 See we are discussing the notion of minimal polynomials we had looked at three numerical examples first so let me recollect this first example was that of a linear transformation which did not have eigenvalues second example the matrix the linear transformation has eigenvalues but it does not have enough eigenvectors third example it has eigenvalues in the field and it has enough eigenvectors that is it is diagonalizable okay we are trying to calculate the minimal polynomial when we were trying to calculate the minimal polynomial we observed that for the first case the minimal polynomial coincided with the characteristic polynomial second case again not diagonalizable the minimal polynomial was not the characteristic polynomial third case I am sorry second example minimal polynomial was equal to the characteristic polynomial third diagonalizable case minimal polynomial is just a product of distinct linear factors okay and it appears as though the buck stops with the characteristic polynomial okay it appears as though we could do with characteristic polynomials probably okay. The answer can be do is yes it is always we will prove that today this is called the Cayley Hamilton theorem okay that is the characteristic polynomial of any linear operator over a finite dimensional vector space is an annihilating polynomial that is Cayley Hamilton theorem okay once you prove we will prove the matrix case not the linear transformation case once you prove this what follows is that the minimal polynomial divides the characteristic polynomial it follows that the minimal polynomial divides the characteristic polynomial. Let us also remember this one important connection between characteristic polynomials and minimal polynomials the zeros of both these polynomials are the same the zeros of these polynomials are the same okay before I prove this Cayley Hamilton theorem I also want to give a little implication which I did not mention when I was doing examples this could have been useful the implications are following suppose T is diagonalizable and let me say as before that lambda 1, lambda 2, etc lambda k be the distinct Eigen values be the distinct Eigen values of T. So remember I am not saying all the Eigen values are distinct I am only saying that the distinct Eigen values have been collected and I am calling them lambda 1, etc lambda k each one goes with a certain multiplicity lambda 1 goes with d 1, lambda 2 goes with d 2, etc lambda k goes with d k where d 1 plus d 2 plus d 3, etc up to d n is equal to the dimension of the space okay. If T is diagonalizable then what I want to prove it is a very simple proof the claim is that remember I am using small m for the minimal polynomial okay the claim is that the minimal polynomial is the product of distinct linear factors that is it is T minus lambda 1 into T minus lambda 2, etc T minus lambda k. If you go back to example 3 I simply ask you to verify that the minimal polynomial is a product of distinct linear factors the reason why I told you was this result in my mind. So let us prove this first the third example was that of a diagonalizable operator okay I want to show that okay for one thing for one thing you cannot have the minimal polynomial of degree okay I want to prove the minimal polynomial is this just look at this polynomial the minimal polynomial cannot be of degree less than the degree of the right hand side polynomial do you agree with that that is because if it were less than the degree of the right hand side polynomial that is k then it would mean what is the meaning of that it would mean one of these factors will be missing okay but if one of the factors is missing then I am in a situation where I have an eigenvalue it is not a 0 of the minimal polynomial. So to begin with this is a possible candidate for the minimal polynomial it cannot be less than this the degree of the minimal polynomial cannot be less than the degree of this right hand side polynomial which is k then the only thing that I need to show is that this is an annihilating polynomial I will show that this is annihilating polynomial then it follows that this is the minimal polynomial this is of course monic the power of t the coefficient of t to the k is 1 so this is a monic polynomial I will only show that this is an annihilating polynomial okay what is it that we must show we must show that m of t is a 0 operator annihilating polynomial so m of t is a 0 operator I want to show that an operator is 0 I will show that the action of this operator on a basis is 0 then it follows that the operator is 0 agreed okay t is diagonalizable that means there exists a basis b with the property that each of the basis vector is an eigenvector of t let b be a basis of v such that each x element of b is an eigenvector of t corresponding to some eigenvalue I am not too worried about that corresponding to some eigenvalue but I know that there is at least one such basis because t is diagonalizable I will show that m t of x is 0 it follows that m t must be 0 do you agree m t is a linear operator m of capital T that is a linear operator I want to show that that is a 0 operator I will show that m of t on the basis vectors is 0 so for every x in b I will show m t x is 0 so I will start with an x in b and consider m t x I must show that this is 0 okay but what is this m of t this is the expression it is t minus lambda 1 I t minus lambda 2 I etc t minus lambda k I operating on x I want to show that this is 0 this is 0 vector but observe that the property that these factors have is that any two of them commute any two of them commute t minus lambda j I into t minus lambda s I this is equal to t square minus lambda s t minus lambda j t plus lambda j lambda s I this is t minus t square minus lambda j t minus lambda s t plus lambda s lambda j I this is nothing but t minus lambda s I into t minus lambda j I remember that in general if s and t are operators s t is not equal to t s okay that is matrix multiplication is not commutative but this is about special matrices special linear operators operators of the form t minus lambda I these commute what is the advantage you go back here this x is x belongs to b and each vector in b is an eigen vector so this x is an eigen vector corresponding to some eigen value lambda I push that to the end operate on x you get 0. So let me say that I know that x is in b and each vector is an eigen vector so there is t x equal to lambda I x for some eigen value lambda I so I know that this equation holds for x that is t minus lambda I of x equals 0 now I use commutativity of these factors to write m t of x as t minus lambda 1 I lambda 2 I etc the ith factor I can take t minus lambda I minus 1 t minus lambda I plus 1 t minus lambda k is the last one but I can push t minus lambda I to be the last one to be the last factor because of commutativity now I see that t minus lambda I x is 0 make use of that the rest of the factors are linear so that is again 0 okay this is really easy if t is diagonalizable then the minimal polynomial is a product of distinct linear factors at the end of next lecture we will be able to show that the converse is also true if the minimal polynomial is a product of distinct linear factors then t is diagonalizable okay which will then give another characterization of diagonalizability okay so the converse will be proved little later we need the notion of invariant subspaces annihilating polynomials I will do that later but this is one result which could be used in those examples if you know that it is diagonalizable then calculating minimal polynomial is easy if you know the eigenvalues okay as I told you we will discuss the Cayley-Hamilton theorem and its proof for matrices okay I just need one little observation before proving the Cayley-Hamilton theorem if I have a matrix with polynomial entries then the matrix can be written as a polynomial whose coefficients are matrices I will just give an example if I have matrix with polynomial entries let us say 1 plus t square t 2 t minus t cube 1 and let us say 1 minus t cube I have a matrix with polynomial entries then I am claiming that this matrix can be written as a polynomial whose coefficients are matrices as a polynomial whose coefficients are matrices what is the meaning simply collect the coefficients of like powers look at the constant for example constant is 1 here 0 1 1 so I can first write it as 1 2 1 sorry 1 0 1 1 times 1 plus corresponding to t there is only one term 0 2 0 0 corresponding to t square so okay let me write this into t this is a constant term of the polynomial for a the polynomial is in terms of 1 t t square t cube the coefficients are matrices this is one coefficient this is a second coefficient into t plus for t square I have only one term 1 0 0 0 t square plus finally for t cube I have 0 minus 1 0 minus 1 t cube is it okay please calculate verify this right hand side is correct I have 1 plus t square is the first term second is 2 minus t 2 t minus t cube the third goes with 1 in this entry the fourth one is 1 minus t cube okay so I have written a matrix whose entries are polynomials this matrix has been written as a polynomial in t the same variable t but with matrix coefficients let us remember this now let us look at Cayley Hampton theorem I will give the statement for an operator and also write down the statement for the matrix and prove the matrix case t is a linear operator on a finite dimensional vector space let okay I am using what p let p of t let p be the characteristic polynomial for t then Cayley Hampton theorem states that p of t is capital T that is the 0 operator that is okay what this says is that the characteristic polynomial is an annihilating polynomial but we know that among all the annihilating polynomials the minimal polynomial is the one with the least degree so the characteristic polynomial must be a product of the minimal polynomial and another polynomial okay in other words the minimal polynomial divides the characteristic polynomial this is an important information and you must compare this with the result that we proved right in the beginning when we looked at the definition of the minimal polynomial so I remember having made this statement that to define the minimal polynomial first of all you must know that there is an annihilating polynomial to show that there is an annihilating polynomial remember we took the operator's identity t t square etc and went up to t to the n square and show that this polynomial must be and showed that these vectors these operators are linearly dependent that gives rise to an annihilating polynomial okay now look at that estimate for that annihilating polynomial it is a polynomial of degree n square okay what this theorem says is that you could do with much less you do not have to go beyond n that the degree of the characteristic polynomial is n this theorem says you do not have to go beyond n at most the characteristic polynomial is that clear okay proof as I told you I will prove the matrix case let me just write down the statement for matrices let a be the matrix of t relative to some basis then we would like to show that p of a is a zero matrix this is what we want to show see for proving this I will make use of of course this observation and also another non trivial result on determinants which I will not prove this non trivial result on determinants for the proof we will need the following this non trivial result is what I will write down next we use the following for any square matrix B for any square matrix B if you look at the matrix B into the adjoint of B for any square matrix B the matrix B into the adjoint of B what is this B times I yes that is what I want this is determinant of B times I this is not A inverse remember that that see the determinant of B could be zero okay that is quite possible so there is no inverse coming here if you know the determinant of B is non zero then you can remember this is an equation involving matrices first B is a matrix adjoint of B is a matrix on the right hand side determinant B is a number times identity matrix so this is an equation involving matrices and then if determinant of B is not zero we could divide by that number and then write this as 1 by determinant of B into adjoint of B call that as some T okay T is a linear transformation let us call it C then BC equals I you can also show CB equals I so it follows that this is a inverse okay but this is more general than this determinant B equal to zero non zero case so I will make use of this result this is again a non trivial result using the property of determinants this is not easy but it is not difficult again so I will make use of this result and then prove Kale Hampton theorem I will apply this identity for the matrix A minus lambda I in place of B okay before that let me call let me call determinant A minus lambda I I know that this is a polynomial of degree n so let me call this as alpha not lambda to the n plus alpha 1 lambda to the n minus 1 etc alpha n minus 1 lambda plus alpha n this is a polynomial of degree n I am writing it like this now this alpha not could be 1 or minus 1 depending on the order but let us keep it as it is what is that we need to show we need to show that P of A is equal to 0 so we must show that when I replace lambda by A when I replace lambda by A I get a polynomial in A I must show that this polynomial is the 0 polynomial I am sorry I get a matrix I must show that this matrix is a 0 matrix okay. So we must show that alpha not A to the n plus alpha 1 A to the n minus 1 etc alpha n minus 1 A plus alpha n identity we must show that this is 0 matrix okay let us now make use of A minus lambda I in place of B and then see what we get so I am using this formula applying A minus lambda I for B okay in this I am applying A minus lambda I so I have this A minus lambda I into adjoint of A minus lambda I is equal to determinant of A minus lambda I into identity matrix I will substitute determinant of A minus lambda I the expression that I have written down just now here and keep this right hand side and then look at what I have from the left hand side so let me write the right hand side the right hand side is alpha not lambda n I is that correct yeah times identity plus alpha 1 lambda to the n minus 1 identity etc alpha n minus 1 lambda identity plus alpha n identity okay just to remind myself the right hand side is a matrix determinant of A minus lambda I into I that is a matrix okay now look at the left hand side look at the second factor adjoint of A minus lambda I how do you compute the adjoint of a matrix the adjoint of a adjoint of a matrix B let us go back to this is computed by using minors cofactors taking transpose first compute the minors what is a minor of an entry the minor of an entry is the determinant of okay if you have an n cross n matrix the minor of an entry A ij is a determinant of the n minus 1 cross n minus 1 sub matrix obtained by deleting the I throw jth column okay okay we are calculating minor of A minus lambda I okay A minus lambda I along the diagonal I have A 1 1 minus lambda A 2 2 minus lambda etc let us take any arbitrary element calculate its minor I will not calculate its minor I will only say what this minor the form must be can you see that the okay I am deleting I am deleting a row a row and a column and then I am looking at an n minus 1 cross n minus 1 sub matrix I calculate the determinant of that now that is the determinant will be a polynomial of degree n minus 1 that is what I want that is a polynomial of degree n minus 1 also it is a determinant of a matrix whose entries are polynomials so I can write this polynomial like I will make use of this observation I can write this polynomial in terms of in the variable lambda but the coefficients are matrices so this is the point that is probably the most crucial in this apart from this formula adjoint of A minus lambda I I can write this as it is a polynomial of degree n minus 1 but the entries are the coefficients are matrices this is the point. So I will use a similar notation let us say B 1 lambda to the n minus 1 plus B 2 lambda to the n minus 2 plus etc plus B n minus 1 lambda plus B n but I must observe that where B 1 B 2 etc B n are matrix coefficients so they are matrices adjoint of A minus lambda I adjoint of A minus lambda I is a matrix whose entries are polynomials in lambda adjoint of A minus lambda I is a matrix whose entries are polynomials in lambda but the maximum degree of each polynomial in the adjoint of A minus lambda I is lambda to the n minus 1 because you are calculating the determinant of n minus 1 cross n minus 1 sub matrix. So these are matrices whose entries of course depend on A that is not important it is mentioned only for completeness these entries of these matrices depend on A now substitute this into the left hand side equate coefficients of like terms you get you do a one more operation and then you get P of A equals 0 okay let us substitute to calculate the left hand side left hand side is A minus lambda I into adjoint of A minus lambda I the expression is available here I have B 1 lambda to the n minus 1 plus B 2 lambda to the n minus 2 etc B n minus 1 lambda plus B n this is a left hand side let me collect the coefficients and write down the terms accordingly there is only one term corresponding to lambda to the n that term is minus lambda B 1 sorry minus B 1 there is only one term corresponding to lambda to the n that is this there are two terms corresponding to lambda to the n minus 1 one coming from this minus B 1 the other one I am sorry one coming from this and the other coming from this A B 1 minus B 2 lambda to the n minus 1 let me maybe write down one more term for lambda to the n minus 2 you can verify A B 2 minus B 3 A B 2 there is a B 3 to the lambda n minus 3 into that will give lambda n minus 2 so minus B 3 so it follows this order plus coefficient of lambda coefficient of lambda will come from contribution from second term in the last term last but one and the first term A B n minus 1 minus B n lambda and constant term is just A B n okay these are the terms corresponding to the degrees lambda to the n lambda to the n minus 1 etc so I must compare these two so maybe I will call this equation 1 right hand side formula 1 left hand side formula 2 okay look at right hand side I have formula 1 left hand side I have this expression these two must coincide so I get the following equation coefficient of lambda to the n on the right hand side is alpha not I minus B 1 is alpha not I A B 1 minus B 2 is alpha 1 I A B 2 minus B 3 is alpha 2 I A B n minus 1 minus B n that corresponds to what lambda right so that is alpha n minus 1 times identity the last equation is A B n equals alpha not sorry alpha n identity so I have these equations n plus 1 equations polynomial of degree n so there are n plus 1 terms these are n plus 1 equations multiply the first 1 by A to the n minus 1 this by A to the n minus 2 this by this by A this by A square and then add you get okay so let us say I multiply this by A to the n this by A to the n minus 1 n minus 2 etc this will be multiplied by A square and I will just leave it right so no this will be multiplied by A this is multiplied by identity so that yeah I must go from A to the 0 to A to the n multiply and add A to the n okay maybe I will just do that quickly minus A to the n B 1 pre-multiplying matrix multiplication is not commutative pre-multiplying minus A to the n B 1 plus A to the n minus 1 B 1 minus there is already n A minus A to the n minus 1 B 2 plus A to the n minus 1 B 2 minus A to the n minus 2 B 3 plus etc plus this one is multiplied by A A square B n minus 1 minus A B n plus A B n on the one side on the other side alpha not A to the n plus alpha 1 A to the n minus 1 etc plus alpha n minus 1 A plus alpha n identity you can see that the left hand side is 0 these two get cancelled these two get cancelled etc you take any term the first term each term has two terms you take any term the first term gets cancelled with the second term or the previous term etc so the left hand side is 0 the right hand side is what you have here that is Cayley Hamilton theorem the left hand side is 0 matrix right hand side is P of A okay let us do one final problem in this section which is purely for fun maybe I want to look at this 4 by 4 matrix say remember calculating the characteristic polynomial is not difficult alright but calculating the Eigen values is difficult but in certain problems it will be very clear. So I want you to look at this problem let us look at the linear operator T on R 4 whose matrix with respect to the standard basis is this 0 1 0 1 1 0 1 0 1 0 1 1 0 1 0 I would like to calculate the characteristic polynomial of this as well as the minimal polynomial this matrix has some nice properties which we will make use of symmetry really okay look at A square say I am not trying to advocate this procedure for all matrices for this matrix it is nice to see how the minimal polynomial characteristic polynomial can be calculated very quickly okay what is A square? A square is 0 1 0 1 into 0 1 0 1 it will be 2 0 2 0 0 2 0 2 0 2 0 2 0 this is A square okay still there is no pattern look at A cube A cube can be shown to be see it is a product of this into this that is 4. So you will have 0 4 0 4 4 0 4 0 mistake no problem this is correct what is your question I am calculating A square here is that correct is something wrong here 2 0 2 0 0 2 0 2 okay this is fine this into this okay and A cube instead of this I will go back to this pattern see before coming to the class I will work it out and come okay that is called preparation so this is correct please check this this is equal to 4 times A that is what happens in this example okay this is 4 times A which means I have caught hold of an annihilating polynomial right A cube minus 4 A is 0 so I will look at I look at the polynomial I will call it f this time f of t is t cube minus t cube minus t that is t into t square minus 1 sorry t cube t cube minus 4 t that is t square minus 4 that is t into t plus 2 t minus 2 whichever way if I call f as this polynomial then f of A is 0 that much I know I know that f of A is a 0 matrix okay I know that minimal polynomial is a polynomial of least degree among the annihilating polynomials okay so the minimal polynomial is see the degree of this is 3 the degree of this is 3 can the minimal polynomial be a linear polynomial can the minimal polynomial be a linear polynomial linear polynomial is a polynomial of the form t minus lambda minimal linear polynomial is t minus lambda okay so can this be the minimal polynomial for this matrix can it be linear if if the minimal polynomial for this matrix is linear then I must have something like A equals lambda times I A equal to alpha times I for some scalar alpha obviously not so A is not alpha times identity A is not which means can you see that all these 3 individual factors do not constitute a minimal polynomial that is the minimal polynomial is not a linear factor do you first agree it is not linear it is all that I want to say is that this is a minimal polynomial okay I want to rule out the possibility that it is not quadratic nor is it linear linearity is immediate the minimal polynomial is not linear what is the reason if it were linear then A would be a multiple of identity and so that is if the minimal polynomial is linear then the minimal polynomial is of the form t minus alpha for some alpha but if that were the case a minimal polynomial must be annihilating polynomial so if I replace t by A I must get A minus alpha I is 0 that is not the case so it is not linear can it be quadratic if it is quadratic what are the choices if it is quadratic remember minimal polynomial must divide this polynomial so this is like a benchmark if it is a quadratic then it is either of the form t into t minus 2 or t into t plus 2 or t minus 2 into t plus 2 okay if it is of the form t into t minus 2 then t square minus 2 t must be an annihilating polynomial that is A square equals 2 A that is not the case if it is of the form t into t plus 2 then I must have A square plus 2 A equals 0 that is not of the form the last possibility is t square minus 4 equal t square minus 4 is a possibility for the minimal polynomial in which case A square equal to 4 I that is again not the case t square minus 4 is the third quadratic polynomial that is not a minimal polynomial because that is not the minimal polynomial because A square minus 4 I is not 0 A square is not equal to 4 I and so we have already seen that this is an annihilating polynomial so this must be the minimal polynomial okay so please give the reasons for the quadratic case it is not quadratic also and so the minimal polynomial for this matrix is t into t plus 2 into t minus 2. Now the minimal polynomial has the property that the zeros of the minimal polynomial are the characteristic values so the characteristic values of the matrix that we started with are 0 plus 2 minus 2 also 0 2 minus 2 are the Eigen values of the matrix A there is a relationship between the characteristic polynomial and the minimal polynomial they have the same set of zeros the dimension is 4 I have 3 Eigen values so one of them comes twice which one one of the Eigen values must come once more the algebraic multiplicity the algebraic multiplicity of one of these Eigen values must be 2 which one is that if I give the answer you must also give the reason Indians invented 0 so let us take 0 first I want to know the dimension of the set of solutions of A x equals 0 x that is A x equal to 0 what is the dimension of the set of solutions of A x equal to 0 A is given to you it should be 2 is okay but I want you to give a reason why is it 2 way quickly rank of A is 2 this row is the same as this row this row is the same as this row these two are obviously independent rank of A is 2 nullity of A must be 2 so dimension of set of all solutions of A x equal to 0 is 2 so 0 comes twice rank of A is 2 and so the Eigen value 0 appears twice as an Eigen value of this matrix and so what is the characteristic polynomial so I will simply say appears twice so P is the notation we have for the characteristic polynomial so P of T is T square into T minus 2 into T plus 2 the characteristic polynomial should be of this form T square into T plus 2 into T minus 2 is the matrix diagonalizable see the answer is yes there are at least two reasons the answer is yes the matrix is diagonalizable there are at least two reasons one reason if you assume my result that a matrix is diagonalizable if and only if the minimal polynomial is a product of distinct linear factors that has not been proved but you could have taken the statement the other thing is distinct Eigen values of linear independent Eigen vectors I am assured of three independent Eigen vectors 0 minus 2 plus 2 three independent Eigen vectors come alright do I have one more Eigen vector which is independent with these three coming from A x equal to 0 okay the problem lies the question is 0 deficient that is do I have two independent Eigen vectors for the Eigen values 0 but that is easy to see in this problem A x equal to 0 there are two independent solutions rank is 2 nullity is 2 that is what we determine just now the matrix is diagonalizable okay the matrix is diagonalizable and it is similar even if you view this as a matrix over rational field the field of rational numbers the matrix A is similar I will use this the matrix A is similar to this 4 by 4 matrix let us say I follow this order then it is this matrix okay this matrix I will come back this matrix will come that was your doubt I think 0 0 2 0 I am taking it in this order 0 comes twice 2 and then minus 2 so A is similar to this that means P inverse A P equals this matrix how do you get P? P is a matrix whose first two columns are the independent solutions of A x equal to 0 third column is the only solution of A x equals 2 x fourth column is the only solution of A x equals minus 2 x only non-zero solution that is a matrix P that is invertible and P inverse A P equals a diagonal matrix okay I think I will stop here I want to discuss the notion of invariant subspaces annihilating polynomials and how it helps these two notions help in proving the converse statement that if the minimal polynomial is a product of distinct linear factors then the matrix is then the operator is diagonalizable I will prove this in the next class okay let me stop here which argument? See if you want to show that the matrix is diagonalizable then you must show that it has four independent eigenvectors three independent eigenvectors are already there because of 0 minus 2 plus 2 they are distinct so I look at a non-zero solution of those eigenvalue equations I want one more that one should one more should correspond to the eigenvalue 0 because the characteristic polynomial is this that should correspond to 0 so do I have two independent solutions of A x equal to 0 is what the question is but the rank of A is 2 we observed just now the rank of A is 2 so nullity is 2 so there are two independent solutions take those two independent solutions put them in the matrix P you get P inverse A P as this diagonal matrix okay so this is see this is just to illustrate how certain problems can be approached do not apply this for example in the exam you will waste time okay we will meet on Monday.