 So, now in this module we shall see how we can go about finding that unique monic generator to say that it exists and all is one thing, but if you actually want to find out let us say even for a test case a 3 by 3 matrix it may be quite another might not be very happy with just the mere existence you should ask for what is the constructive way to go about like at least even though we did not delve into determinants much at least you know that this other object this characteristic polynomial there is a way to get around and find out what it is take the determinant of a minus x times I yeah it is a monic polynomial and that is going to give you the characteristic polynomial, but this minimal polynomial we know it is an object that exists somewhere up there how do you get around to finding what this is okay so that is going to be very crucial okay so that is our next goal as it turns out the way I have laid out the foundations you might think this is a huge degree polynomial right n squared plus 1 we might have to go up to because at least that is when we went up to n squared plus 1 that is when we assured ourselves that such a thing actually exists but it turns out you have to do no such thing so a constructive way would be thus I am going to define another ideal as if this was not enough already so I am going to define another ideal and that is why I actually erase that subscript earlier and you will see now so suppose vi is a vector sitting inside the vector space v and a is an operator mapping from v to itself I am going to define the annihilating ideal of a with respect to vi so that is a subscript as what this is no longer going to be related to an operator but of course it is still an ideal of set of polynomials such that fx sitting inside this commutative ring of polynomials is such that whenever f of a acts on vi so f of a is an operator yeah I am not asking for f of a to be 0 unlike in the earlier case but I am saying whenever f of a acts on this given vector vi that is why this vi is to be specified a priority then it belongs to so this is not the 0 operator anymore remember this is a 0 sitting inside v yeah so this 0 is sitting inside v so this is again I am saying this is an annihilating ideal of a except that I am not just adding a few more terms with respect to vi okay let me just write it annihilating ideal of a with respect to vi so what is the end game here of course I must convince you first that this is indeed an ideal are you convinced this is an ideal take f1 and f2 belonging to the annihilating ideal of a with respect to vi let us take f1 plus f2 acting on a acting on this vector vi that is nothing but f1 a acting on vi plus f2 of a acting on vi so each of them individually gives you the 0 vector so of course the first property of an ideal is satisfied okay second property f belongs to the annihilating ideal of a with respect to vi and g is just a polynomial see the good thing about all of this is that any power of a commutes with itself that is the one of the most fundamental observations so I do not care about whether it is fa into ga or ga into fa you see the point so what can I say about f times g of a this is fa into ga yeah so let us let this act on vi so this is on vi and then I can tweak the order so I can just say this ga times fa times vi yeah which of course by virtue of this term going to 0 it is a 0 vector so 0 vector when acted upon by any operator has to be 0 so this goes to 0 therefore the whole thing goes to 0 so it is an ideal right what do you think is going to be the largest degree of any polynomial in this I mean again is this a meaningless pursuit or of course you know that n squared plus 1 is ideally 0 so it nullifies any vector so for vi also but when I specify this vi my hope is that I should be able to do better my search should be somewhat easier is that unfounded why am I suddenly looking for this annihilating ideal of a with respect to vi instead of the general annihilating ideal of a I somehow guess that this should make my job easier and why and how it should make my job easier is the question that I am posing now so for that you need to look at what what do you think i times vi a times vi until a to the n squared do I need that what are the sizes of these vectors these are vectors in v so I need go no further than this this is how many objects n plus 1 and sitting inside each of these objects is in v so they are sitting inside an n dimensional vector space and they are n plus 1 in number so they cannot be linearly independent very basic ideas we have developed early on in this course so therefore again there exists alpha 0 alpha 1 to alpha n such that alpha 0 vi plus alpha 1 a vi plus dot dot dot till alpha n a to the n vi is equal to 0 so of course these are not all 0 clear on this so far okay next what does this mean just like the earlier case that means we can say that alpha 0 plus alpha 1 x plus dot dot dot till alpha n x to the n definitely belongs to the annihilating ideal of a with respect to vi is that clear just give you some time to absorb this look at the definition of the annihilating ideal of a with respect to vi it does not have to mean that the operator itself is 0 but that whichever vi you have it must be in the kernel of f of a but this is precisely telling you that if everything else fails at least this will always work yes any question why a to the these are objects inside v so this is the collection of n plus 1 vectors sitting inside the n dimensional vector space v so they must be linearly independent dependent so they linearly dependent then there must exist a non-trivial linear combination such that it vanishes so then I pull out the v as you said rightly I pull out the v and then I am left with a polynomial in a in whose kernel vi belongs therefore that polynomial definitely is a candidate sitting inside the annihilating ideal of vi right if not anything else at least that fellow is sitting inside there yeah so that is going to be my member inside the annihilating ideal of a with respect to vi provided we have found this excuse me how does this help us in narrowing our search for overall minimal polynomial remember that some minimal polynomial is what we wanted to wrap our heads around as I said we know how to find a characteristic polynomial we know only in theory that a minimal polynomial exists but how to actually get it we do not know we do not have a clue right so how do you do it how do we get around to that business let us see here is the idea consider a basis for v given by b is equal to v1 v2 vn straightforward enough up until this point look at the annihilating ideal of a with respect to vi for i going from 1 to till n okay so for the annihilating ideal of a with respect to v1 you get a polynomial like this let us call it p1x or let us okay let us just call it by a rather cumbersome thing okay for annihilating ideal of a with respect to v2 again you carry out this similar operation and you get this polynomial mu sorry this is one right mu v2 of x likewise you carry this out for all of those members in that basis set what have we now got we have got potentially n polynomials one generated from each of these or one obtained from each of these annihilating I should not use the term generated so loosely so one obtained from each of these annihilating ideals with respect to members in the basis not the overall annihilating ideal ideal right but the annihilating ideal with respect to vi is this clear now I am going to define something called the least common multiple or the LCM and you will see it follows exactly the same way as we defined LCM of integers these are after all just commutative rings exactly like integers so polynomials also can have LCM so the definition of an LCM and why is it necessary probably I have already guessed what we will be claiming is that the minimal polynomial is going to be nothing but the least common multiple of these fellows that I have tabulated here okay and that is going to be the central idea okay so what do we do let us say the definition for a set of polynomials g1x till gkx this is g2x gkx gx is said to be the LCM if two things first it is of course a common multiple which means that gi divides g for all i of course i going from one through k i going from one to k that is not enough right that is just a definition of a common multiple if it has to be the least common multiple just transfer your thought process from magnitudes of numbers to degrees of polynomials what do we say so we do not go for degrees explicitly but you know how the division is carried out here for any h here such that gi divides h what do we have we must have g also must divide h so anytime and you find go and find any other common multiple this is the least common multiple so this least common multiple must divide that other common multiple any other common multiple yeah g must divide h so that is the definition of the least common multiple of a set of integer sorry a set of polynomials I keep saying integers because that is so similar right in algebra we do not make a distinction no it is not really very complex I mean if you write in terms of degree how do you have to say this you would have to say for any h of a degree greater than g such that it is also a common multiple how many words are you spending there on the other hand this for any other h okay of course this h is also so it is a given that these are all polynomials so that goes without saying writing it a little loosely here okay so first is it is a common multiple and second is it is a least common multiple now the claim is and this is a big claim here so you see this I hope this definition is okay so I can erase this I do not want to really erase that part because that is more important it is okay so the claim is that the minimal polynomial okay how have I denoted this now okay oops what did I write I did I write a here oh that lands me in trouble did I write a in subscript oh great thank you so the minimal polynomial of a maybe I should also write somewhere this a should come here right double subscript perhaps okay yeah that makes sense so the minimal polynomial is nothing but the LCM of mu a v1 mu a v2 until mu a vn if this is true then I have actually outlined a constructive process for you to go ahead and find the minimal polynomial okay this is just a statement we have not proved anything yet so I will take another module a short module in which we will just prove this okay and then we will be done okay.