 See we are discussing diagonalizability in today's lecture we will derive one necessary and sufficient condition for diagonalizability of a linear operator. So let us go back and look at this problem diagonalizability of a linear operator T this means that I can write down the matrix of T relative to some basis as a diagonal matrix okay in writing down this diagonal matrix let me assume that by the way is it clear that the eigenvalues of this diagonal matrix must be the eigenvalues of T the eigenvalues of the diagonal matrix so what is the meaning of saying that T is diagonal this means T ui so I am just writing down the equation that I gave yesterday instead of alpha let me write lambda i ui these are the numbers that will figure here these lambda i's are the numbers that will figure here okay. Now some of we have seen example the second example where the eigenvalue repeats so some of these could repeat so what I will do is write down the distinct eigenvalues and take care of multiplicity when I write down the diagonal matrix in other words let me say lambda 1, lambda 2, etc lambda k be the distinct eigenvalues be the distinct eigenvalues of T with multiplicities I will use this notation different from what I used yesterday in the last lecture with multiplicities let me say D 1, D 2, etc D k that means lambda 1 comes as an eigenvalue of the operator T D 1 times in other words the characteristic equation that is characteristic polynomial is determinant of T minus lambda i is 0 characteristic polynomial is determinant of T minus lambda i characteristic equation is determinant of T minus lambda i equal to 0 this is a polynomial of degree n when I write down this what I mean is that lambda 1 appears D 1 times as a root of that characteristic equation lambda 2 appears D 2 times etc lambda k appears D k times okay so what is clear is that D 1 plus D 2 etc plus D k is equal to n the degree of the polynomial okay if this is the case then this can be rewritten then I can write this matrix diagonal matrix as lambda 1 now appears D 1 times so the first block will have lambda 1 appearing D 1 times is it clear that I can write it as lambda 1 i 1 lambda 2 i 2 etc lambda k i k all other entries are of course 0 with the convention that i 1 is the identity matrix of order D 1 cross D 1 i 2 is identity matrix of order D 2 cross D 2 etc i k is identity matrix of order D k okay so let me confirm where i i is the identity matrix of order D i so I can first this is the first observation the diagonal matrix D can be written in this manner after listing the distinct Eigen values so let us observe that the statement that I made just now D 1 plus D 2 plus etc D k must be the dimension of the space that I started with I will always assume the dimension to be n okay in the rest of the discussion okay if this is the case what is the characteristic polynomial of T the explicit formula for the matrix of T has been written down we know that the characteristic polynomial does not change when I change the basis okay because it is basically the determinant of some matrix which does not change under change of basis okay so the characteristic polynomial so I am using P for the characteristic polynomial I will use P here after for the characteristic polynomial so the characteristic polynomial of this operator T can be written as it will be lambda will be a variable lambda 1 lambda 2 etc are the zeros so I have lambda minus lambda 1 to the D 1 into lambda minus lambda 2 D 2 etc lambda minus lambda k to the D k is it clear that this is the characteristic polynomial of T okay see in general this cannot be done I am assuming that if T I am assuming that T is diagonalizable in that case I can do this for example look at the operator first operator that we considered yesterday the rotation matrix when the angle is 19 the characteristic polynomial there is lambda square plus 1 okay and lambda square plus 1 cannot be factorized as lambda minus lambda 1 into lambda minus lambda 2 for lambda 1 lambda 2 real because this is an irreducible polynomial over the polynomial ring f x R x okay irreducible polynomial over R lambda square plus 1 is one example we cannot factorize there that is an example of an operator which is not diagonalizable so this cannot be done always if it is diagonalizable then this can be done if the operator is diagonalizable then the characteristic polynomial can be factorized into products of powers of linear factors let me mention there is a product of powers of linear factors okay okay so this is one information that the characteristic polynomial is a product of powers of linear factors is one information I also have the other information look at look at the eigens look at the dimension of the subspace null space of T minus lambda I I I want to calculate the dimension of the subspace null space of T minus lambda I I I I want to calculate the nullity of T minus lambda I okay can you tell me what this is what is this subspace first see I want T minus lambda I I I have T here I can treat like matrices for being specific let us take lambda I to be lambda 1 I is 1 lambda 1 T minus lambda 1 I this first block is 0 all the other entries will remain why because they are distinct this will be lambda 2 minus lambda 1 times I 2 this will be lambda k minus lambda 1 times I k I am doing T minus lambda 1 I so these entries will not be 0 distinct eigenvalues so lambda 1 will not be equal to lambda 2 etc lambda k but this block is 0 I want the set of all solutions of the matrix equation let us say some some A x equal to 0 where A is a matrix whose first block is 0 all other diagonal entries are not 0 what is the dimension of the solution space is it clear it is D I the rank of T minus lambda 1 I will be the rank coming out of these non-zero entries so nullity corresponds to just this so null the nullity of T minus lambda I is D I please check this so what is it clear first the null space has this dimension by the way what is the null space of T minus lambda 1 I can you see that it is the it is the eigen space corresponding to the eigenvalue lambda I this I did not mention in the last lecture this is the set of actually the space the set of all eigenvectors of T corresponding to the eigenvalue lambda I this is the set of all eigenvectors of T corresponding to the eigenvalue lambda I first observation this is a subspace I did not make this point yesterday the set of all eigenvectors corresponding to an eigenvalue forms a subspace because it is a null space of a certain linear transformation null space of T minus lambda I so this subspace has dimension D I remember there are there are two objects coming here one object is a polynomial the other one is the dimension of some subspace if T is diagonalizable these two numbers are the same this is an important observation because we will see that you can go back to example 2 example 3 no example 2 we discussed only 2 examples you can go back to example 2 look at the dimension of the subspace corresponding to the eigenvalue 1 the dimension is 1 there whereas the multiplicity of 1 as an eigenvalue is 2 in general these are different okay for diagonalizability to crucial that these two numbers are the same okay this is the second important point first point is that the characteristic polynomial is can be written as a product of powers of linear factors second fact is that the dimension the number of times lambda I appears as an eigenvalue of A that is D I that number is the same as the dimension of the eigen space corresponding to the eigenvalue lambda I okay okay let us proceed this is important okay as you will see this is important for diagonalizability okay as I told you I want to look at a characterization let me give you 1 or 2 results before I prove that result first I want to demonstrate the following suppose T is an operator on a finite dimensional vector space V and F of lambda okay F of T I will use small t okay let me just say F be a polynomial over F V is a vector space over F F is a polynomial over F by which I mean that this polynomial has its coefficients coming from F okay so if F is R this is a real polynomial if T X equals lambda X for lambda in the underlying field then F T of X is F lambda of X we will need this little result then F T of X is F lambda of X what this means is that if lambda is an eigenvalue for the operator T and if X is eigenvector then look at F of T F of T is also a linear operator see little f is a polynomial F of T is another linear operator for this linear operator we want to show that F of lambda is an eigenvalue with the same eigenvector X with the same eigenvector X okay prove straight forward you just write down F of T let us say it is A naught plus A 1 T plus A 2 T square etc let us say A S T to the S where the coefficients come from the underlying field then what is F of capital T wherever little T comes I must replace it by capital T the first term it is T power 0 that is identity operator okay so my F of T is this polynomial A naught I plus A 1 capital T plus A 2 capital T square etc A S T to the S remember T is an operator on V it is a linear transformation from V to itself so this T square T cube etc makes sense composition this is F of T what do I want to verify verify this so let me write F T X this will be A naught I X plus A 1 T X plus A 2 T square X plus etc I need to calculate each term okay but it is easy to see that T okay let us calculate T square X for example T square X is T of T of X T of X is lambda X so this is T of lambda X this is lambda T of X again T of X equals lambda X so this is lambda square X so T square X is lambda square X by induction T power R X is equal to lambda power R X so this can be rewritten okay so this is a simple result so if you look at F T X it is A naught X plus A 1 lambda X plus A 2 lambda square X plus etc plus A S lambda to the S X X is a vector take that outside all the others are numbers coefficients A naught plus A 1 lambda plus etc plus A S lambda to the S into X this is a this is a number coming from the field but this is precisely F of lambda instead of T I have lambda so this is F of lambda X okay so remember that F T is a polynomial in T if lambda is an eigenvalue and X is a corresponding eigenvector then we have found out an eigenvalue for F of T that is F lambda for the same eigenvector X okay okay so that is a simple computation I want one more result before I prove the main theorem so the framework is the same as before lambda 1 lambda etc lambda k are the distinct eigenvalues of a linear operator T on a finite dimensional vector space V let me say that W I let W I be the eigenspace corresponding to the eigenvalue lambda I so I have these k eigenspaces corresponding to the eigenvalue lambda 1 etc lambda k now these eigenspaces are subspaces I can talk about the sum of these subspaces let W equal to W 1 plus W 2 etc plus W k I take the sum of these k eigenspaces now remember that dimension of W 1 plus W 2 in general is not equal to dimension W 1 plus dimension W 2 you need to remove the dimension of W 1 intersection W 2 okay but in this case the dimensions add up in this case the dimension will add up that happens if the subspaces are eigen space that is what we are trying to do here so what is the conclusion then dimension W is dimension W 1 plus dimension W 2 etc in particular what this means is that eigenvectors corresponding to distinct eigenvalues are linearly independent okay this is this statement is encoded in this eigenvectors corresponding to distinct eigenvalues are linearly independent okay so let us prove this then we will use this in characterizing diagonalizability all that I will do is take a basis for W 1 take a basis for W 2 etc basis for W k show that the union is a basis for the sum W is that okay I will take a basis for W 1 base for W 2 etc W k take the union I will show that that is a basis for W then it follows that dimension of W is the number of elements in B 1 plus the number of elements in B 2 etc number of elements in B k that is precisely this okay so let us write down now basis explicitly let B 1 equal for the subspace W 1 corresponding to lambda 1 so B 1 I will call it U 1 1 U 1 2 etc U 1 L 1 the first subscript denotes that it corresponds to that eigenvalue first subscript denotes it corresponds to the eigenvalue lambda 1 okay so this corresponds to eigenvalue lambda 1 B 2 is U 2 1 U 2 2 etc U 2 L 2 this corresponds to lambda 2 etc B k U k 1 U k 2 etc U k L k this corresponds to the eigenvalue lambda k let these be basis ordered basis for W 1 W 2 etc W k I must show that I must show that this is a basis for W okay spanning set we will disposed of quickly linear independence is what is a little difficult in this problem little more involved than the other one spanning set I want to show that this union this is a spanning set for W okay take anything in W little W that is W 1 plus W 2 etc W k each W 1 in turn can be written in terms of these so it is clear that this is a spanning set clearly this is this spans W linear independence is see to show that this is a basis for W I must show that it is a spanning set and it is a linearly independent set the fact that this is a spanning set is straight forward for the following reason take any little W in W then by definition that W is some W 1 plus W 2 etc W k but look at W 1 the first term that is a linear combination of these etc so this W is a linear combination of these vectors that is those vectors in script B and so this spans W linear independence consider so we need to show linear independence so we need to consider a combination so consider alpha 1 1 U 1 1 plus alpha 1 2 U 1 2 I choose the scalars also according to the superscripts etc alpha 1 L 1 U 1 L 1 plus alpha 2 1 U 2 1 alpha 2 2 U 2 2 plus etc plus alpha 2 L 2 U 2 L 2 plus etc last term alpha k 1 U k 1 alpha k 2 U k 2 etc alpha k what is that last one L k U k L k suppose this is 0 I must show that each of the scalars is 0 okay consider this combination I must show that each scalar is 0 it would then follow that these vectors U 1 1 U 1 2 etc U 1 L 1 U 2 1 U 2 2 etc U 2 L 2 U k 1 U k 2 etc U k L k they are linearly independent okay let us this is this vector is a 0 vector f of t easily take any polynomial for any polynomial f of t I will consider f of capital T f of capital T is also linear I will apply f of capital T on this of this whole thing okay alpha 1 1 U 1 1 plus alpha 1 2 U 1 2 etc alpha 1 L 1 U 1 L 1 this time I will just write down okay does not matter alpha 2 1 U 2 1 etc alpha 2 L 2 U 2 L 2 plus etc the last one the last one is alpha k 1 U k 1 plus etc plus alpha k L k U k L k f of t is also a linear operator so any linear operator has a property that it is action on the 0 vector this is a 0 vector so 0 is this f of t is linear because t is linear I will apply this to each term okay so can you see that the first term will be alpha 1 1 f t U 1 1 use the previous lemma it is f of remember U 1 1 is an eigenvector corresponding to the eigenvalue lambda 1 all these U 1 1 U 1 2 etc U 1 L 1 see they come from they come from the eigen space corresponding to the eigenvalue lambda 1 so each of this is an eigenvector corresponding to the eigenvalue lambda 1 apply the previous lemma f t of x equals f of lambda x if lambda is the eigenvalue so the first set of terms is it clear that it is alpha 1 1 f lambda 1 U 1 1 plus f lambda 1 U 1 2 etc U 1 L 1 first terms will go with f lambda 1 plus the second ones will go with f lambda 2 the last one will go with f lambda k do you agree with this I forgot the constants here also okay is it clear the first set of terms coming from this bracket the first one yeah the first set of terms here they will go with f lambda 1 because each of those vectors U 1 1 U 1 2 etc U 1 L 1 each of those vectors is an eigenvector corresponding to the eigenvalue lambda 1 and I am appealing to the previous lemma okay this is true for any polynomial f this is true for any polynomial f I will make now particular choices of f and then show that this is 0 another choice I will show this is 0 etc okay suppose I have one choice for which the second set of terms third set of terms etc the last set of terms vanish this only remains does it follow that those those scalars those scalars will be 0 okay I will choose that polynomial in such a way that this is not 0 so it will follow that alpha 1 1 alpha 1 2 etc alpha 1 L 1 is 0 apply the next polynomial I will choose it in such a way that f lambda 2 is not 0 it will follow that these coefficients are 0 etc I will go back and substitute into this equation I am sorry I want to conclude each scalar is 0 so it will follow that these are independent okay now what is that polynomial for the first one I will do it for the first one the rest is similar for the first one I will choose f of t to be say I want these to be 0 I want a polynomial I want the polynomial to have lambda 2 etc lambda k to be 0 give me one choice t minus lambda 2 t minus lambda 3 etc if you choose this then f of lambda 1 f of lambda 1 is lambda 1 minus lambda 2 lambda 1 minus lambda 3 etc product lambda minus lambda k these are distinct so that is not 0 okay so f of lambda 1 is not 0 f of lambda 1 is not 0 but f of lambda i equal to 0 for all i greater than or equal to 2. So I will take this polynomial apply it to this equation then I get okay I take this polynomial apply this polynomial to this equation then I get the second set of terms are 0 etc I have only these terms remaining I can write it as f lambda 1 outside into the rest of them alpha 1 1 u 1 1 etc alpha 1 k l 1 u 1 l only the first set of terms will remain f lambda 1 is not 0 so this can be cancelled look at the rest that is 0 but these are linearly independent right they form a basis for w 1 these are linearly independent from this it follows that the first set of coefficients alpha 1 1 alpha 1 2 etc alpha 1 l 1 first set of coefficients must all be 0 so you apply the second polynomial which is t minus lambda 1 into t minus lambda 3 t minus lambda 4 etc t minus lambda k then you can show the second set of coefficients 0 etc so it follows that each of these scalars started with this equation follows that each of the scalars is 0 and so these vectors are linearly independent so this is the basis for the space w okay as I mentioned earlier in particular this means that Eigen vectors corresponding to distinct Eigen values are linearly independent we have proved something more okay let me prove this theorem then theorem that characterizes diagonalizability the framework is as before t is a linear operator on a finite dimensional vector space with dimension n over f v is defined over f I have lambda 1 lambda 2 etc lambda k as the distinct Eigen values let w i be the Eigen space corresponding to the Eigen value lambda i then the following statements are equivalent I will give two conditions that are necessary and sufficient for t to be diagonalizable first statement is t is diagonalizable second is a condition involving the representation of the characteristic polynomial characteristic polynomial I am denoting by p I will write p of lambda it is lambda minus lambda 1 to the d 1 lambda minus lambda 2 to the d 2 etc lambda minus lambda k to the d k where d 1 plus d 2 plus etc plus d k equals n I am assuming that n is a dimension of the space v the last condition is in terms of sums of the Eigen spaces if you look at the sums of the Eigen spaces call it w then this w the subspace is the whole of the space v this is the last condition we have already observed that a implies b if t is diagonalizable then we have seen that the characteristic polynomial has this form from b implies c proof a implies b already observed b implies c follows because okay you can set w to be the sum of these subspaces then by the previous lemma it follows that dimension of w is summation i equals 1 to n dimension w i this comes from the previous lemma because these are Eigen spaces corresponding to distinct Eigen values but what is the condition on look at condition b the condition on these numbers d 1 d 2 etc d k is that their sum is n so dimension w is summation i equals 1 to n d i that is n so I was 1 to k there are k subspaces yeah so this sum is n so but we know that the sum of two subspaces again a subspace w is a subspace of v having the same dimension so w must be the whole of v that statement c the sum is equal to w w is equal to v we have shown so b implies c also holds c implies a c implies a follows from the definition of what we mean by diagonalizability is that clear I want to show t is diagonalizable that is I want to show that there is a basis script b of v such that each vector from b is an Eigen vector of t that is diagonalizability we saw this yesterday t is diagonalizable if and only if there is a basis b for v each of whose vector is an Eigen vector for t I know that v is the direct sum of is a sum of these subspaces look at the construction that we did earlier take a basis b 1 for w 1 b 2 for w 2 etc b k for w k each of the basis has a property that each of these basis for the subspaces b 1 b 2 etc b k have the property that their elements the vectors in b 1 for instance is an Eigen vector corresponding to lambda 1 the vectors in b 2 are Eigen vectors corresponding to the Eigen value lambda 2 etc the combination is a basis for v the union of these basis for the subspaces is a basis for v and so each vector of this basis is an Eigen vector of v is an Eigen vector of t so t is diagonalizable so can I just say c implies a follows from really the definition which we saw yesterday really the definition that there is a basis b for v each of whose vectors is an Eigen vector for t okay so this is one characterization that is necessary sufficient condition for t to be diagonalizable one is that the characteristic polynomial can be written as a product of powers of linear factors second is that the whole space v can be written as a sum of these subspaces the subspaces being the Eigen spaces okay this is one characterization we will also look at another characterization involving the so called minimal polynomials okay that I will do in the next one or two lectures but before I conclude I want to at least mention what is a minimal polynomial we want to know when precisely a linear transformation is diagonalizable okay one answer has been given here two answers really look at the Eigen spaces take the sum verify if that is the whole space v the other thing is look at the characteristic polynomial verify if it is a product of powers of linear factors then you know that t is diagonalizable there is another answer as I told you which comes in terms of the minimal polynomial okay so let me at least give the definition of the minimal polynomial remember that we are looking at an operator a single linear operator t and we are analyzing the operator t in this study what is important is to identify classes of polynomials which have the property identify classes of polynomials let us say f of t such that f of capital T is 0 what are all polynomials f of t such that f of capital T is 0 now why is this statement true that will be clear only a little later we will get the connection between minimal polynomial and the characteristic polynomial then it will be clear as to why these polynomials are important okay but let me at least give the concept of the minimal polynomial coming from this okay remember f is a polynomial its coefficients come from the underlying field okay now if you look at f t call that script a look at f t call that script a then see what is this f t f t is a set of all it is it has an algebraic structure it is a set of all polynomials in a single variable t in a single real variable t I should actually write f but for the sake of convenience I am writing r single real variable t so I should actually write r of t okay let me go back and change this to f so it is either for our discussion let us say it is a set of all real polynomials over a single variable t single real variable t the coefficients are real the variable t is also real I am using script a for that it is what is called as an algebra you know that it is a Euclidean domain set of all polynomials it is a Euclidean domain it is a commutative ring where you can do Euclidean algorithm it is a commutative ring where Euclidean algorithm can be applied it is an Euclidean domain which has a property that you know the concept of an ideal concept of an ideal in a sub ring okay does not matter if you do not know you will learn it this semester sometime what can be shown is that an ideal is a sub ring of see this is what is called as an algebra an algebra is an algebraic structure where you can do multiplication of vectors okay so an algebra is something more than a vector space where there is also a possibility of multiplying vectors now multiplication here is multiplication of polynomials multiplication of polynomials you know term by term multiplication of polynomials is term by term so one could do multiplication of polynomials it also has one or two little axioms but let us not worry about that this is an algebra this is a Euclidean domain where I can do where product of vectors make sense in such a Euclidean domain the notion of an ideal comes an ideal is a sub ring an ideal is a sub ring which has a property that if I take an element from the ideal and an element from outside the ideal that is f of t then the product will belong to the ideal okay then the product will belong to the ideal ideals in a Euclidean domain have the property that they are generated by a single unique polynomial ideals in a Euclidean domain are characterized by the property that they are generated by a single unique element which means anything in the ideal is a multiple of a specific polynomial anything in an ideal given an ideal anything in that ideal is a multiple of a unique element coming from that ideal okay I need this property one can also do without this property for the minimal polynomial I wanted to find the minimal polynomial first of all the question is given a linear transformation t does there exist a polynomial f such that f of t is 0 okay I will give two answers one for a specific example linear transformation matrices they are equivalent so I will take this matrix 1 1 1 1 I want you to consider this polynomial f of t equals t square minus 2 t f of t is t square minus 2 t how did I get this at the end of the next two lectures you will also be able to write down such polynomials t square minus 2 t then you can verify that f of a is a 0 matrix a square equal to 2 a a square is equal to 2 a for this matrix that is the reason why I chose t square minus 2 t then f of a equal to 0 so given a linear transformation this makes sense the question does there exist a polynomial f such that f t equal to 0 makes sense I have given one example I will actually prove it I will actually prove that given a linear transformation on a finite dimensional vector space there is a polynomial which has a property that f of capital T equal to 0 tomorrow and then maybe define the minimal polynomial and how it is related to the notion of diagonalizability okay let me stop here.