any other Eigen value that is what I claimed. So, now it is time to try and show why that must be so. So, suppose we assume that so we are going to talk about a hat here after ok the normalized positive matrix. So, suppose the absolute value of lambda is equal to unity and A x is equal to lambda x and of course A is a positive or rather a hat is a positive matrix and R of A hat is equal to unity ok. So, we are trying to argue the contrary that on that spectral radius you can have potentially other points also which correspond to the maximum modulus or maximum magnitude of the Eigen value right. So, this we will have to rule out why this cannot be the case right ok. So, suppose we take the moduli of entry wise this is a vector this is a vector we can take that. So, then this becomes the modulus of lambda which is just unity. So, this just becomes this right right, but what can we write this as what is this fellow going to be? See this is going to be again we have A hat i j x j or rather 1 j x j A hat 2 j x j A hat n j x j each fellows individual moduli like this on the left hand side right and on the right hand side we have x 1 x 2 this must be true, but what can we say about if we had taken the moduli of A out first yeah can we not say that the moduli of A hat x is less than of course the moduli of A 1 matter because A is a positive matrix. So, I can just write this without the moduli it is a real matrix positive matrix what difference can the moduli make can I not write this just think about it because now I am restricting the entries of x to be only positive positive reals yeah earlier this x need not just contain positive numbers know it can contain anything complex number is anything whatever we like now I am just restricting it to be just the moduli of those numbers. So, this is true right. So, if this is true what can we conclude if we are able to show that this is true and not just is this inequality true, but it must be equal then we will be done because we started with the assumption that lambda is not equal to unity the mod of lambda is absolute value of lambda is equal to unity. So, therefore, we are led to this conclusion and this is what true. So, if we can now show that this is actually an equality then we will be done. So, so far let us carry forward this inequality what we have is A hat acting on x minus x what does this remind you of something we have just seen if the action of A on a vector this is also a vector you agree it is a real vector the action of A on a real vector has the effect of increasing its entries yeah then what can we say A hat raised to the higher powers this must also be true, but now recall that we have this different object here now let us say let us define a new vector let us define this vector as p. So, we have on the one hand from this we have A hat acting on p is this and of course, we have A hat acting on x also greater than 0 yeah. So, we have this and of course, we have A hat acting on x also greater than or equal to 0. If you have two positive vectors just like with positive numbers what can we say both positive numbers I do not care which is bigger than which I can scale up the smaller number sufficiently until it becomes bigger than the bigger number if I am given two positive numbers 2 and 10 I can always multiply 2 by 5.1 and it becomes bigger than 10 same exact idea holds for these positive vectors right if entry wise every vector in one vector is smaller than the corresponding entries of the other vector then you just look at the biggest difference there is scale it up suitably and you can make the smaller vector to be bigger than the first one. So, I do not care essentially I can always choose it means that there exists gamma such that A acting on p it is less or which one will help me yeah yeah is greater than gamma times A acting on mod x what the I mean for the abuse of notation I can always find some gamma right to result in this right. So, what happens now substitute for this p here. So, I have A acting on actually I think I should call the lower fellow as p maybe perhaps I have noted it down wrong no I do not think so yeah no it is fine it is fine yeah. So, this is A hat I have missed the hat in patient here. So, A acting on A hat p p I can replace by x minus this x is greater than gamma times A hat acting on this x right from which yeah I think it is right from which I can just pull out this A acting on A hat x is greater than 1 plus gamma times A hat acting on x yeah that should be alright let me bring the gamma to the other side yeah that should work and let us call this new object as B. So, this is A hat. So, you agree that of course, gamma positive means this is also positive matrix. Now, what is it what does it tell me a positive matrix acting on a vector is greater than this. Therefore, B to the k acting on A hat x is greater than A hat x sorry yeah it is all A hat here we have forgot have I missed the hat hat notation somewhere. If I have then you put it there yeah it is all hat we have normalized it we are not only going to talk about normalized positive matrices now. So, that we can just deal with the number 1 instead of having to carry R all the time. So, you agree that this is true now limit k attending to infinity what do you think is going to happen to B to the k what is 1 plus gamma it is a positive number greater than 1 sitting in the denominator raised to the kth power. So, this is going to be 0 right. So, in other words you will have A hat x is equal to 0, but that is impossible right it is a contradiction. So, where does this contradiction stem from we assumed that this inequality was a strict inequality that is what led to this contradiction eventually. So, then this must be an equality rather than an inequality and if this is an equality then what have we got in one shot something very beautiful not only is 1 exactly the number 1 equal to an eigenvalue of this positive matrix the normalized positive matrix, but the corresponding eigenvector is a positive vector because after all this fellow is a positive vector once I have taken entry wise absolute values that is a positive vector. So, the largest eigenvalue is real there can be nothing else on the circumference of that unit circle apart from the real positive eigenvalue and the corresponding eigenvector is exactly going to be equal to some positive vector of course, you might argue oh hang on I can just multiply it by a scalar which is negative and make it negative of course, you can. So, it is either all positive or all negative nothing in between right because it is eigen space is the span of that vector right. So, you can also scale it in a negative direction. So, in technical terms you might say it is either all positive or all negative, but we will only be interested in the all positive solution because most practical problems where we encounter positive matrices then solutions also demand that we give some positive numbers of solutions of such systems like the Leon T. F's model. So, at least this much we have seen that there is only one root on the unit circle for the normalized positive matrix equal to the spectral radius and that is real corresponding eigenvector being positive all right. Next we are going to try and check out as to why this eigenvalue that we have now figured out the real one is going to be a singleton only they can be not it is non repeating in other words it is algebraic multiplicity is equal to its geometric multiplicity is equal to unity. I am going to make some use of the Jordan canonical form, but I am not going to explicitly write it down I will just tell you what the result is I am going to ask you to think about it and maybe it might help in your exams who knows. So, think about the proof, but we have done Jordan canonical form in sufficient detail so that you should be able to prove it on your own right. So, of course do I need to really show that this a x I mean this a x is equal to a x is equal to x. So, even if this x to begin with was non negative you know that it is in the image of a anything in the image of a positive matrix any non negative matrix in the image of a positive matrix must be all positive. Therefore, indeed the eigenvector is all positive like I claimed yeah that is just an aside, but yeah let us get to the important business. Actually we have just shown you that there is this eigenvector eigenvalue equal to the spectral radius, but we have not technically really shown you that there can be no other complex number sitting over there even that I will leave to you to complete as a proof the proof is not very difficult there is just one key observation. So, you see we had this summation a I had i j x j suppose there is another fellow for whom this is true. So, right. So, then what would you have you would have some lambda x I right you would have some lambda x I this is true and then if you take the absolute values on either side what would you come up with you would come up with summation of the absolute values of a i j x j is equal to the absolute value of x I because the lambda I lambda would vanish if lambda was a complex number of unit absolute value unit module I then this would be true what does this tell you what can you conclude from this see this is the sum of the of each individual things here and if you know sum up these fellows as well what what happens to the sum on both sides. So, look at this when is this equality possible like we have seen earlier this is less than or equal to what you take the absolute values of this and the absolute values of this. Now because of that spectral radius and this fellow having a magnitude equal to unity that equality still has to hold. So, you are basically asking for a hat i j x j is equal to summation a hat i j x j, but now the only way that this is possible is it is a complex Eigen value then the Eigen vector is also complex. So, these are all complex numbers. So, what sort of complex numbers will adhere to this equality if you had another complex Eigen value of unit magnitude then this would still be true like in the previous case. Now if this has to be true what is this this is the sum of certain complex numbers complex numbers are like vector center 2 d space. If you are taking a complex number scaling it up by a certain real number because a i j's are real and adding them unless they are all aligned up together can you have an equality. If they are all aligned up together what can you say about these x j's do you follow what I am saying you take a complex number you take a complex number along this direction you take another complex number along this direction you take the sum that sum is never going to have a magnitude which is equal to the sum of the individual complex numbers unless they are all aligned up together. So, they have to be all in the same line because complex numbers are like the 2 d space. So, if this equality has to hold then all these x j's are along the same direction a plus i b gamma i plus gamma i b 2 gamma i whatever alpha i plus alpha i b so on which means what that these numbers are all just scaled versions of each other. So, you can just take out that complex number and cancel out on both sides and you will have to the real part of the equation only which means that Eigen value is just 1 you started with an Eigen value which is potentially complex complex number of unit moduli. But now if this has to hold and this equality has to hold like we proved earlier if this equality has to hold then all these complex numbers must line up together if they lined up together they are scaled versions of one another if they are scaled versions of one another real scaled versions of one another. So, those real numbers can be pulled out and that will be the corresponding vector right and the complex part gets cancelled the common complex number let us say the unit complex number along that direction you just pull it out as common and the other others remain. So, the common complex part gets cancelled so it is real number equated with another real number and if that holds then this lambda must be the real number 1 not just some complex number of magnitude 1. So, that is actually the proof. So, I kind of did not delve into that, but the lambda has to be unity it cannot be a complex number yeah I mean I am going a little quickly on this because again we want to get to the Leontief's model and answer that important question at the end, but if you are interested these are kind of optional material for further studies like Perron-Frobenius theory and also it is pretty interesting, but if you are specifically interested then you should definitely read up what we are covering are some properties of positive matrices do read up on non- negative matrices that is actually the contribution of I believe Frobenius Oscar Perron first worked with only positive matrices ok right. So, we have a kind of established that this is true. Now, let us try and check out this positive matrix and say that the algebraic multiplicity of 1 is equal to 1 and suppose the geometric multiplicity of 1 is more than 1 oh sorry it is the opposite right yeah the geometric multiplicity is 1, but the algebraic multiplicity is greater than 1 which means that it is repeated if you look at the characteristic equation then there are more factors of lambda minus 1 than just 1 lambda minus 1 squared lambda minus 1 cube, but there is only one geometric multiplicity support in which case actually I can just choose it to be any number smaller than the algebraic multiplicity the same argument it means that it has a Jordan canonical form if it has a Jordan canonical form and if I keep raising it to higher and higher powers can you imagine what is going to happen at some point you are going to have a combinatorial term when it raised to the mth power m plus whatever m minus k plus 1 something choose k minus 1 yeah times I think lambda to the m sitting in the corner most diagonal what do you think is going to happen to this as m tends to infinity I leave it to you as an exercise it is a bit of application of limits lopital see lambda is more than 1 suppose ok or let us say lambda is unity lambda more than 1 is it is blown up lambda is unity then what happens to this if the modulus of lambda is unity then it is just this term right this is a combinatorial term that blows up so what happens in the Jordan block as m tends to infinity if you keep raising a to higher and higher powers of a Jordan block do you see that the matrix because ultimately when you have a it is just some tj t inverse so a to the m is also similar so now if we let m tend to infinity this fellow would blow up which means a to the m must also blow up yeah but can a to the m blow up because what do we know eventually about a to the m there is at least one eigen value which is unity so if you take that eigen direction and if you keep hitting it that eigen vector by the way picks out every column see here's the argument very loosely speaking if it had a Jordan canonical form which is not diagonalizable then we are in trouble because as m tends to infinity this blows up why is that a problem because if this blows up then this operating on any x any vector would also blow up not any vector precisely that vector which picks up that deadly combinatorial terms but we have to have some vector in fact the precisely the eigen vector corresponding to one will pick out every column and if it picks out every column then it's definitely going to pick out some term like this which blows up so therefore a to the m acting on x entry wise blows up so of course it's not blows up but can it's not blow up because we know that ultimately along that direction if you keep hitting it what happens it must result in equality you know so it must be equal to the original vector so you will come to the conclusion that the original vector is already an finite vector cannot be the case you understand this fellow must blow up everything in along every direction a vector which picks out every column of this has to blow up eventually I know that the eigen vector corresponding to one the one which led to this yeah the eigen vector the original eigen vector not the general eigen the original eigen vector is definitely all positive so it picks out every column which means it has the effect it captures the effect of every entry the largest possible m that there is that entries also the corner most entry the north eastern corner most entry of the Jordan block is also going to be picked up which means as m tends to infinity this that fellow must blow up and if that fellow blows up the eigen vector then it violates the very property that the eigen vector is left invariant see the point so therefore you cannot have a Jordan block at least if you have greater if you have algebraic multiplicity greater than one the geometric multiplicity had better also be greater than one so this is the part I said I won't prove I will ask you to check this out so this this point is actually kind of ruled out so the only other possibility is that greater than one and geometric multiplicity is equal to algebraic multiplicity which is when we have a diagonal block for the eigen value corresponding to one now apparently we can have the case because these terms won't appear now the nilpotent part is not there anymore in the Jordan block for one so it's just the diagonal entries that are getting hit repeatedly and those are the only things we need to worry about so why is that not possible okay so that is possible suppose that at least that the eigen space corresponding to lambda is equal to one has exactly the requisite number of eigen vectors right if there are multiple let's say R let's say K is the geometric and algebraic multiplicity of the eigen value one that means there's a K by K K by K block of all diagonal ones that is sitting somewhere in the overall Jordan canonical form but this also means that there are exactly K eigen vectors linearly independent eigen vectors corresponding to one so suppose X and Y belong to kernel of I minus a yeah such that X and Y are linearly independent right then what happens actually let's do it a little differently let's say X belongs to kernel of I minus a transposed and Y belongs to kernel of I minus a in two different directions let's say so what can we then say now this we will not require we can actually do with a transposed I believe yeah I think I just I was right the first okay let let it be this so let yeah I think it should be okay let X or rather let P is equal to X minus X I by Y I times Y we already seen that if these are both eigen value eigen vectors corresponding to one they must would be positive and this is just the ith entry of X and ith entry of Y what do you think is going to be the ith entry of P 0 right so P I is equal to 0 but what is P I what is rather what is P P is also an eigen vector is it not because it is a linear combination of eigen vectors for eigen value one but P belongs to kernel I minus a right if P belongs to kernel I minus a implies P must be a positive vector like we argued but P cannot be a positive vector if we choose it like this so you cannot have diagonalizability either so either way if your multiple eigen values either it's a proper block diagonal form in which case this contradicts it or it's a Jordan form in which case it blows up and therefore it blows up along the direction of the eigen value eigen vector as well that is also impossible so not only the fact that you have exactly one eigen vector eigen value equal to the spectral radius but also you can have no other complex eigen value whose magnitude is also equal to the spectral radius and the multiplicity of that real eigen value equal to the spectral radius is exactly the algebraic multiplicity is exactly equal to one so it's this is a unique fellow yeah like it has no other parallel is what this means so what does all this mean in the context of what we have seen so far in the Leon T. F's model you can try an exercise I'll not prove this but yeah I have noted that down as an exercise not relevant to the Leon T. F's model which we shall cover now in the next module but before we since we have done this much the interesting deal is the eigen vector corresponding to lambda is equal to one which is the spectral radius is all positive if you take any other eigen value of a positive matrix the corresponding eigen vector cannot be all positive okay sounds very weird right so not only is the eigen value unique in that sense that it sits there in the real axis but the eigen vector is also saying it's the only one in the positive octant or whatever you call it when you have all its entries positive in three dimensions we say it's in the positive octant in higher dimensions it means that all the components are positive so there can be no other eigen vector in that entire you know octant so to say or the generalization of the octant there off that's the thing that you have to prove so you try and assume that there is another eigen value of course its magnitude is less than unity and for that eigen value assume that there is an eigen vector all those entries are positive and then try to contradict try to see why that cannot be possible whatever little bit we have covered of this we're on Frobenius theory we should be in a position to get back to our Leon T. F's model and do some interesting things so what did we have we had the fact that Lx is equal to x was required to have a solution as your friend said if everyone is thinking about profit can this have a solution if everyone is thinking about profit what does it mean so there is no external buyer so will there be an incentive this is very common human psychology right we don't generally tend to be philanthropical that's why philanthropists are rare we have nothing to gain out of this no profit then why should we keep manufacturing just for the greater good people don't do that normally right very grim view of the world but that's true so everyone would try to minimize or kind of get to a column sum which is less than one if that happens if every column sum is less than one if every column sum is less than one can you have a solution for this that's the question suppose every column sum is less than one yeah suppose every call sum of a L is less than one let me hit this one with a one transposed on either side right that'll also be zero if I hit this one with the one transpose what happens then I have summation x i is equal to one transposed L x right what happens to this this is all summation row sum right because this one transposed is taking a column sum and the column sum is what so li so let's say gamma i is equal to summation l column right ji summation over j yeah yeah so this is gamma i x i yeah I think that's going to lead to the contradiction somewhere because now you see what I have here is one minus gamma i x i is equal to zero what do I know about the x i's yeah that's going to lead to the contradiction what do I know about the x i's they're all positive numbers because they correspond to the eigenvector for possibly the largest eigenvalue right so these are all positive and these are all negative numbers can the sum of all negative numbers be zero because this gamma i or rather it's positive no gamma is positive so these are all positive numbers basically anyway whether it's some of all positive numbers or some of all negative numbers they cannot all be zero right so this is the contradiction yeah good good yeah so finally after a bit of struggle life come up with the conclusion so in other words if everyone's trying to be profitable in this in this game so that's the lesson we have proof that if everyone's trying to be profitable unless there is other incentive you will not be able to sustain such a system of economics right just out of goodwill just out of the sense of cooperation or fellow feeling there has to be some incentive what is that incentive that incentive is going to be external demand so currently every industry is supplying exactly as much as is required to cater to its own needs as well as the needs of its other fellow industries and you cannot profit out of this if one fellow decides to profit out of it immediately it goes kaboom right explodes but if there are external buyers who also want your product then remember what happens then what you'll have is this l x yeah it's going to be equal to x plus some external demand vector d yeah then you will require i minus l or whichever way you like like with a minus sign whatever x is equal to d now what do you need essentially if everyone's breaking even but just about creating enough for each other's needs this is not going to be invertible because one is going to be an eigenvalue if l has column sum equal to all ones if all the column sums of l are equal to one then one is an eigenvalue therefore this is singular so any arbitrary demand that I give you you'll not be able to meet it unless it is in the current it is in the image of i minus l only those demands which are in the image of i minus l can be met you cannot meet other demands right on the other hand now if everyone starts to think of profit then column sums become less than unity eigenvalue cannot be unity this cannot be singular this is non-singular this always has a solution right so in in presence of an external demand for your products the same situation the same profiteering tendency that sort of did not allow your industry to sustain now if you're trying to profit out of this external demand you'll see that that will allow you to have a feasible so now when there is external demand people will be willing to cooperate and that's what people do why would Boeing try to manufacture aircrafts if people are living in stone ages and don't want to fly in aircrafts right so people have to have some interests so when they're interplays between corporates or industries unless they are each convinced about their dependence on each other and the fact that the end product that a related industry is supposed to sell has some demand in the market of course it might happen a little late and then they realize a little later so collapse is sort of delayed but it will eventually happen so external demand is what keeps the market driven if the demand goes down everything else will collapse right because everyone will try to profit profit here out of it and eventually no demand means you're only supplying for yourselves and that will not be able to sustain your industry because your money will run out right so that's what this famous Leontief model is there's not more to it in fact it turns out if you know the theory of non-negative matrices you can prove a stronger result here we have assumed that everyone is trying to profit here and in the presence of demand that allows you to sort of cater to any sort of demand actually don't need all of them to be profiteering even if one industry thinks of profit the rest n-1 do not even care about profit even then your economy will be able to sustain so that means unless all your economic sectors are kind of you know messed up at least one has to survive if none of them survive if all of them sort of collapse then you're doomed but at least one fellow has to profit that means the one fellow is at least producing more than it consumes right in that case you'll still be able to sustain that cannot come from this theory that we have developed or talked about today that will come from the theory of what provenius contributed to this which is the theory of non-negative matrices right so with that we have come to the end of this thank you