Lec 31 MIT 18.085 Computational Science and Engineering I

chiefly in mind here uh two are coming from last time uh I wanted to say just a little more about this neat theorem of Max flow Min cut this was a this was a do an example of Duality and a particularly nice one and uh um chapter 7 of the book has uh examples and and more discussion of that then um you were kind enough to get involved in my new uh interest which was well the specific question was uh when can I complete a matrix given some of its entries when can I finish it out to have rank one and I have some news so of course I'm eager to tell you about it and uh and then the I I realized that it might be a pleasure to talk about Game Theory as another uh two person this is twers game theory created by uh V noyman and uh uh it's another example of Duality uh but uh just coming from somewhere from such a different source that it wasn't recognized uh for a little while that that it fit the framework okay with Max flow Min cut I just have this uh uh example to mention uh suppose all the capacities are one on the edges so on every Edge we can send one um uh unit how well in this example let's say how much could I get from the source to the sink uh um well along those edges okay uh oh have I made it so I could get four could I get four through on that yes I think I could right because yeah so let me let me uh make it a little harder let me add another Edge there so hopefully I could get more through but I think I can't I think uh that how much could I now get through from the source to the sink there are 1 two 3 four five edges leaving the sink and then they go different places uh and there could be as we had last time uh uh the the bottleneck could be in the middle of the of the path or it could be right here at the uh sink maybe it is maybe in this case I can only get four I could even uh yeah I could I think maybe the maximal capacity is four agreed yeah and the cut that that shows me that proves it is actually well in this case sort of a degenerate case the sink is all by himself in the cut so there's a cut of one 2 three and four edges getting through and that's the maximum flow also and what I let let me draw the the paths that that it could follow one set of paths that it could follow uh let's see I could send one along this path and I could send one along this this path and one around here and one around straight through there so those notice that those paths are disjoint separate paths uh the and they have to be right with the with with this kind of flow I'm not going to gain anything by sharing by having uh by by sharing an edge with two flows so what I'm trying to say is that for this example uh it the max flow equal Min cut can be expressed in a really in a different way the ma so what I want to do is restate this for the special case here as the maximum number of disjoint pass S2 s Prime source to sync so there's an example four disjoint paths is equal to one so I'm looking for how many independent paths go from here to here and it's equal to this the capacity across the cut and I would like to find a different uh set of words for that uh so what happen this cut suppose I actually make such a cut so I'll actually use the Eraser to to cut the the uh Network so what does a cut mean it's a it's a a separation of the nodes into one group with the source and another group with a sync well sorry that in this example the group with a sync is only the sync but nevertheless that's that's a cut and what's what has that cut done that cut has separated the source from the sink it's it's literally cut the two apart so the and the capacity of course is the the across the cut is the number of times we clipped an edge so here here's what I'm trying to to say the maximum number of disjoint paths that measured how much flow we could do equals the minimum number of cut of cut of cut edges to split may I use it use a strong verb there to split s from S Prime the minimum number number of edges that it took to to to uh uh separate this from this you you see that that this is a a restatement and maybe we could think why why I mean this is as usual we should see weak Duell so what would what did weak Duality mean weak Duality which was usually the easy thing to to to to see so if I just made suppose I had just made this statement that I draw a network and I look at how many separate paths there are that don't don't share any edges that's what disjoint means from source to sync how many paths uh and I looked at the minimum number of edges of cuts that I had to make to separate the two now it should weak should be maximum less or equal minimum should be easy and remember it's slightly strange that it it's that direction that's easy of course it's the maximum of one thing and the minimum of another why is that easy how do I know that um that if I have uh the minimum number of edges to to if I find it if I find some edges that split yeah H how do I know that if I if I find any cut any any any bunch of edges I could snip through that would separate Source from sync and I count those edges and I get K how do I know that the number of PA can't be more than K yeah yeah if if exactly if each Edge if uh if I manage to cut the thing with k then there couldn't be more than K disjoint paths because then I wouldn't have cut one of them let let me say that again just easy uh suppose suppose K Cuts split s from S Prime then can't be more can't be greater than K disjoint paths because if there were more than K disjoint paths then I didn't cut them all and then I haven't separated the source from the sink so that that would be the reasoning that I take any cut that works I say okay maybe it took k a separate edges to clip but then the number of disjoint paths couldn't be larger than K because then if it were one of those paths isn't getting cut and and uh uh we haven't we haven't separated s from S Prime so that that says that any any cut that's acceptable is large is at least as large as any count of disjoint paths so we increase the count of disjoint paths up to its maximum we decrease the number of cuts to the do the best cutting we can and we have less or equal and the point is we actually have equal and maybe here's a way to here's a proof of equal suppose we have uh well the idea of the proof is that if if uh um that suppose we have a maximum flow yeah I I um this is in the book and and it's hard to take in verbally but if we had a maximum flow suppose we had the flow as large as possible then I want to see why some cut of that size is holding us up so we have a Max a flow as big as possible then I keep with s some edges might some some with s some edges I might be able to send more to some uh along some edges I might be able to send more for example along that Ed Edge I could send more ha that's interesting okay now I'm going to I may find another cut here so I I keep I group with the S all the nodes to which I could still send one more okay let me see I need I'll use this so with this I keep th this note because I could still send one more to that who and then oh yeah this is going to be tricky because I could s now send backwards on this Edge so having sent these forward I could sent one backwards I think I'm not going to get into this uh uh uh because I don't want the whole lecture to be this rather specific topic what I wanted my my point is that the we have these uh special uh dualities for some special network problems and that they are special cases of linear programming I mean what are the equations yeah so here is one final question on this what are the equations that govern a flow I'm trying to maximize a flow what are the equations that govern a flow I I want to fit it into linear programming I want to maximize the flow so I have my unknowns are the flow along every Edge then I have some inequalities and I have some equations and if you tell me what those so so I'm I've I've so what what describes an okay flow so so so so Ma mat to match to match linear programming I want to ask you what are the what are the inequalities for a flow so I so I'll maximize a flow what are the inequalities and what are the equations what I mean a flow is is um the the what are the unknowns here the unknowns are let me say what the unknowns are uh x i j is the flow from uh I to J okay so do you remember that in linear programming you're maximizing something but you have certain inequalities now what were the inequalities in this in this flow problem the inequalities were that the flow couldn't exceed the capacity we we put a capacity of five seven nine one Whatever on those edges so on every Edge there was a upper bound and now I want to say what are the equations that flows have to satisfy what what's the equation I claim that at every node we've got an equation at every at every interior node every node other than SNS Prime I have an equation and and what is it it's Kirk's law right Kirk Hof's law so we have the the equations are Kirk off's current law that's that's my little point that that we we're coming around again to the to the uh most fundamental most fundamental equation in Applied Mathematics I I guess I dare say that um if you're using current law then you have to allow negative flows how does that change if I well let's see uh so I'm interpreting Kirk off's current law as at every node the flow in equals the flow out you're right so so there'll be some pluses and minuses doesn't that change your inequality so now you have a nonlinear inequality the magnitude of XI has to be less than C well let I'll just I'm not even sure in this instance but but suppose we had what looks like a nonlinear inequality turn it into linear inequalities just that's a very good example of how flexible this is how could I turn that which certainly looks nonlinear into linear inequalities double the number I could double the number right I could leave it as XI J and add to that minus XI J is what would it be greater than or I forgotten minus CI J what what would what would it be if I want if I want XI to be small maybe it would be that would it right and then uh yeah yeah uh so or is that no is it is that right I think you need to I I I don't get rid of that minus no on C on C maybe minus XI greater equal C J yeah okay okay my point is we could do it and we'd still have linear programming uh double the number of uh constraints but we we'll handle those and the point I wanted to make was this key equation kof's current law I don't mean to say that when we look at all of Applied Mathematics kiru's current law jumps out uniquely except what jumps out is balance equations conservation equations in equals out plus possible sources that that's that that so much of Applied Mathematics is that's one of the equations to look for then there'll be the equation that describes the particular uh the the physics of the problem you know what what is the flow how does the flow connect to potentials or something but that that balance equation is always there okay so that's I I'm I'm going to stop there with a hope well with a hope that you find some of those um uh ex special cases appealing and then also with the hope that you'll enjoy Game Theory but I'm really Keen to tell you about if you'll help me again as you did last week you kept me going through this question let me remind you what that question was the question was uh so let me put the question here the question was I have a matrix and I'm wondering about which positions could I give arbitrary values and then complete the Matrix to have rank one so this is my so I'm giving any nonzero values and let's take an example could let's let me take the example that I had last time well let me take this example could I give any any nonzero value values in those six places and aim for complete to rank one is that possible and your answer is no because we have here if if rank one will mean what will rank one mean it'll mean that all 2x two determinant are zero and that won't be true here so I better take one of those out there or let me let me take let me make it a little L obvious by taking that one out could I now complete so now I'm I'm I'm less ambitious I'm I'm putting in five entries and I'm putting them in this position and could I now complete those to complete this to rank one and the answer is yes right I could this this would this 2x two would now tell me what that number had to be this 2 by two would tell me what this number had to be if I do it two just this 2 by two would tell me that number this 2 by two would tell me that number well and you might say wait a minute this two 2 by two told me that number also so maybe it's less uh uh um maybe it's uh needs a little bit more thinking so right the way I did it was not completely convincing because I would said okay do that one then do that one then do that one then do this one and then you say wait a minute those three were specified that determined that one so uh I would need a more convincing what would be more convincing can can you can you give me the a correct uh a better uh well I could go back to that and do it more carefully but how how do you see that that could be completed to rank one rank one will mean that every row is a multiple of every other rowo so right right we see that row already set just enough for each row not too many right yeah that's right we want to know that just enough and not too many okay so after playing around with examples like that the conjecture was that the right number to specify if this is K by K complete a k by K so 3x3 in this instance the right number is five that's our guess and and so uh the general guess was 2K minus one so that's what I want to prove that's what I want to prove that that that would be the theorem that uh it's completion rank one completion is possible if and only if every and I have to check the submatrices too every submatrix of every submatrix of let's say of order let me call its order K has less than or equal to 2K minus one specified entries specified entries and I'm going to specify them nonzeros so that that's that's like a hope a new theorem and and uh um uh it sort of partly happened because of the examples that we played with so so why so does this does this example pass that test yes moved it over to to where here the as it is yes but of course when I complete it when I complete it I I can hopefully get it to have rank one and and as you say if I looked at igen values they would be zero and zero and whatever that Trace is yeah so yeah I put in numbers so let me let me take a specific case can could I give and you'll see by the way I write these numbers in 1 4 7 2 and 11 now the question is can I complete that Matrix to rank one yeah this is very thank you for every question you ask improves the way I think about it so so okay so we would say yep we can do it uh all right I could do it 2 by two you know I would see okay what do I need there I need a half is that right and really what I need is that for that row 147 for this row to be half as much so this would be seven halves and for this row this row should be in the ratio 11 to 7 or whatever yeah sorry when I said arbitrary numbers I didn't mean that arbitrary uh yeah but anyway I could do it and you realize why just just to be sure that I can't do it for any five suppose I specify those five now I can't complete to rank one because I'm lost at the start here if I've made those arbitrary then that's a 2X two that isn't determinant that isn't zero and no way I can fix it I I could make those numbers if if you want it right uh so so that's why I have to in the test has to go not only the big guy but all the smaller ones because if a smaller one fails my test the bigger one might still have only 2K minus one five specified entries here but the smaller if a smaller one messes up then I'm it syns me before I can I've got no way to save it okay so uh uh let me let me take a 4x4 just uh so 4x4 I should be allowed seven entries and uh okay now here think through how suppose I have seven entries how can I complete how can I prove that I can complete so I'm I'm asking um what what I'm really talking about now is that math step from doing a bunch of examples and getting convinced and getting a proof that's absolutely convincing see what I'm trying to say that I now have to think seven entries any any seven subject to this constraint how can I show that it's possible all right can I can let me so actually strictly in this proof I have two directions to go but let me let me suppose that this test is passed and I have a 4x4 with seven entries that passes that test and I'll conclude that I can complete and then there will be another step of the proof that that shows that that condition is necessary but let's do the first one way that that's okay I've got seven entries in a 4x4 4x4 seven entries okay and now let's see so my here's my reasoning one row has to have and all the rows couldn't have two entries right right that would make eight so one row has to have no entries or one okay so I'm covering all cases so so so one row has to have either no entries or one entry right so we can reorder the rows and make it the first row in fact then we could reorder The Columns and make it the First Column so here's a so so up in the first row is only one entry and I'll reorder The Columns that won't change anything just to keep our our we're looking like that right there has to be a row with only one entry because there's only seven entries all together and four rows okay there has to be a row either with one or or with or there could be a row of all zeros uh with what do I mean by this zero I mean unspecified should I put U for unspecified okay you for unspecified or I could possibly have a there could be I could have seven in the whole thing but there would be but let's let's kill this possibility first here okay now now how how could I show now I've got seven the seven are still specified but they're only in three rows okay so what do you think now uh let's see oh I've got four columns here now the wonderful thing about linear algebra is the relation of rows to columns that's what you got to keep remembering okay so I still have four columns and I've got seven specified so again I have a column that has only one or or none right because there were four columns I've got to put seven x's in here some column is only going to have like One X and the and the other six are going to be in here okay so I claim that this totally VI that this can't happen because I'm I'm assuming this right I'm assuming this let's get the logic straight I'm proving that if this is true I assume it's true and then show this show this that's this is the direction I'm going okay so I assume that every not only the big Matrix the 4x4 had seven nonzeros but every submatrix of order K didn't have more than 2K minus one but here I have a submatrix it's 3x3 and it's got six so this this that couldn't have happened let me come back to here again I'm going to shoot the same go the same way here um now what do I want to do I'd like to okay now I have one specified guy that's more reasonable unspecified unspecified now I look at the columns okay I've got four columns so what is what's possible here uh this column might have two guys it might have this specified and one more and unspecified unspecified right that that could be or I better have another or out here let's see let's see so I'm starting with these let's see what are the other well let me follow this case through first suppose this is my situation suppose remember I figured out okay one row does have to look like that well don't you have to have point where you're ring that's what I have to show and and this is so in this this is the good case where I do all right now what's the what's the deal here I have uh it looks like this and then how many are specified in here the remaining five so what's what's up with this case I apply the theorem for smaller size and I know I can complete this I can complete the five because it's satisfied all its little submatrices of course were part of the part of the part of the assume and it's got five which is the right number so I can complete it so somehow I do complete it now of course I have to just check having completed um some completed the this to have rank one uh can I now finish the job so now how do I finish the job here uh knowing that those rows are all in the same direction those three rows are multiples of each other in the small one now I have to achieve that in the 4x4 okay well I guess I see one row completely set and so now how do I fill in this row just the ratio whatever I mean that row set this guy sets the ratio and I can fill it in and how do I fill that one in again this was in this was in a certain ratio with this one so the last entry has to stay in the same ratio and similarly here this was in a ratio with this because this was rank one in the in the 3X3 and then when I'm given that I'm I'm given that so actually there's a unique way to finish so you could say yep that that was no anybody see that I've got one more case to worry about I well this this was x u u and then I said okay what if my row what if this column has uh let's see uh yeah what right that's a case I have to worry about what if nothing else was specified there why am I how do I know that that couldn't be because then I've got six more in the 3X3 which which I assumed I didn't have okay now what okay I think I've got another case to worry about mathematicians are awful worriers all right what else do I have to think of maybe the what if what if this had more like three or or four now what do I deal how do I deal with that case what I'm hoping for is to find a 3X3 that's that's okay uh now what's up here in this case one of those other columns only have one one of those other columns will only have one right one of those other columns will only have one because as I've used up three there's four four more to specify and only and three and three columns so so some other column has got got one easily right okay so suppose a second another column only has one all right now what do that like now yeah now the good columns to look at are this this row and that column this row and that column I I'll move that column over or you see out of this row and that column I've used two so I'm remaining with a 3X3 I have a 3X3 remaining with five you see that if I look at this remaining 3 by3 because this row and column have only used two only specified two I have five unspecified ones in this 3x3 I can fit I can complete those by induction by the is that right y right because whatever some some of those were specified but but a total of five were specified and I can finish the job on the 3X3 and then uh again I'm uh then I can finish the complete job because now I have a row completely set and that will set the remaining rows I can am I okay all right I I've partly convinced you by shouting right but uh but uh that's how the the the proof uh didn't turn out to be too messy and I probably could tighten it up a little bit in in what I've described but that that's sort of the the way the proof would have to argue by induction that that there has to be a row with only one specified then we check the columns we look at the possibilities not too many we eventually we either rule out some of those possibilities or we identify a smaller Matrix that obeys the rules and then by the induction step we can complete that Matrix and then we finish okay so that's that's that and let me tell you about if if you're willing to go one more minute on this question what about completing to rank two because of course the trouble is another bad thing about mathematicians once you finish right you get the proof happy you're only happy for about 30 seconds because then you think of the next case and it's always harder so now suppose we're trying to complete a matrix to rank two all right I'm jumping now to a new question how many would I expect to specify say in a 4x4 say in a 4x4 how many entries would I expect to specify here it was 2K minus one seven how many now if I'm going to complete to rank two that's easier right easier to rank two because then I can have two rows that are quite different and the others have to be combinations of those so I could have two rows specified completely differently any completely I'm using X right for for specified I could specify two rows arbitrarily now I could specify two entries in the third row and then do you do you see that I could I could finish the job and keep rank two I I'm what I'm trying to do first is guess the right number and I'll do that by assuming that they're in a nice pattern and and and see okay how many could I expect to do in that pattern okay suppose I'm given two complete rows then if I'm given this row and this row that number and that number that tells me what combination of this row and this row it takes to get these two guys right right I've got two coefficients in that combination so I can get I can take a combination to get these two right and then I better have those unsp specified because they're going to be whatever they come out and similarly here I can I can uh specify two and uh some combination of those two rows will match those two and then those unspecified ones will be filled in I'm discovering in a 4x4 case I can I would think I can spe specify 12 and but that's the question here I can those 12 I can specify and of course the right so the question is can I alwayss like your 2 by two case with rank one it yes it's it's okay so I went away thinking well all right got this proof got the right number here 12 can't be that hard but then um I came up with a case that I didn't know how to do it so so this I decided this is this is so I don't know the I I don't have a proof yet I I should say we I'm doing this with a really sharp young guy named David ingerman he's he's the source of ideas here okay so now can I specify any 12 well let me show you what I got hung up on could I specify this is 4x4 I think I can specify 12 I was pretty sure but now suppose my unspecified are the diagonal and I specify all 12 of these guys so I specify those 12 numbers and these four I could leave blank as a sort of way to see that those are not specified and I want to fill this in to have rank two fill complete to rank two well can you interchange rows and columns I get in of the second well if that's yeah if so done but the trouble is there's a row with only two guys in it if I move move it elsewhere it's still I got only two if I move it if I move yeah but think I mean this is exactly the right question so I move that row up or down still has two if I change columns what happens on that row those two are still there they just in different columns so that's why this one was so hard because this has got three in every Row three in every column and I can't with just simple reordering get out of that fix so I couldn't see how to do it frankly you know you where do you start I didn't have a starting Place somehow and um let me call these numbers like uh A and B that we would like to be able to find a b c d okay now yesterday afternoon David convinced me I could we could do this one so I can show you how he what he sold me on here okay how he did it and then that's only so at this point so this is again a progress report uh our progress report I think is he figured out how to do this one this particular one so I still have we still have the question of any 12 in a 4x4 and then the more general question is so completion so let me complete the the the the theorem completion to rank two the number here becomes everything should is 4 K minus 4 that's the right number which when K is 4 that's 16us 4 which is our 12 so this was the this is the the guess conjecture you could say completion to rank two is possible if and only if every submatrix of order K has no more than 4K minus 4 let let me check 3x 3x3 what would that say I I'm saying that I can complete to rank two uh and suppose it was 3x3 how many am I specifying just just I'm specifying eight now is it true that if I had a 3X3 Matrix and I specify eight well how many have I got left only one left can I at rank two just just that that one if we can't do 3x3 well just go and shoot ourselves so so uh so 3x3 3x3 so this is rank two I'm still I'm still into the rank two thing if 3x3 and with eight specified eight X's well put them wherever that's eight only one left only one one Freedom left but I want to get I only I'm aiming for rank two how do I do Itor yes that's right uh so in terms of linear algebra so this is unspecified the some combination of that well another a quick way to say it is make the determinant what Zer zero make the determinant zero which the whole the 3X3 if the 3X3 determinant is zero What's the rank less than three two take two right does does it have to be exactly r two no no no no less or equal rank two yeah less or equal rank two yeah yeah yeah I have to achieve rank two but of course if I made if I specified stuff you know if I specified that row exactly in the same direction as that row then and and all sorts of stuff then I could maybe even get lower rank but but I'm shooting here for rank two so I'm saying if you give me eight I can certainly fill in that ninth what one way the way we saw it was some combination of that row and that row will make those two right and it'll tell me what that has to be and just I'm just saying to say the same thing in another language some number there will make the determinate zero some number there will make the determinate zero so I'm okay but so 3x3 completing to rank two no problem but 4x4 with 12 specified completing to rank two big headache okay so can I show can I come back to this one now this is my last case that I know about and then the goal is to get the theorem to work for for rank two completion where this is for okay here here was the idea suppose I look at this this and this so a 3X3 and I and I I I I certainly have to complete that to rank two so what'll it take well let's look at the determinant so I'd like to complete this so what's the determinant I have to make the determinant zero right if I have a 3X three and I'm trying to get lower rank rank two that surely means it's singular it's determinant is zero so so uh so the determinant of that will have to be zero so what's what is the determin it's a * B time a * B * something times times uh something and then what else how many you remember a 3X3 determinant what's it going to look like it only involves there's an AB term and then there's will be a an a time something q a * some specified and a b times something specified and uh and other constant and other specified terms that don't involve A and B has to be zero that's the determinant that's the 3X3 determinant okay but the thing is it involves both A and B so I mean that's only one of the many equations many 3x3 determinants but it's the one I want to look at and the other one I want to look at is this one can I I'll Circle the another 3x3 that's interesting in yellow take this I just want to use a and b can I I want to use this this and this [Music] and no let me see what a where's it gone I haven't got David to help me here I'm looking for and and this is that right I'm looking for a three no no here this this and this okay there's another 3x3 so it's determined also has to be zero and it involves it's independent because it involves these three numbers and not these three that were that appeared here so for this determinant to be zero okay what's it what's that determinant well again I have an AB it looks just like this AB times some other specified thing plus a * some other specified thing Q Plus B B time some other specified thing R plus some other specified thing has to be zero well out of this I can here's the punch line then out of this I can determine A and B they're they're nonlinear equations that's a little scary but I can I can uh uh subtract write multiple of this equation from this one and kill the ab and then I have a linear equation connecting A and B so I can solve for a and I can plug in then I get a quadratic equation for a which I can Sol it's two solutions and I don't yet know whether they're both okay and then back to find B in other words this leads to a and b and a similar discussion starting sort of symmetrically upwards would would lead to CND D and I believe that would they would then be right I I haven't even completed that that part but I think it would be so anyway that's a progress report that we now have equations that determine the ab c and d and we believe that when they determine the whole thing will have rank two and then we're uh I can't say often run because uh I don't have a general pattern there so still uh much to much to think through here I do want to so that that that's that's uh just I I just thought you might like to see three things about this problem in the rank one case you saw how a final theorem looks how how a proof could do do do the whole job and what a proof consisted of in this case you see uh theorem in progress where one case is clear we've now puzzled out another case but but in our heads we have the we know the job is all cases and we're not going to get all cases by by uh exhaustion the only way we can get it is eventually to see a pattern here and we don't see it yet and now thirdly I just want to say because it wouldn't have happened except for this giving this course that do you remember that I made the remark that I about something I had not known that if I have a triagonal matrix and let's assume these are nonzeros so it's a triagonal matrix do you remember my remark about its inverse probably not I didn't so in the beginning of the course we had particular numbers twos and minus ones and we computed the inverse those those special matrices k and t and so on we we we actually found the inverse so I thought that was interesting then those inverses had a special form and then I read that that form is always there that and what was that form that form was that above the on and above the diagonal so for these 12 guys that that was rank one that was so that was the thing that the inverse of a triagonal matrix has low rank above the diagonal so above the diagonal here it's of the form UI * VJ this is the the the in the IJ entry of the this is T for tri diagonal it's inverse the igj entry has its form UI VJ and let's suppose it's symmetric then it has low rank above the diagonal it doesn't have low rank overall so this is this is a different question from from the completion we were trying to complete to have low rank for the whole Matrix here the thing's invertible the rank is full the rank is n but but half of it has low rank and the other half has has uh entries by by symmetry they must be ujv and the total rank is n but the the interesting part was that and new to me was that half of it had rank one yeah I jumped in a question Lex in his work on noise showed that I'm not sure where he got it from okay that if you take a correlation the absolute okay take a correlation which is like this all right yes on the right okay then you get the inverse when you do the inverse the correlation comes out as triagonal okay so there was a specific case of importance right and and these cases that we did were specific cases of importance and I had no idea that sort of this was always happened that uh that uh uh if you had rank one here then the on the other side you had Tri diagonal if you have rank two here above the diagonal what do you figure then this one will be five diagonal so I guess now I'm I'm I'm sort of looking for a general law there I I I I guess I'm not looking so the the general law is then known that if we have rank five here then we then I mean it's a big Matrix say rank five above the diagonal completed by symmetry so it'll have the thing will have full rank but the other side the inverse will if I said rank five there then it would have one two three four five diagonals above the main diagonal and Five Below okay fine what do you mean by rank above the Diagon well good question that's what led to that's what led us to this crazy stuff what do I mean by rank above the diagonal well I certainly mean that that it if it has the form UI VJ do you who asked the question yeah uh I mean I I identify immediately a matrix that's what I mean by having rank one above the diagonal if it if it has a form UI VJ that means it has the form of column you know it it it uh um it's that's one column of of uh U's U1 U2 u3 times a row of V's V1 V2 V3 the entry in that would be UI VJ and it's a it's a rank one it's a column times a row so I I recognize that as a rank one Matrix and I'm seeing it here in different uh notation so that's what I mean by rank one Matrix if I completed the whole Matrix by this pattern I would have rank one but symmetry completes it with the opposite pattern and and gives me full Rank and yeah yeah so that would be the rank one case yes all right here here's my idea here's my idea okay algebra is saying that if we have a certain number of diagonals here we have a certain rank here fine can't beat that okay now but I think there must be behind that somehow a more general rule that if the if the thing is approximately a band Matrix now what does that mean if it tails off fast you see what I want to do now is generalize this to when these are not necessarily strictly zero and this has strict rank whatever but I wanted to see it's sort of like Heisenberg somehow if this tails off fast then this has pretty close to rank one there's my there's my conjecture that if I if that that if these are little then the rank here is near one okay what does that mean I suppose it means that I have a rank one Matrix plus something small so there's my conjecture uh unwi ly stated here and imprecisely stated but you see that the you see it's reasonable right and if it's reasonable it's probably true and and if it's probably true it's probably been proved that's the bad point but uh um so that'd be my guess that if I have a matrix that's almost a band Matrix then its inverse is almost um uh a rank almost a low rank uh above the uh above the diagonal okay stated that way it sounds like it's pretty clear so I have no idea uh uh whether that that idea whether whether that concept will be it it may be not not difficult to to pin down but it's uh it's it's sort of the step from algebra to analysis it's algebra when you're talking about strict zeros and strict rank analysis is when you get you get things like near zero and near rank then then it's estimates of uh inequalities you're bounding things and that's sort of the difference between those two parts of mathematics so and most people are sort of on one in one side or the other algebraists are just just have that Knack of playing with um with um combinations and Analysis is more about uh if you're given certain bounds on something what else what what bounds do those imply on other things okay I'm I've taken more than an hour of on this lecture and uh let's make this then can we take a break and I'm happy to to uh and and also so I'm going to call this then lecture 32 will be concentrate on Game Theory and um apologize to for for for a work in progress like that but maybe at Lector 31 when I feel I know you all so well uh you forgive me for uh describing something that's incomplete but uh um but connected to what we've been doing okay so uh comput uh okay you're going at this very hard does it have app why why okay why all right fair question right and and um what's my answer well just it's really you're really seeing what a mathematician who sort of on the border of pure and applied what happens I mean uh so so I yes I do hope it has applications and and I guess in this final minute here I was describing applica my my coming back to Applications cuz cuz a a lot of matrices a lot of matrices in in applications are big on the diagonal and and tail off talking being bad limited yes that that's right yeah that that exactly I mean that that's another word that falls into this thing so diagonally dominant is a is a and and people so you meet a lot of diagonally dominant matrices that they make your life easier and um because somehow they're they're better matrices to deal with they might they're probably positive definite because that diagonal is so tough that it that it uh makes a thing positive definite but then you want to say why what what what's the consequence in the inverse yeah so so maybe at this point I'm thinking of applied here but over here it's pure fun right I mean yeah who knows there could be an application but I don't require that to get excited about it that's that's the that's the reality I just require a puzzle that I can't see the way the end to yeah all right good okay