 We are discussing the notion of minimal polynomial right and see the objective as I stated before is to characterize diagonalizability. We already have one characterization you compute the Eigen values compute the dimensions of the Eigen spaces if the dimensions add up to the dimension of the vector space then it is diagonalizable okay dimensions of the Eigen spaces add up to the dimension of the vector space then it is diagonalizable we will look at another characterization in terms of the minimal polynomial okay this is another cornerstone one of the cornerstones in linear algebra that we have seen is a rank quantity dimension theorem okay. Now to prove this theorem you need certain notions today we will discuss two of them one is the invariant subspace invariant subspace of a linear transformation and then the notion of a T conductor okay so let me make these notions precise so I will first discuss the notion of an invariant subspace. T is a linear operator on V a subspace W is called an invariant subspace of the transformation T of the operator T if this condition holds T W is contained in W that is an invariant subspace okay. That is a meaning that is for all x element of W it must follow the T x belongs to W okay let us dispose of the trivial examples of subspaces W equal to singleton 0 and W equal to the whole space they are obviously invariant so we will be interested if there are proper subspaces that are invariant under T okay and you will see how this plays a role in characterizing diagonalizability. Let us look at some examples may be before examples before numerical examples let me also tell you that the range space and the null space are invariant under T these are easy to verify so I will leave the proof for you the range space and the null space are invariant under T okay we also have another set of subspaces that are invariant under T let S be a linear operator on V such that S T equals T S S is an operator that commutes with T suppose S is an operator that commutes with T then the range and the null space of S are also invariant under T okay remember for us T is fixed for us T is fixed S is an operator that commutes with T then range and null space are invariant under T all these are little exercises for you okay to verify. In particular what is relevant to the present context is the eigen space corresponding to any eigen value these are invariant subspaces for any operator T because of this result look at look at T minus lambda I for any eigen value lambda and this commutes with T T minus lambda I commutes with T so in place of S I have T minus lambda I where lambda is an eigen value so by the previous theorem a previous result which we did not prove null space of T minus lambda I is invariant under T but null space of T minus lambda I is precisely the eigen space of the operator T corresponding to the eigen value lambda so eigen spaces are invariant under T eigen spaces of an operator T are invariant under T okay this is the most important consequence for us okay let us now look at one example where the operator does not have an invariant subspace if you want operators for which there are invariant subspaces just look at any operator which has an eigen value this theorem says the eigen spaces are invariant okay so I want to give an example of an operator which does not have an invariant subspace this goes back to the example that we have seen before okay I want you to look at T from R 2 to R 2 whose matrix is okay I am really looking at the rotation operator so the matrix of this okay I will straight away define T, T is defined by the rotation operator T of X is minus X 2 X 1 this is the rotation by 90 degrees rotating the vector X by 90 degrees this is the linear operator I want to show that this operator does not have any non-trivial invariant subspace okay 0 and R 2 R trivial is invariant subspaces so if I take a subspace which is non-trivial which is invariant under T the subspace must be of dimension 1. Let W be a one dimensional subspace invariant under T I look for a one dimensional subspace and see if it is invariant under T so I am assuming there is one such it is one dimensional so the basis consists of one element so let me take W to be span of a particular vector X star X star is not 0 okay X star is not 0 suppose T W is contained in W if T W is contained in W, T W is in particular for X star I will have T X star also as an element in W T X star should also belong to W but W is span X star so it is a multiple of X star this must be alpha X star for some scalar alpha W is span by X star so anything is a multiple of X star but this means a alpha is an eigen X star is not 0 this means alpha is an eigen value for T but we know that T does not have eigenvalues this is a contradiction so this T does not have an invariant subspace this is a contradiction to what we had seen earlier and so this T has no proper invariant subspace okay so much so for invariant subspaces let us keep this aside and then look at the notion of the T conductor of a vector into a subspace is this example clear this operator does not have a proper invariant subspace we had looked at this example before and we have shown that this operator does not have an eigenvalue if lambda is an eigenvalue we had seen that 1 plus lambda square is 0 but this is an operator on R R 2 to R 2 so the eigenvalue must be real so there is no root for 1 plus lambda square equal 0 so it does not have an eigenvalue if it had an invariant subspace it would have had okay next is a notion which is rather similar to the minimal polynomial okay but little more general than the minimal polynomial that is another notion we need for characterizing diagonalizability of operators the notion is the following I am having a subspace let W be an invariant subspace of the operator T let W be an invariant subspace of an operator T see for us the framework is T is an L of V V is finite dimensional we are trying to characterize diagonalizability okay W is an invariant subspace of T let Y belong to V but not in W the T conductor the T conductor of Y into W that is the name the T conductor of Y into W I will define this to be the following subset of F of T is defined by the notation is S T Y W the T conductor of Y into W W is an invariant subspace Y is an element that does not belong to W T is the operator that we started with S T Y W this is the set of all polynomials G in F T set of all polynomials with coefficients coming from the underlying field for us either real or complex over the single real variable T we know it is a principal ideal domain so collect all those polynomials that satisfy the property that G T Y belongs to W Y does not belong to W but G T Y must belong to W G is a polynomial G of capital T is a linear operator on B so G T Y make sense I collect all those polynomials that satisfy this property for a fixed Y fixed subspace W and T is of course fixed collect all the polynomials that satisfy this property for a fixed Y for a fixed invariant subspace W that will be called the T conductor of Y into W okay this is the collect set of all polynomials I am going to leave it to you to prove that this is an ideal this is a subspace and has a property that if G belongs to this and F belongs to capital F then the product belongs to this is an ideal so it has a generator that is each element each polynomial in this is generated by a unique polynomial that polynomial will also be called the T conductor of Y into W but before that is this non-empty first question we have defined something similar the minimal polynomial is there any relationship between the minimal polynomial and the T conductor of Y into W what is the property of the minimal polynomial one important property it is an annihilating polynomial M T of any X is 0 so this Y also M T of Y is 0 what is the problem the minimal polynomial is an annihilating polynomial so M of capital T of any vector must be 0 W is a subspace so that must belong to W so to begin with the minimal polynomial belongs to this okay but there are more general elements that is the important point here there are more general polynomials but they cannot be annihilating polynomials remember that because if they were annihilating polynomials of T different from M of T their degrees will have to be greater they cannot be smaller there are annihilating polynomials agreed but the degrees of those annihilating polynomials will be greater than the degree of the minimal polynomial minimal polynomial definitely belongs to this so this is not M T okay this is non-empty ideal of the principal domain F T so this is generated by a single unique element that will also be called the T conductor of Y into W okay so I will simply say that the unique we will be interested in the monic generator the unique monic generator of S T Y W will also be called the T conductor of Y into W it is also called the T conductor of Y into W so depending on the context we will know whether we are talking about the subspace or the unique monic polynomial okay see we have seen that the minimal polynomial belongs to this so can you immediately conclude that this T conductor divides the minimal polynomial what is the meaning how does it happen let us denote G let us denote let us denote the T conductor by G that is we have S T of Y W this is generated by the single polynomial G that is what I mean G is the unique monic polynomial which generates this ideal then what we know is that since M belongs to S T Y W it follows that G must divide M G must divide M G must divide M but this G cannot be an annihilator because of G divides M then the degree of G is less than or equal to the degree of M in general this cannot be the this cannot be an annihilating polynomial of the operator T I will give you an example of this G okay but before that let us prove the following result so remember G divides M so this is more general than the minimal polynomial the T conductor of a vector into a subspace is more general than the minimal polynomial of the operator T okay then we have the following result see our problem is to characterize diagonalizability but what we would right now do is to settle with something less is to settle with the triangulizability so let me give the definition an operator T is said to be triangulable is said to be triangulable if there exist a basis script B of V such that the matrix such that the matrix of T relative to B diagonalizable it is a diagonal matrix triangulable if it is a triangular matrix such that this is triangular, triangular means either the lower off diagonal entries are 0 or the upper off diagonal entries are 0 we will stick to the lower being 0 so we will say that T B is it does not make a difference we will say that the lower triangular part is 0 so it is upper triangular an operator T is said to be triangulable if the matrix of T relative to some basis of V is upper triangular let us first characterize upper triangular matrices we will need the following result it tells you something about the structure of G I want to state this lemma before characterizing triangulability this lemma will be useful there T is a linear operator on V and W be a proper invariant subspace of T let lambda 1, lambda 2, etc lambda k be the distinct eigenvalues of T and let us suppose that the minimal polynomial M stands for the minimal polynomial let us suppose the minimal polynomial can be written as a product of powers of linear polynomials that is T minus lambda 1 to the R 1, T minus lambda 2 to the R 2, etc T minus lambda k to the R k the minimal polynomial is a product of powers of linear polynomials then we have the following first there exists x which does not belong to W condition 2 is T minus let us say lambda i x there exists an x that does not belong to the subspace W T minus lambda i x belongs to W for some eigenvalue this is what we will show there exists an x which does not belong to the subspace W but which has the property that T minus lambda i of x belongs to W for some eigenvalue lambda okay we will prove this lemma and then use this to characterize triangulability and then characterize diagonalizability see remember already the minimal polynomial appears here okay so the minimal polynomial will play a crucial role in the diagonalizability which is what we will at least state today the proof is as follows see before that I told you I told you that we will have some information about the polynomial G the information that is provided in this theorem is as follows under the very mild condition that the eigenvalues of the operator T exist under the very mild assumption that the eigenvalues exist if the eigenvalues exist then you can write the minimal polynomial in this form okay so except for operators like the rotation operator this condition is satisfied under this condition what this says is that this T conductor of the vector x into W is a linear polynomial can you see that the T conductor see we are interested in the T conductor of x into W from this can you see that x does not belong to W but T minus lambda i x belongs to W so all you have to do is look at the polynomial little T minus lambda this polynomial has a property that x does not belong to W means the G is not the constant okay G is not the constant polynomial from constant you go to linear polynomials this says that G must be a linear polynomial there exists an x which has a property that the T conductor of x into W is a linear polynomial that is second condition really okay that it is not constant is a first condition okay as I told you I will give a numerical example but let us first look at the proof of this result see W is not equal to V W is proper subspace W is a proper invariant subspace so this is not the whole of V so there exists x in V such that x does not belong to W I will look at let me call it y there exists y such that y does not belong to W I want to now look at let G be the T conductor of y into W W is not the whole space V it is a proper invariant subspace I pick one element y which does not belong to W and then construct in principle the T conductor of y into W I am calling that T conductor as G can you see that G cannot be constant G cannot be a constant why because if G were a constant then what is the property that G satisfies G T y belongs to W okay this property satisfied by G in fact among all those polynomials that satisfy this condition G is the one with the least degree coefficient of the highest degree is 1 etc coefficient of highest degree is 1 this is a unique monic polynomial that satisfies this condition so if G were a constant then this would be a constant constant times y belongs to W means y belongs to W but y is something that we started with does not belong to W so G is not a constant G is not a constant polynomial so it must be T minus lambda J into H of T where the degree of H is strictly less than degree of G so H T y does not belong to W also if you look at T minus lambda J I x it is T minus lambda J I H T y that is G T y which belongs to W coming from this okay G is not a constant so it must be at least a linear polynomial so G is T minus lambda J into some H H could be a constant but H definitely cannot be a constant because otherwise you will get a contradiction here so the degree of H is less than the degree of G and so H T y cannot be in W I am calling H T y as x now this x satisfies the second part x does not belong to W alright but look at T minus lambda J x that is T minus lambda J I into H of T into y which is this operating on y that is G T y G T y belongs to W so we are through okay so this is the x that satisfies these two conditions so what follows is that G is a linear polynomial that is the consequence so to summarize there exists x which does not belong to W such that such that the T conductor of x into W is a linear polynomial okay okay I want to give a quick example you please verify the details yeah yes H T cannot be a constant actually oh we are proving H T is a constant yes we are proving H T is a constant we are proving H T is a constant H T is a constant is also consistent with this statement H T is a constant this is a multiple of y y does not belong to W so this does not belong to W H T is a constant is what we are proving yes okay the T conductor is a linear polynomial. Let us use this in characterizing triangulability okay I told you I will give an example you please verify the details I just have the information this example is that of the second example of an operator which is not diagonalizable I want to look at the operator over R 3 whose matrix is this 3 1 minus 1 2 2 minus 1 2 2 0 the characteristic polynomial of this matrix is we are using P for that P of T is T minus 1 into T minus 2 the whole square that is what I remember T minus 1 into T minus 2 whole square I will call lambda 1 as Eigen value 1 lambda 2 equals lambda 3 equals 2 this is an example of an operator which is not diagonalizable let me take let me take W as W 1 the Eigen space corresponding to the first Eigen value set of all x in V such that R 3 such that A x equals x A x equals lambda x lambda 1 x that is A x equals x so this is the Eigen space corresponding to the Eigen value 1 the first Eigen value remember that in this example we do not have enough Eigen vectors for a second Eigen value the second Eigen value has only one Eigen vector span by 1 0 2 okay the second Eigen value has only one independent Eigen vector the Eigen space corresponding to second Eigen values of dimension 1 I will take W 1 this is an invariant subspace any Eigen space is an invariant subspace of the operator T so this W 1 is invariant I want to give an example of the Y that is constructed in the theorem so let me give this Y as 1 1 2 take this Y what is the T conductor of this Y into W 1 verify it is T minus 2 for this Y please verify it is T minus 2 okay now remember that for in this example we have verified that the minimal polynomial is the same as a characteristic polynomial okay but the T conductor is a linear polynomial so this is something more general than the minimal polynomial okay but it pertains to only a particular vector Y so please verify the details here this should sort of consolidate what we are doing there okay I want to characterize the triangulability it is just one factor no see minimal polynomial is T minus lambda 1 to the yeah here T minus lambda 1 to the R 1 etc G is just one of those factors what is the problem M is a multiple of G G divides M G divides M so M is a multiple of T multiple of G so one factor T minus lambda J for some eigenvalue rest of them are here one more factor no we are proving it is a constant if it is a constant then how it is divided that is my question that constant is 1 yeah the constant is 1 that constant is 1 it is T is 1 pardon we do not know presently we do not know in this proof we do not know what H T is the only thing that I know at this stage see it is not a constant so it could be a linear polynomial quadratic polynomial whatever okay but at this stage I can this right this much there must be a linear factor forget about H but H now has a property that its degree is 1 less than G so H T Y cannot be in W that is the property we are exploiting okay remember G divides M the degree is less than or equal to the degree of M but G in general is not an annihilating polynomial okay so I need to yeah use this lemma to yeah use this lemma to characterize triangulability let us see how far we can proceed so I want to characterize triangulability this is the general theorem an operator tree is triangulable if and only if the minimal polynomial is a product of linear factors an operator tree is triangulable if and only if the minimal polynomial of the operator is a product of linear factors there are possible powers okay but it is a product of linear factors okay so I do not think I have enough time to prove it but I will at least make this observation that if an operate if the eigenvalues of an operator lie in the underlying field then the operator is triangulable if an operator has all eigenvalues in the underlying field then it is triangulable okay but not all operators are diagonalizable okay we have already seen operators examples of operator at least one which is not diagonalizable but all operators are triangulable provided you only meet this minimum condition mild condition that the eigenvalues must belong to the underlying field okay.