 We are looking at invariant direction decomposition characterization of diagonalizability. We are to prove the converse part right. Let me write down that statement of the converse there is a slight modification to the statement that I gave in the last lecture. So let me write down all the conditions. Conversely suppose that the numbers, suppose that there exist I want to say suppose that there exist distinct numbers lambda 1, lambda 2, etc lambda K and K non-zero linear operators, K non-zero linear operators I am calling them E1, E2, etc EK these are operators on V such that the following conditions are satisfied. First condition is T is lambda 1 E1 plus lambda 2 E2 etc plus lambda K EK. T is a specific linear combination of these operators. Condition to identity is just the sum of these K operators such as second condition. Third condition I will write EI EJ equal to 0 for I not equal to J. This is not the third condition that I gave yesterday. Third condition I gave yesterday is EI square is EI. Suppose I have these numbers lambda 1, etc lambda K and operators K operate non-zero operators that is important such that these conditions are satisfied. Then EI square equals EI this is really a consequence not an assumption. It is a consequence of these three conditions. This happens and Phi also holds. Now what is Phi? Phi is range of EI equals the Eigen space corresponding to the Eigen value lambda I. So I will simply say Eigen space for lambda I. This is another consequence. Condition 4 which was assumed yesterday is really a consequence of 1, 2, and 3. Also what is more important is that really this statement I want to prove earlier than 4 and 5 lambda 1, etc lambda K are the Eigen values of T. These are precisely the Eigen values of T that is what this means. And T is diagonalizable. These are the four consequences. We will first show that lambda 1, etc lambda K are the Eigen values and then show that EI square is EI then show that T is diagonalizable and finally show that range spaces are the Eigen spaces, okay. So this is a slight change from what I had given yesterday, okay. First thing is this. That is easy. I is E1 plus E2, etc plus EK. So I multiply by EI. EI is EI into I that is EI summation J equals 1 to K EJ. I is fixed. The subscript I is fixed. The running index is J. When J takes a value I have condition 3. I have condition 3 that the product is 0 when the subscripts are different. So this goes inside and I have the product summation J equal to 1 to K EI EJ. When J equals I it is EI square. All the other terms are 0. So this is just EI square. It started with EI. I have EI square. So condition 4 holds. That is an immediate consequence of 1 and 3, sorry consequence of 2 and 3. That is condition 4. I will show that range of EI is the Eigen space for lambda I not entirely. I will show that range of EI is contained in the Eigen space corresponding to lambda I. Later I will show that the Eigen space for lambda I is contained in range of EI. So this shows 4. Next we show that okay let me write this. We show that range of EI is contained in the Eigen space for lambda I which is null space of T minus lambda I. This is the Eigen space of the Eigen value lambda I. I will show that range of EI is contained in this. Let X belong to range of EI then X equals EI X. Consider T X minus lambda I X. I want to show that this is 0. I want to show that this is 0 right. If I show this is 0 then it follows that X belongs to null space of T minus lambda I. I started with X in range of I. So that would prove this okay. This can be written as T I will use representation 1 and then lambda I identity X I will use representation 2. So I can write this as summation J equals 1 to K for T I will have lambda J EJ X. For T I have summation lambda J EJ. For the second term minus so I will use a bracket here. Second term goes with an I. I is fixed. So this goes with lambda I minus lambda I into identity X. Identity is summation EJ. Identity is summation EJ. So this is EJ X. Do you agree? Lambda J EJ X lambda I EJ X. Lambda J EJ minus lambda I. So this is lambda I. I can write this as this is 0. I want to show this is 0 okay. Let me write down. This is equal to summation J equals 1 to K lambda J minus lambda I EJ X is what I wrote. X is EI X okay. So when again J is running index I is fixed. J is running index I is fixed whenever J is not equal to I these terms are 0 whenever the only term that is left is when J takes a value I. When J takes a value I this is 0. So this is 0 okay. So what we have shown is that T minus lambda I X we have shown that this is 0. Yes and so this holds. So range of EI is contained in null space of T minus lambda I. What also what this also means is that see each of these operators is non-zero. Each of these operators is non-zero which means range must have at least one non-zero element one non-zero vector which means that what is the consequence? Since EI is not equal to 0 there exists X not equal to 0 X element of range of EI that is from what we have shown just now there exists X not equal to 0 such that T X equals lambda I X that is these numbers are eigenvalues of this operator T right. So lambda 1, lambda 2, etc lambda K these are not the eigenvalues these are eigenvalues presently these are eigenvalues of T the question is does it exhaust all the eigenvalues is it possible that we missed some eigenvalue which is not counted because of the way we have done the way we have got this inclusion have we missed other eigenvalues have we missed other numbers which are possible eigenvalues of T then we will show that these are the only eigenvalues so it will follow that these are the eigenvalues is the problem clear these are eigenvalues alright are there other eigenvalues which are not accounted here let me prove that that would prove this statement then diagonalizability will follow after that we will prove condition 4 condition 5. So let me take a number which is an eigenvalue and then show that this number must be one of these one of these lambda I yes why am I using I suppose that is clear okay I want to show that if lambda is a number if lambda is a number for which T minus lambda X is 0 for some X not equal to 0 then lambda is one of these okay. So let lambda be such that T X equals lambda X with X not equal to 0 I want to show that this lambda is one of these lambda I's then it will follow that this is a complete list of eigenvalues till now we have not shown that this exhausts all the eigenvalues of T we have shown that these numbers are eigenvalues we will show that any other number that satisfy this equation must be one of these okay so that this statement would be proved okay lambda such that T X equals lambda X X not equal to 0 so look at this equation this means 0 equals T minus lambda I of X I will now use the representation once again summation j equals 1 to k lambda j E j minus lambda E j of X lambda j E j is T I is summation lambda I is lambda summation E j this is summation j equals 1 to k lambda j minus lambda into E j X lambda j minus lambda E j X this is 0 I will use this part okay now I will have to make use of the fact that this is an equation like U 1 plus U 2 plus et cetera U k equal to 0 where U 1 is in range of E 1 U 2 is in range of E 2 et cetera but we know that these are these are eigenspaces because range we have shown range of E I is contained in null space of T minus lambda I so this is an equation involving eigenvectors really is that clear what I have is 0 equals U 1 plus U 2 plus et cetera U k where what is U I U j may be U j is the j th term that is lambda j minus lambda E j X I am calling the first term U 1 second term U 2 et cetera this is 0 now this U j belongs to range of E j and range of E j we have just now shown any vector in range of E j if it is not 0 then it is an eigenvector so these spaces are independent range of E 1 range of E 2 et cetera are independent because they correspond to distinct numbers lambda 1 lambda 2 et cetera lambda k this is an equation involving a sum where the each term comes from a subspace the subspaces are independent so this means each must be 0 U j equal to 0 for all j that is lambda j minus lambda E j X equals 0 for all j now is this clear the fact that the fact that U j equal to 0 from this equation follows because I am now really looking at eigenvectors corresponding to distinct eigenvalues I am looking at subspaces which are independent that is the reason why each U j is 0 I have this equation for all j suppose E j X is 0 for all j then what you know about X from this V j X equal to 0 for all j what you know about X X must be 0 because this equation tells me that X is equal to E 1 X plus E 2 X is by E k X so if each E j X is 0 then X is 0 but we started with X not equal to 0 we have X not equal to 0 so if E j X equal to 0 then X equal to 0 a contradiction and so E j X cannot be the 0 vector for at least 1 j this cannot be 0 for at least 1 j I go back to this equation this number into this vector is 0 for at least 1 j I am sorry this is to for all j for at least 1 j this vector cannot be 0 so this number must be 0 for that j so lambda I want to really write lambda equals lambda j for that j so what we have shown is that these numbers exhaust all the eigenvalues lambda 1 etcetera lambda k exhaust all the eigenvalues of the operator T is that clear we started with lambda being an eigenvalue so the equation T X equals lambda X with X not equal to 0 must be satisfied the rest is consistent with what we have as assumptions okay so what we have shown is that these are precisely the eigenvalues now lambda 1 etcetera lambda k are precisely the eigenvalues of T why is T diagonalizable that is almost tautology it does not need much explanation probably any vector X in the space V can be written as sum of vectors X 1 X 2 etcetera X k X i coming from E i but I know that range of E i okay I know that anything in the range of E is an eigenvector so long as it is not 0 all that I will do is take a basis for range of E 1 take a basis for range of E 2 etcetera combine them take the union that will be a basis for V this basis for V has the property that each vector is an eigenvector so from this equation and from the fact that anything in range of E i belongs to the null space of T minus lambda i it follows that T is diagonalizable is that clear what do I want to show I want to show that I want to show T is diagonalizable so I want to show that there exist a basis for V each of whose vectors is an eigenvector okay all that I do is take a basis for range of E 1 take a basis of range of E 2 etcetera take a basis for range of E k the fact that the sum is equal to identity means that I have exhausted the number of elements in a basis for E 1 plus the number of elements in a basis for E 2 etcetera the number of elements in a basis for E k that number must be equal to the dimension of the space because of this condition so T is diagonalizable okay so it is an immediate consequence of 2 I just write that by taking basis for range of E 1 range of E i okay let me just give it for the sake of completeness namely B 1 etcetera B k and by setting script B to be the union of these basis it follows that the matrix of T relative to B is diagonal that is each element in this basis is an eigenvector for T each element in the basis B is an eigenvector for T because each element in the basis B must belong to one of the B i's anything in B i is basically a subset of range of E i anything in range of E i these are basis vectors so they are non-zero vectors so each of them is an eigenvector is that clear so T is diagonalizable that is the most important part but it follows as an easy consequence of what we have done earlier. The final part is to show that this eigen space is contained in range of E i that is the last part all other things have been proved okay we have shown that range of E i is contained in the eigen space we must show that the eigen space is contained in range of E i null space of T minus lambda i we will show that this is contained in range of E i okay let us take an element I will call it U let U belongs to null space of T minus lambda i then U is T minus lambda i T minus lambda i U is 0 I will again use that representation for T and i summation j equals 1 to k this time I will write down the simplified expression this will be lambda j minus lambda i E j U lambda j E j U that is T U lambda i U is lambda i E j U i is summation E j so I have this now this is 0 again this is an equation involving vectors that are in range of E i range of E j so each term must be 0 because these are independent subspaces so I have lambda j minus lambda i E j U this is 0 for all j not equal to i because when this is 2 for all j is okay we will exploit this for all j this is true all you have to do is yeah conclude from this that when j is not equal to i look at this equation when j is not equal to i it will tell you that E j U so just tell me if this is clear so j not equal to i implies E j U is 0 when j is not equal to i see I have this product to be 0 about the ith equation I do not know about the ith equation I cannot make any conclusions but about all other equations what I know is that this is not 0 so this must be 0 okay I have equation from 1 to k i occurs somewhere that ith equation is lambda i minus lambda i into E i U equals 0 so there I cannot conclude E i U equal to 0 by if I look at the other equations it follows that E j U is 0 for all j not equal to i okay what is the meaning of this E j U equals 0 for all j not equal to i it means again any vector can be written as any vector x can be written as E 1 x plus E 2 x etcetera all terms are 0 except ith term so x belongs to range of E i ith term comes from range of E i so sorry not x U belongs to range of E i so from this the last step is U belongs to range of E i is fixed so I started with U in null space of T minus lambda i I have shown that U belongs to range of E i already we have shown that range of E i is contained in null space of T minus lambda i so these subspaces are the same that is the range of E i is the eigen space corresponding to the eigen value lambda i okay now that is the complete theorem with its proof probably we look at an example numerical example okay I want to look at a numerical example where we will calculate all these an example where the matrix is diagonalizable the operator is diagonalizable so let me quickly see if this works by the way before I work out this example there is a fact that I will state without proving we can discuss it if you want the proof later outside the class it can be shown that this the operators the projections E j's are in fact polynomials in T in particular maybe I will give this as an exercise let me define P j of T to be the product of all these polynomials P j of T so the product is over I let us say I does not take the value j look at T minus lambda i by lambda j minus lambda i I am defining a polynomial in this manner this polynomial for one thing has a property that P j of T i is delta i j this polynomial has a property that P j of T i is delta i j P j of lambda i is delta i j I am defining for the numbers lambda 1 is lambda k that are distinct for every fixed j here the product the index is i the product index is i these are these are really Lagrange interpolating polynomials we have encountered these before these are Lagrange interpolating polynomials what follows what can be shown is that E j is P j of T what is the meaning of this it says there is a the formula for E j's can be obtained from the formula for the polynomials here okay why this is true is an exercise for you really I will freely use this in the example okay what is an example let me take the matrix A there are relationships between these are called Newton's formulas there are relationships between the entries of the matrix and the eigenvalues for example the sum of the eigenvalues is the trace of the matrix the sum of the product of two eigenvalues taken at a time is the sum of the 2 by 2 principle minors the product of the eigenvalues is the determinant of the matrix okay I will use these three properties I do not prove any of these again these are exercises I want to write down the characteristic polynomial for this matrix I use P right for that so P of T I claim is 2 plus 3 5 plus 2 7 lambda cube minus 7 lambda square I want to look at the 2 by 2 determinants 6 minus 2 is 4 6 minus 2 is 4 what I want is really 4 minus 1 is 3 that is 11 it goes to the minus this goes with a plus the final term is a determinant what is a determinant determinant is 5 my plus 7 plus 11 lambda minus 5 this is the characteristic polynomial okay this can be shown what are the factors 12 minus 12 1 is a factor verify that this is lambda minus 1 the whole square into lambda minus 5 5 is also a factor okay of lambda is this diagonalizable how can I repeat what is lambda equals to 1 the null square sub T minus I is 2 if you see this for lambda equals 1 you substitute there is only one equation 1 2 1 1 2 1 1 2 1 only 1 row the rank of that is 1 so nullity is 2 so this matrix is diagonalizable A is diagonalizable A is diagonalizable I want to compute see all that I want to do is to illustrate that this previous theorem by means of this example so I will compute E J's and then verify sum of E J equals identity and then T can the product is 0 T can be written as lambda 1 E 1 plus lambda 2 E 2 that is enough really okay E 1 is P 1 of T P 1 of T there is only one factor left out P 1 corresponds to the first eigenvalue T minus 5 identity by what is fixed is 1 is fixed so 1 minus 5 minus 4 this is E 1 5 I minus T by 4 let me write this maybe 1 by 4 could be taken outside 5 I minus T 5 minus 2 3 5 I minus T minus 2 minus 1 minus 1 minus 1 minus 2 5 minus 3 is 2 5 minus 2 is 3 there are many minuses let us take minus 1 by 4 outside minus 3 2 1 1 minus 2 1 1 2 minus 3 please check that my calculations are correct what is E 2? E 2 will be T minus I by 4 I am sorry I minus T minus I is correct yes so this is 1 by 4 T minus I 1 2 1 1 2 1 1 2 1 is that okay yes when you add do you get what we want okay this 1 goes with a plus sign 3 plus 1 4 4 by 4 1 yeah this goes with a minus sign the previous 1 would have been better I will do it with the previous 1 0 this is 0 this is 0 this is 4 by 4 is 1 this sum is 0 this is 0 this is 0 this is 3 plus 1 by 4 so sum is identity E 1 plus E 2 equals identity see I could have once I got E 1 I could have got E 2 by using the fact that E 2 is I minus E 1 but I wanted to really verify that so I calculated E 2 independently by this formula okay finally you check that T that is A in this case lambda 1 E 1 plus lambda 2 E 2 okay that is 5 by 4 plus 3 by 4 8 by 4 2 that is the first entry please check this is satisfied A is E 1 plus 5 E 2 please also write down the range spaces of these two operators they will be precisely the Eigen spaces for example it is immediately clear that range of E 2 range of E 2 is 1 1 1 spanned by 1 1 1 range of E 2 is spanned by 1 1 1 and E 2 range of E 2 is 1 dimensional range of E 1 is 2 dimensional because lambda for lambda equals 1 the null space has two independent vectors okay so this is just to illustrate as I told you there are certain properties I have just stated I have not used the relationship between the roots of the equation P of lambda equal to 0 that is Eigen values and the entries and why is this formula correct you please check this there is also another result which at least I will state maybe let me state this result and then approve it in the next lecture but let us kind of a summary of what we have done is the following we discussed the notion of diagonalizability we characterize diagonalizability in terms of the minimal polynomial okay and operators diagonalizable if and only if the minimal polynomial is a product of distinct linear factors. There was another characterization if the dimensions of the Eigen spaces sum up to the dimension of the space then the operator is diagonalizable there is really another characterization that we have got just now this involves invariant direct sums invariant direct sum decompositions if T is diagonalizable and if I can find out certain operators then I know that operators will satisfy these conditions and conversely if there are certain operators and certain numbers related in a specific manner then T is diagonalizable if you look at the general problem then in all these cases what is important is to realize that the minimal polynomial can be written as okay where if T is diagonalizable then R 1 R 2 etc R k they are all 1 if T is not diagonalizable then sum of these will be at least two at least one of them will be two for instance but if T is not diagonalizable we know that it is always triangulable if the minimal polynomial is of this form any operator is triangulable provided all its Eigen values lie in the underlying field which is the same as saying that the minimal polynomial is of this form okay so this kind of encompasses all these both the case of diagonalizability and triangulability what happens if in the case of R for instance the irreducible polynomial the degree of an irreducible polynomial can be two in the case of R in the case of C the degree of an irreducible polynomial is precisely one for instance T square plus 1 is an irreducible polynomial in the field in the space of polynomials over R in the case of C these are linear polynomials okay what happens to a direct sum decomposition in this case that is what we will refer to as a primary decomposition theorem in this general case that is in the case when the Eigen values lie in the underlying field what is the representation what is the direct sum decomposition okay I will write down the statement and prove it tomorrow this is called the primary decomposition theorem in other words this is a kind of a most general theorem that one has all the other results are particular cases all the other results that we have proved can be shown to be particular cases of this V is finite dimension suppose that the minimal polynomial is written as P 1 to the R 1 P 2 to the R 2 etc P k to the R k I am writing an expression which is more general than what I wrote down here suppose that the minimal polynomial can be written in this manner where remember little m is the minimal polynomial for us that is a notation where R i's are positive integers R i's are positive integers what are these P i's P i's are monic irreducible polynomials over the underlying field F now this F is general R C rationals it could be a finite field it could be any other field then we have the following let me write like this let me use this notation this is what I want to say V is the direct sum of the null space of P i t power R i V is the direct sum of the null space of P i t power R i each of these subspaces is invariant under T each of these subspaces I could have called them W i each of these subspaces W i is invariant under T W i is contained in W i this is called the primary decomposition theorem this is the most general that one could do for any operator why is it called primary primary decomposition is clear it is called primary decomposition theorem because see it is a monic polynomial the degree of the coefficient of the highest degree is 1 that is monic irreducible what is irreducibility it is because of this irreducibility that this is called the primary decomposition theorem if a polynomial is irreducible over a field then it is called a prime polynomial that is if a polynomial is irreducible when do you say that a polynomial is reducible if it can be written as a product of two or more polynomials each has degree at least 1 each has degree at least 1 okay x square minus 1 is x minus lambda square minus 1 is lambda plus 1 into lambda minus 1 so lambda square minus 1 is not irreducible over R lambda square plus 1 cannot be written as lambda minus lambda 1 into lambda minus lambda 2 where lambda 1 and lambda 2 are real numbers so lambda square plus 1 is an irreducible polynomial over R lambda square plus k where k is a positive number is an irreducible polynomial over R such polynomials are called prime polynomials if a field is such that the prime polynomials are only linear such a field is called an algebraically closed field this is the same as saying that any polynomial equation has all its zeros in the field any polynomial equation has all its zeros in the field is the same as saying that the polynomial can be factored into a product of linear factors okay so in general if F is not R or C an irreducible polynomial is called a prime polynomial so these are prime polynomials they cannot be factored into a product of polynomials of lesser degree just like how you have composite numbers and prime numbers so this is called primary decomposition theorem for any operator this is the most general that one could have okay the decomposition corresponds to null space of P i t R i I will prove this theorem in the next lecture let me stop here.