 Okay we are discussing the minimal polynomial okay minimal polynomial is a polynomial which has a property it is a polynomial f of t which has the property minimal polynomial is a polynomial f such that f of capital T equal to 0 okay I will give the precise definition but the question is whether there exists such a polynomial okay I had given an example yesterday example of a matrix 2 by 2 matrix and a polynomial so let me recall that I had given this matrix I am looking at the linear transformation whose matrix relative to the standard basis is this linear transformation on R2 and I defined f of t to be t square minus 2 t okay then we notice yesterday that f of a is a 0 matrix okay so at least for this matrix we have shown that there is a polynomial f such that f of a equal to 0 in the general case let me give an argument and then give the precise definition we are seeking a polynomial f which has the property that f of t is 0 given a transformation given an operator t okay in the general case let us look at the following t is a linear operator on a finite dimensional vector space what is the dimension of v as usual it is n what is the dimension of L v n square dimension of L v is n square so if you look at and see L v is the space of all linear operators on v so consider these operators consider the following operators starting from I t t square etc operators on v these are elements of L v etc I go up to t to the n square consider these operators these are in L v so these are elements these are vectors in this vector space these are n square vectors in this vector space n square plus 1 rather they are n square plus 1 vectors in the vector space L of v the dimension of L of v is n square so these must be linearly dependent so there exist scalars not all 0 at least one of them is non-zero not all 0 there exist scalars not all 0 such that scalars I mean alpha I not all 0 such that alpha not I plus alpha 1 t alpha 2 t square etc alpha n square t n square equals to 0 operator these vectors in this space L of v are linearly dependent so there are scalars at least one of which is not 0 all you have to do is pick up the polynomial from this equation define let us say f of t by alpha not plus alpha 1 t alpha 2 t square etc plus alpha n square t to the n square the polynomial coming from this left hand side this polynomial has a property that f of t equals the 0 map okay so we have proved in the general case also for any linear operator on a finite dimensional vector space there is at least one polynomial in fact there are infinitely many there is at least one polynomial f that satisfies f of t equals to 0 operator what we do is look at see as I described yesterday we look at the collection m t this is the set of all polynomials f so I am using f in R t for instance set of all polynomials f with real coefficients over the real variable t such that f of t equals 0 operator what we have just now shown is that this is not empty for any operator this set is not empty this is a subset of R t this is a sub ring this also has a property that it is an ideal and this is an ideal in R t R t is a nucleotid domain it is also called a principal ideal domain f x over the variable x is a principal ideal domain principal ideal domain means that every ideal is generated by a unique element so there exists m in R t such that such that this capital M sub t is generated by this m is generated by this m that is every polynomial in m t is a multiple of this little m every polynomial in m t is a multiple of little m what are the properties of this little m little m has a property that it is a monic polynomial it is a monic polynomial the coefficient of the highest degree of m is 1 what is the other property m has a property that m of t is 0 right it comes from this what is the other property that m has degree of m is less than or equal to degree of f whenever f is such that f of t is 0 okay so let me just write down these properties m satisfies the following the first condition is that it must be such that m of t is 0 such a polynomial is called an annihilating polynomial so I will just write down on this side annihilating polynomials it annihilates t it destroys t that is it takes t to 0 this is called an annihilating polynomial so m of t any polynomial f that satisfies f of t equal to 0 is called an annihilating polynomial in particular m of m must be an annihilating polynomial second property m is monic monic means the coefficient of the highest degree is 1 remember if the coefficient of the highest degree is not 1 you can always divide by that number the coefficient and you will get it so m is monic that is another property property 3 we are calling it a minimal polynomial minimal in what sense minimal in the sense of degree if f belongs to R t and f of t equals a 0 polynomial if f belongs to R t and f of t equal to 0 0 operator then so I told you that m t is generated by little m which means I also told you that anything in m t is a multiple of this little m okay so I have this condition degree of m is less than or equal to degree of f I must have this condition satisfied by m for example f could be a multiple of m in which case the degree of f and degree m are the same okay but it cannot be less than the degree of m any annihilating polynomial must have degree less it cannot have degree strictly less than the degree of the minimal polynomial okay so this m is called the minimal polynomial such an m is called the minimal polynomial the minimal polynomial for t for the operator t it depends on t different operators will have different minimal polynomials okay in the worst case when you do not know what an algebra is what a sub ring or what an ideal is you can take it for granted that this definition is sensible there is a minimal polynomial for each operator okay let us look at some examples but maybe before that I need to tell you the relationship between the minimal polynomial so we are discussing eigenvalues eigenvectors what is the relationship between the minimal polynomial and the eigenvalues let us first settle this so let me just prove the following following result t is an L v and v is finite dimensional and so I will always reserve this little m for the minimal polynomial for an operator and the characteristic polynomial I will reserve p okay so m for me will always be the minimal polynomial for t and p will be the characteristic polynomial p will be the characteristic polynomial any other annihilating polynomial I will use f the zeros of the minimal polynomial and the characteristic polynomial for an operator p are the same okay this is the connection the connection between the minimal polynomial and the characteristic polynomial is that they have the same zeros means that is m of lambda equals 0 if and only if p of lambda equals 0 if lambda is a root of the equation m of t equals 0 then lambda is also a root of the equation p of t equals 0 okay how do you prove it can you see that there is a connection now between p is the characteristic polynomial so any root of the equation p t equal to 0 lambda is an eigenvalue so the zeros of the minimal polynomial are the eigenvalues of the operator proof we need to show that let lambda be such that T x equals lambda x for some x not equal to 0 that is lambda is an eigenvalue of t x is a corresponding eigenvector okay we are actually proving the sufficiency part first yeah if lambda is an eigenvalue then I am showing that lambda is 0 of the minimal polynomial okay is this okay m t x equals 0 do you agree with this first m is a minimal polynomial so it is an annihilating polynomial so for any x m t x is 0 in particular for this eigenvector 0 is m t x but you remember this result that we proved m t x if lambda is an eigenvalue this can be written as m lambda x T x equals lambda x we saw this result in the last lecture this is m lambda into x m lambda is a number x is a vector x is an eigenvector so it is not 0 m lambda must be 0 so one part is very simple if lambda satisfies p lambda equal to 0 then we have shown m lambda is 0 okay we have only used the fact that m t x is m lambda x conversely let us suppose that m lambda is 0 we must produce a vector we must show that there exists a vector let us say y such that T y equals lambda we must show that there exists a vector y not equal to 0 such that T y equals lambda y it would then mean that lambda is an eigenvalue and so p of lambda is 0 okay m lambda equal to 0 means that if I use T as the variable for the polynomial m do you agree that I can write m of T as T minus lambda into q of T if lambda is a 0 of m of T equal to 0 then T minus lambda is a factor of m of T T minus lambda is a factor of m of T this is this is by definition what do you know about q of T the degree of q is at least one is precisely one less than the degree of m so q cannot be an annihilating polynomial where q cannot be an annihilating polynomial that is q of capital T cannot be the 0 polynomial the reason since degree of q is strictly less than degree of m and any polynomial of degree less than the minimal degree of the minimal polynomial cannot be an annihilating polynomial so q of T is not the 0 operator q of T not the 0 operator means that there is one nonzero at least one nonzero vector x such that q of T x is not 0 there exists x not equal to 0 such that q of T x is not 0 that is the meaning of saying that an operator is not 0 this is this will be my eigenvector so I will call this y call y as q T x then for one thing y is not 0 also I will start with this 0 m T x that must be 0 because m T is a minimal polynomial 0 equal to m T x this is written as T minus lambda i times the polynomial q into x this is T minus lambda i q T x is y into y so what I have is what I have is T y equals lambda y with y not equal to 0 so I have shown that y is an eigenvector corresponding to the eigenvalue lambda in terms of P this means P of lambda is 0 okay now this is an important connection between the minimal polynomial and the characteristic polynomial they have the same roots except for multiplicities okay remember that we have not shown we have not shown that the multiplicities are the same okay but they have the same roots this is an important point maybe we should now look at examples okay some numerical examples my first example will be one that exemplifies this diagonalizability I realized I have not given an example so for diagonalizability let me first do an example and also consider the examples that we have discussed earlier we have discussed at least two examples the first example of an operator which does not have an eigenvalue so it is not diagonalizable second example of an operator which has enough eigenvalues but not enough eigenvectors again it is not diagonalizable third example that I am going to discuss now will be diagonalizable this will also tell you what diagonalizability means for matrices if it is not already clear to you so I am going to discuss examples and also by the side discuss the minimal polynomials okay okay the first one let me consider the operator T whose matrix relative to the standard basis is given by this 3 by 3 matrix I will take this matrix this is the matrix of a linear operator relative to the standard basis let us say I will leave it to you for verifying that the characteristic polynomial of this matrix we are using P the characteristic polynomial of this matrix please check that it is over the minus sign so I will say characteristic equation for this problem the characteristic equation will be please verify that it is lambda plus 1 the whole square into lambda minus 3 this is the characteristic equation of this matrix okay expand this 3 by 3 determinant A minus lambda equal to 0 the eigenvalue minus 1 comes twice eigenvalue 1 comes once what I know is the eigenvectors corresponding to 1 eigenvector corresponding to minus 1 and the eigenvector corresponding to 3 are independent I must verify if minus 1 the eigenspace corresponding to minus 1 has dimension 2 I must verify if the eigenspace corresponding to the eigenvalue minus 1 has dimension 2 okay what is lambda equals minus 1 I need to go back to this look at the case lambda 1 is minus 1 I am looking at the number of independent solutions of this equation Ax equals minus lambda 1 x that is A plus lambda 1 I A plus I x I must solve this is that correct minus 1 yeah I meant lambda 1 x I meant lambda 1 x okay so I have Ax so A plus lambda 1 lambda minus 1 what is going on yeah minus lambda 1 yeah so I need to solve this A plus I x equal to 0 so what are those equations these equations are minus 8 4 4 minus 8 I must add 1 4 4 minus 16 8 8 into x the row reduced echelon form of this matrix will have only 1 nonzero 2 0 rows the first row is minus 2 1 1 or if you want you can say it is 1 minus 1 by 2 minus 1 by 2 in any case the number of solutions of this equation is 2 the rank of this matrix is 1 the row rank of this matrix is 1 because second and third rows are just multiples of the first row so the row rank of this matrix is 1 which means the null space will have nullity is 2 so null space has dimension 2 so there are 2 independent solutions of this equation okay let us compute for completeness I have there is only one equation there is only one equation that I must solve I must write down 2 independent solution this is one equation in 3 unknowns I can fix 2 of them so I will take x 3 to be 1 for the first case and x 2 equal to 1 so I have these let us say x 2 equals 2 so x 1 is 1 is that okay minus 2 plus 2 plus 1 so that is not 0 so I am taking x 3 I can take x 3 to be 0 x 2 is 2 x 1 is 1 so that gives me 1 vector let me write x 1 the superscript 1 that is 1 vector for me 1 2 0 I will write down another vector for the same equation let us say that comes from I will write that here another choice is x 3 x 1 sorry x 2 is 0 x 3 is 2 then x 1 is 1 that gives me another independent vector which is I think I have not written correctly here 1 2 0 x 2 yeah 1 0 2 right is that correct 1 0 2 x 2 and x 3 behave similarly so I can interchange their roles that is what I have done so this is another this is one this is another obviously they are independent so I have 2 independent eigenvectors for the eigenvalue minus 1 which comes twice as an eigenvalue okay let us also calculate an eigenvector for the eigenvalue 3 I need to solve Ax equals 3x A minus 3x so I must solve minus 12 4 4 minus 8 A minus 3 0 is that correct 4 minus 16 8 7 minus 3 4 I am looking at A minus 3 minus 12 0 4 all the other entries are the same so this into x I need to do row reduced okay may be just by observation let us do this quickly minus 2 1 1 let us say 2 minus 1 1 that is the first equation second equation is 2 0 1 I divide by 4 4 minus 2 minus 1 I divide by minus 4 minus 2 minus 1 that is correct second row I am dividing by 4 minus 4 2 0 minus 1 I am dividing by minus 4 3 I am dividing by minus 4 3 minus 1 minus 1 okay 1 minus 1 by 3 minus 1 by 3 2 0 minus 1 4 minus 2 minus 1 and then minus 2 times this plus this 1 minus 1 by 3 minus 1 by 3 minus 2 times this that is 2 by 3 2 by 3 minus 1 minus 1 by 3 minus 4 times this plus this okay so I have this to be 0 4 by 3 minus 2 that is minus 2 by 3 minus 4 times this 4 by 3 minus 1 minus 1 by 3 1 by 3 okay so please check I can remove the last row it is the same as the second row without reducing it to the row reduced echelon form etc okay now I see that I must fix x 1 x 2 x 3 can be determined uniquely okay same example I can fix x 1 let me multiply by minus 3 3 x 1 minus x 2 minus x 3 multiply this by 3 2 x 2 minus x 3 let me take x 1 to be 3 maybe x 1 to be 1 then I must solve for these 2 x 2 plus x 3 is 3 2 x 2 minus x 3 is 0 3 x 3 3 x 2 is 3 x 2 is 1 what is x 1 x 2 is 1 sorry x 3 x 1 has been fixed x 3 is 2 I will call this a third vector x 3 first coordinate is 1 second coordinate is 1 third coordinate is 2 so please check that this is another eigenvector 1 1 2 minus 12 plus 4 plus 8 0 minus 8 plus 8 minus 16 plus 8 plus 8 okay so this is an eigenvector for the eigenvalue 3 now look at all these 3 vectors I will use this portion now script B equals x 1 x 2 x 3 these 2 vectors x 1 and x 2 obviously are independent and x 3 must be independent with x 1 x 2 because x 1 x 2 correspond to the eigenvalue minus 1 x 3 correspond to 3 so these 3 are linearly independent vectors so this forms a basis for R 3 such that each vector is an eigenvector so the operator T that we started with must be diagonalizable in particular this matrix A is diagonalizable what is the meaning of A being diagonalizable let us call P as a matrix whose entries are the columns x 1 x 2 x 3 I define a new matrix P whose columns are the eigenvectors taken in this order first correspond to minus 1 second corresponds to 3 is this P invertible P is invertible because the columns are independent so the homogenous equation P x equal to 0 has x equal to 0 as the only solution this P is invertible and P is invertible and look at A P we have done this calculation before A P is A into x 1 x 2 x 3 this A can be got inside A x 1 A x 2 A x 3 but x 1 x 2 x 3 are the eigenvalues so this is minus 1 right minus x 1 minus x 1 3 x 3 minus 1 minus 1 are the eigenvalues for the first 2 vectors for the third one 3 is eigenvector right. Can I write this as x 1 x 2 x 3 into minus 1 0 0 0 minus 1 0 0 0 3 I can please check this I can write it in this manner in our notation this matrix is P let me call the other matrix as D D stands for the diagonal matrix yes this is P into D okay where D is that is A equals P D P inverse you can also look at P inverse A P as D this is what diagonalizability for a matrix means this is what diagonalizability for a matrix means I have also mentioned this when we wrote down the matrix of a linear transformation A is similar to the diagonal matrix D A is similar to the diagonal matrix D and so A is diagonalizable D is diagonalizable if and only if A is diagonalizable A is diagonalizably means that A must be similar to a diagonal matrix okay so this is the first example of a matrix that has been shown to be diagonalizable what is the minimal polynomial for this matrix we have written down the characteristic polynomial let me recall the characteristic polynomial I am using the notation P for this matrix is lambda plus 1 onto the 2 into lambda minus 3 probably it goes to the minus sign but does not matter I am looking at P of lambda equals 0 okay so the characteristic polynomial is lambda plus 1 square lambda minus 3 what is the minimal polynomial for this where do we start with what is the first choice for the minimal polynomial lambda plus 1 into lambda minus 3 okay one choice is lambda plus 1 into lambda minus 3 what is the other choice it must also have another possibility is lambda plus 1 into lambda minus 3 whole square right the number of the zeros must coincide okay but for this matrix I want you to verify that the minimal polynomial is just lambda plus 1 into lambda minus 3 please verify for this matrix that the minimal polynomial is lambda plus 1 into lambda minus 3 the reason why this must be true I will explain a little later but this can be verified by just two matrix multiplication you need to verify that M of A equals 0 that is A plus I into A minus 3 I is 0 okay so please verify in this example that the minimal polynomial is this and observe that the minimal polynomial is a product of distinct linear factors this example the minimal polynomial is a product of distinct linear factors that is it does not have lambda plus 1 whole square or lambda minus 3 whole square it is just a product of these two distinct linear factors okay that is the minimal polynomial for this example let us go back to the second example the second example that we discussed earlier tell me if the entries are correct 3 1 minus 1 2 2 minus 1 2 2 0 can you just check and tell me if this is the second example this is an example of a matrix which is not diagonalizable in spite of the fact that the Eigen values exist are the entries okay okay what is the characteristic polynomial for this matrix we have verified what lambda minus 1 the whole square into lambda minus 2 lambda minus 1 into lambda minus 2 whole square this matrix is not diagonalizable simply because it is defective with regard to the number of Eigen vectors there is no problem with the Eigen values Eigen values are there 1 comes with multiplicity 1 2 comes with multiplicity 2 but it does not have enough Eigen vectors it has only two Eigen vectors that is if you look at the matrix lambda minus if you look at the matrix A minus 2 I we have discussed the example before I just want to consolidate if you look at the matrix A minus 2 I that is 1 1 minus 1 2 0 minus 1 2 2 minus 2 okay you observe that the third row is a same I mean it is multiple of the second so I can delete it the rank is 2 the rank of this matrix is 2 the nullity of this matrix is 1 that is nullity of the linear transformation whose matrix is this nullity is 1 which means the dimension of the null space of A minus 2 I is 1 but dimension of null space of A minus 2 I is precisely the Eigen space the Eigen space is of dimension 1 okay so there is only 1 Eigen vector corresponding to the Eigen value 2 there is 1 Eigen vector corresponding to 1 in any case I do not have 3 Eigen vectors 3 independent Eigen vectors for this matrix so this is not diagonalizable okay so please check that nullity check that the Eigen space corresponding to the Eigen value 2 that is 1 dimension of the Eigen space corresponding to the second Eigen value is 1 and so there are only 2 independent Eigen vectors for this matrix A so A is not diagonalizable what are the choices for the minimal polynomial the choices are lambda minus 1 into lambda minus 2 lambda minus in whole square into lambda minus 2 or lambda minus 1 into lambda minus 2 whole square okay these are the possible choices I wanted to verify in this example that this is the same as the minimal polynomial okay please verify that the minimal polynomial in this example the same as a characteristic polynomial I told you there are 3 choices for the minimal polynomial 1 is lambda minus 1 into lambda minus 2 you can verify A minus I into A minus 2 I is not 0 the other choices lambda minus 1 whole square into lambda minus 2 that is A minus I the whole square into A minus 2 I verify that that is also not 0 finally this is the other choice verify that M lambda is P lambda okay in this example finally if you look at the third example which is the first example the matrix A is 0 minus 1 1 0 this simply does not have Eigen values what is the characteristic polynomial for this matrix characteristic polynomial is lambda square plus 1 if you look at this as a complex matrix this can be factorized as lambda plus I into lambda minus I so as a complex matrix this has I and minus I as the Eigen values okay the minimal polynomial is a product of if you look at it as a complex matrix then the minimal polynomial will have to have both these roots you please also verify that A square plus I equals 0 and so the minimal polynomial in this example is equal to the characteristic polynomial okay the minimal polynomial is a characteristic polynomial for the reason that A square plus I is 0 and that both the both the factors both the roots minus I and plus I must appear in the minimal polynomial okay probably I will stop here.