 See a little preamble before I move on this course is about linear transformations on vector spaces okay. Now concrete realizations of linear transformations on vector spaces are matrices that explains why this indulgence on matrices okay for so many lectures. We will have another two lectures probably including today's lecture on matrices, elementary row operations, properties, etc okay. Today I would like to look at certain qualitative aspects of the elementary row operations, row equivalence, row reduced echelon form, etc okay. But before that I will recollect what we discussed last time okay towards the end of last lecture I remember having discussed this problem R is a row reduced echelon form okay. So this is to make a quick reminder R is a row reduced echelon matrix I am writing down the expanded form of R x equal to 0 okay. I was writing down the expanded form of R x equal to 0 what I know is that small r is the number of non-zero rows of capital R okay this is my notation. So let us look at this equation R x equal to 0 I will have something like this R is a row reduced echelon form. So there are certain zeros I have the first non-zero entry that is 1 and then certain other entries then the second row will be 0, etc let us say up to here and then a 1 here, etc, etc I have zeros here, 1 somewhere here okay. So this is the 1st R rows let me write this is 1st R rows let me write on the left R non-zero rows of capital R, R non-zero rows of capital R that corresponds to this the rest of the rows are 0 okay this is my R, M minus R totally there are M equations in N unknowns, M minus R 0 rows that is the last part okay. Now this R into x is equal to 0 so I will do matrix multiplication that into x, x 1, x 2, etc I will call this some x star this is another x star, etc this is x n this is the 0 vector okay I am already assuming you know matrix multiplication but in any case we will review it a little later. So what does matrix multiplication tell us? The right hand side vector 0 the 1st 0 the right hand side number this 0 is the dot product of this row with this column okay. Now this 1 appears in the column number C 1 this corresponds to certain x I am calling that x C 1 this is column C 1 this 1 appears in column C 2 this corresponds to certain other unknown I am calling that x C 2 etc. So I have what I have done is to effect a relabeling of the unknowns as those corresponding to the leading non-zero entry of the non-zero rows of R those are x C 1, x C 2, etc x C R the other unknowns I am calling them V 1, V 2, etc V N minus R and just relabeling them for the purpose that I will write down the R equations coming out of this homogeneous system and then it gives us a qualitative information. So based on this what I had written is that let V 1, V 2, etc V N minus R denote the other N minus R I am sorry N minus R denote the other N minus R variables okay means not the ones we have denoted by x C 1, x C 2, etc x C R just to make it a little more precise. Then these R equations can be written using a summation formula R x equal to 0 is the same as x C 1 plus summation j equals 1 to N minus R alpha 1 j V j equals 0 that is I am now denoting the first row entries after this 1 as alpha 1 C 1 plus 1 alpha 1 C 1 plus 2 etc this is alpha 1 C this is alpha 1 N etc that is my notation this is the first equation then second equation is written similarly alpha 2 j V j etc x C R the last non-zero row gives me this equation x C R plus summation j equal to 1 to N minus R alpha r j V j equals 0 the last non-zero row gives me this equation. So what this means is that so if N minus R is positive then arbitrary values could be assigned to V 1 etc V N minus R and x C 1 etc x C R are determined uniquely for every choice of V 1 V 2 etc V N minus R I will have unique x C 1 x C 2 etc x C R coming from this set of R equations for every such choice of V 1 etc V N minus R I get x C 1 x C 2 etc x C R that gives me the set of all solutions okay and we had also observed towards the end of last class that if R is strictly less than N like what I have assumed here if R is strictly less than N then R x equal to 0 has a non-trivial solution a non-zero we call it a non-trivial solution. A non-zero solution means at least one coordinate at least one unknown is not 0 at least one unknown is not a 0 okay remember this is important because this is a homogeneous system we know it is consistent it always has 0 as a solution so one is interested in a non-zero solution. So there is a guarantee if R is less than N then the homogeneous system has a non-trivial solution I also left you with a little problem if R is a square matrix and the small R is equal to N I asked you the structure of R so what is the structure of R? Identity matrix. Is that clear to all of you so R is equal to identity of order N identity matrix of order N so just for the sake of completeness I write the identity matrix okay it has only that the non-zero entries the off diagonal entries are 0 the principal diagonal entries are all 1 this is the identity matrix of order N that is what happens if R is a so the only row reduced echelon matrix whose number of non-zero rows equal to the order of the matrix is the identity matrix okay we will need this a little later okay let us now proceed with today's lecture a little qualitative information. So in particular I want to prove 2 or 3 results let me start with the first one I consider a rectangular homogeneous system where the number of equations is strictly less than the number of unknowns a homogeneous system is what I am going to consider the number of equations strictly less than the number of unknowns okay this is a fundamental result of linear algebra that a rectangular system of homogeneous linear equations where the number of equations is strictly less than the number of unknowns always has a non-trivial solution okay the proof will essentially make use of what we have proved what we have stated just now let R be the row reduced echelon form echelon matrix row equivalent to the matrix A okay standard notation capital R is a row reduced echelon form of A that is the row reduced echelon matrix row equivalent to A then the systems A x equal to 0 and R x equal to 0 we call them as equivalent systems that is they have the same set of solutions A x equal to 0 and R x equal to 0 have the same set of solutions. As before let us keep R as a variable that is the number of non-zero rows of R let R equal the number of non-zero rows of capital R the number of non-zero rows of R cannot exceed the number of rows of R so R is less than or equal to m, m is the number of rows and what is given is that m is less than n as part of the data of the problem m is less than n okay so R is strictly less than n by the observation that we made just now it follows that there is a non-trivial solution okay. This R x equal to 0 has a non-trivial solution so is x so is A x equal to 0 so is the system A x equal to 0 because a set of solutions of R x equal to 0 coincide to the set of solutions of A x equal to 0 okay so this is based on the observation that if R is strictly less than n then R x equal to 0 has a non-trivial solution at least one non-trivial solution okay okay. For the square case we can do a little better square homogeneous system when does it have a non-zero non-trivial solution that will be my next result I have a square matrix with the real entries then the system A x equal to 0 has a non-trivial solution if and only if A is row equivalent to the identity matrix of order n has 0 as the only solution, has 0 as the only solution if and only if A is row equivalent to i okay this do not say A is row equivalent to i identity is the row reduced we know it is a row reduced echelon matrix okay okay so now there are two parts unlike the first the previous theorem there are two parts if and only if let us prove the easy one first suppose that A is row equivalent to i then the systems A x equal to 0 and I x equal to 0 by definition have the same set of solutions A x equal to 0 and I x equal to 0 have the same set of solutions that is I am applying this fact that we had observed previously that if R is a row reduced echelon form of A then A x equal to 0 and R x equal to 0 have the same set of solutions I is a row reduced echelon form of A so these two systems have the same set of solutions but I x equal to 0 has only x as a solution this implies x equal to 0 okay that is A x equal to 0 has only the trivial solutions I x equal to 0 implies x equal to 0 so this is the easy part of the proof the homogenous system A x equal to 0 has only 0 as a solution if A is row equivalent to the identity matrix we need to prove the converse we next prove the converse so again we need to look at the row reduced echelon form of A and then show that this row reduced echelon form is equal to identity so conversely suppose that x equal to 0 is the only solution of the system A x equal to 0 so we will look at R x equal to 0 let R be the row reduced echelon matrix row equivalent to A and small r is equal to the number of non-zero rows of capital R okay then what we have is that then R x equal to 0 has x equal to 0 as the only solution x equal to 0 as the only solution because A x equal to 0 has x equal to 0 as the only solution which means the system R x equal to 0 does not have a non trivial solution appeal to the previous result by appealing to the previous result what follows is that R is greater than or equal to n okay so let me write down this state so R x equal to 0 has no non-trivial solutions R x equal to 0 has no non-trivial solutions so R is at least n okay but R is a number of non-zero rows of capital R it cannot exceed n number of rows number of non-zero rows of R cannot exceed the number of rows of R and so R is equal to n that is I have a row reduced echelon matrix which a square row reduced echelon matrix with the property that the number of non-zero rows is equal to the order of the matrix so R is equal to i as we have observed before okay so what we have shown is that if A x equal to 0 has x equal to 0 as the only solution that is if this homogeneous system of homogeneous square system does not have a non-trivial solution then the matrix A is row equivalent to the identity matrix okay. So much for the homogeneous case let us now move on to the non-homogeneous case that is when the right hand side is not the zero vector whether it is consider the non-homogeneous case and remember that the essential difference between a homogeneous system and a non-homogeneous system is that a non-homogeneous system need not have a solution okay homogeneous system always has a solution so this information also should be obtainable when we do the elementary row operations ideally when you do the elementary row operations on a system A x equal to b we must also know whether at the end of the algorithm for instance whether it can tell us if the system has a solution okay. Now that information if you have a row reduced echelon matrix is immediate so that is what I want to illustrate next so I am considering the non-homogeneous system A x equal to b okay so let us recall that A is an m cross n matrix m rows and n columns x is the unknown vector x 1 x 2 etc x n b is a requirement vector b 1 b 2 etc b m I am writing them as column vectors what I will do now is I not only have the matrix A I also have the right hand side vector so I will form a new matrix call it A prime I will adjoin B I will adjoin B as a last column of A prime so my new matrix A prime is A comma B I will apply elementary row operations on this then I will get R prime row reduced echelon matrix now this R prime is R comma D okay what I do is I apply elementary row operations on A I get R and then apply the same set of elementary row operations in the same sequence to the matrix that is a column vector b and then I get the vector D okay that matrix I am calling that as R prime okay. Now what is not clear but can be shown is that if you take this matrix as it is and then do the elementary row operations on this treating it as a single matrix then you will eventually end up with R comma D okay that is there are two ways of proceeding one way is to apply elementary row operations on A to get the row reduced echelon matrix R apply the same elementary row operations on B you will get the matrix D you will get the vector D that is one way of proceeding the other way is to take this as a matrix as a single matrix apply elementary row operations write down you will get it of this form the first part will be row equivalent to the first part second part will be row equivalent to the second part okay. So this is what we have done from what we have discussed earlier you have row equivalence which means if you look at the system A X equal to B I am calling this usually as I I will also look at R X equal to D I am calling that system 2 if R comma D is obtained from A comma B by these elementary row operations then what I know is that any solution of A X equal to B is a solution of R X equal to D and conversely any solution of R X equal to D is a solution of A X equal to B because elementary row operations correspond to linear combinations of equations of that system any equation of system 2 is a linear combination of the equations of system 1 and conversely okay and so we know that this if this happens then these two systems have the same solution which we observed in the homogeneous case that holds in the non-homogeneous case also. In the non-homogeneous case what do these equations represent so I look at the expanded version of this something like R X equal to 0 that I had written down in the beginning let me write down R X equal to D as before small R is a number of non-zero rows of R X C 1 X C 2 etc they correspond to the first leading non-zero entries of the R non-zero rows V 1 V 2 etc V n minus R they correspond to the other variables then I get the following X C 1 plus these are very similar to the previous the homogeneous case alpha 1 J V J this is the now D 1 okay it is 0 earlier X C 2 plus summation J equals 1 to n minus R alpha 2 J V J equals D 2 etc X C R plus summation J equal to 1 to n minus R alpha R J V J equals D R there is nothing much in these equations what happens to the other n minus M minus R equations what are the other M minus R equations these M minus R equations in the homogeneous case do not matter they do not contribute anything they do not give any they do not lead to any constraints they do not give any information but the last M minus R in the non-homogeneous case in fact have the essence of the fact that the system has a solution or does not have a solution so what are the last M minus R equations the non-homogeneous case remember on the left I have on the left I have R say I am writing down R X equal to D okay let me just do it again so that it is transparent R is there are certain 0s 1 here certain entries here some 0s 1 here certain other entries here etc 0 here 1 etc these correspond to the R non-zero rows the other entries are all 0 this is M minus R this is R this into X that is my D my D is equal to this I have written down the first R non-zero equations the last equations the last equations give me 0 times X 1 etc X n so it is just 0 on the right hand side I have D R plus 1 next equation 0 equals D R plus 2 etc D M this is what I must have now these I told you do not appear on the homogeneous case because D is 0 in the homogeneous case D is 0 in the homogeneous case because B is 0 you have a 0 column you have a 0 matrix you do elementary row operations it will remain 0 there will be no change. So B is 0 in the homogeneous case so D is 0 so these equations do not figure in the homogeneous case these things have to be satisfied in order that the system has a solution in the non-homogeneous case okay now that is the condition what is the condition that AX equal to B has a solution this is the condition let me write down AX equal to B has a solution if and only if D I is 0 for all I greater than R this is M for all I greater than R D R plus 1 D R plus 2 these are numbers constituting the vector D these must be 0 this is the condition that must be satisfied for the system to have a solution again if these conditions are satisfied then one will fix arbitrary values to V 1 V 2 etc V n minus R fix determine the other variables X C 1 etc X C R and one can write down the set of all solutions okay so these are the conditions that are necessary and sufficient for the system AX equal to B to have a solution okay okay so let us consolidate by looking out working out one or two examples numerical examples you have any questions I want to illustrate two examples one where the system as a solution one where the system does not have a solution okay if you do not have any questions let me proceed so I want you to consider AX equal to B for some vector B where my matrix A is this 3 minus 1 2 2 1 1 1 minus 3 0 okay this is my matrix I would like to verify by using elementary row operations to verify if this system is consistent if it is consistent can I write down all the solutions okay let us do the elementary row operations we are so familiar now I do not want to write down what operations I am performing so I know that A is equivalent to let us say I interchange these two so 1 minus 3 0 2 1 3 minus 1 2 is equivalent to 1 minus 3 0 minus 2 times this is 0 minus 3 6 plus 1 7 this will remain as it is so please check my calculations minus 2 times this plus this minus 2 that is 6 plus 1 7 this will remain as it is minus 3 times this is 0 minus 3 9 9 minus 1 8 minus this remains the same the next operation is to divide by 7 okay so maybe I will do that here itself 1 minus 3 0 0 1 1 by 7 0 8 2 that is my next operation dividing the second row by 7 then I know this is equivalent to I will keep the second row this entry remains the same 3 times this is 0 3 by 7 I should have done it along with B no problem I just append it okay this is my A comma B A prime this will be interchange the change of these two B 3 B 2 B 1 this time I am keeping this as the pivot row minus 2 times this minus 2 times this plus this B 2 minus 2 B 3 minus 3 times this plus this okay that is equivalent to we will do that here this is fine then I have divided by 7 I can do one more operation so that is I will divide by 8 or 2 is better 0 4 1 I am dividing by 2 I am dividing by 2 this becomes 4 this is 1 1 by 2 times that then yeah I need to use this as the pivot row so that is equivalent to I will do these two simultaneously the second row I keep it as it is the first row is 3 times this this will remain 1 this will remain 0 this is 3 times this 3 by 7 3 times this plus this tell me if this is fine 1 by 7 3 B 2 plus B 3 minus 6 by 7 plus 7 by 7 that is 1 by 7 3 times the second row okay please check the calculations here and this one will be minus 4 times this plus this minus 4 1 minus 4 by 7 is 3 by 7 minus 4 times this plus this minus 4 times this plus this I want to volunteer for that entry minus 4 times this plus this 8 by 7 minus 3 by 2 16 minus 21 minus 5 by 14 you agree with this minus 4 times this plus this so 1 by 2 B 1 remains the same B 2 will be minus 4 by 7 B 2 B 3 only will change minus 4 by 7 that makes it 8 by 7 8 by 7 minus 3 by 2 essentially that 16 minus 21 minus 5 by 14 B 3 agreed okay one last step then I must make this 1 and then reduce this so this is equivalent to I divide this by 1 to get 1 0 3 by 7 1 by 7 3 B 2 plus B 3 0 1 1 by 7 1 by 7 B 2 minus 2 B 3 divide by 7 by 3 divide by 7 0 0 1 then this last entry is 7 by 3 1 by 2 B 1 minus 4 by 7 B 2 minus 5 by 14 B 3 actually I can solve the system as it is from here but I want this row reduced echelon form so I must do two more operations which will make sure that these two entries are 0 okay so let me complete that let me do these two operations also write down the solution is it clear even at this stage that the system has a solution for whatever B okay let me proceed the last row is kept as 0 0 1 7 by 3 etc please I am leaving this I want to make these two 0 so I am going to leave the other calculations also determine those two stars lots of fractions involved so I am going to leave that okay but what is clear is that this system has a solution for all B because the number of see what I have done in the last step is keep the last row as it is second row I am using this I am multiplying this by minus 1 by 7 I am adding it this makes this entry 0 multiply by minus 3 by 7 to add to this this entry will be 0 there will be appropriate changes in the first two rows I am leaving them for you to calculate but what is clear from this example is that the number of yeah what is clear is that R is equal to 3 the order of the matrix A and so you do not have to verify this Di equal to 0 for this this is vacuously true okay and so for this system for this coefficient matrix A whatever be the right hand side vector B this will always have a solution so Ax equal to B has a solution for all right hand side vectors B okay this has a solution for all right hand side vectors B okay please remember that we could make this conclusion based on the fact that these equations are not present these equations are not present if these equations are present then you must verify that these are satisfied but they are not present here the number of non-zero rows is 3 the order of A so I have only D1, D2, D3 there is no D4 okay so that is the first example this always has a solution let us look at another example A this is my A this time I have the vector B okay B is this vector the number of unknowns is 4 the number of equations is 3 okay so let us quickly do the elementary row operation see what happens so I am starting now with A prime okay let me do the first operation at once because this entry is 1 for me as made matters simple 1 minus 2 1 2 the right hand side vector is 1 minus this plus this okay that is 0 3 2 minus 1 1 minus this plus this 0 9 minus 6 minus 3 2 what is going on this I know this row equivalent to the second row I will divide by 3 okay that is the next set of operations this is not 8 in the row reduced echelon form okay but I can make the last entry 1 this is still not in the row reduced echelon for that last non-zero row is not this okay but I can multiply by minus 1 and then remove this okay but this is this is the final row reduced echelon form okay this is the final row reduced echelon form now this is my D my vector D in this problem is 5 by 3 1 by 3 1 R is 2 the number of non-zero rows of capital R which is row equivalent to A please remember not this this has 3 non-zero rows let me confirm R equal to is equal to the number of non-zero rows of R which is row equivalent to A okay which means I must verify is this true is this true R is 2 so I must verify that D 3 is 0 it is not true so A x equal to B has no solution is that okay number of non-zero rows of capital R which is row equivalent to A what we have done here is A prime this is what I am calling as A prime right from the beginning this A prime has 3 non-zero rows A has only this A prime is row equivalent to a matrix which has 3 non-zero rows A is row equivalent to a matrix which has only 2 non-zero rows when there is a difference of this that will lead to inconsistency okay so this problem this system does not have a solution that is clear okay if I change this B to B 1 B 2 B 3 we can also determine the condition under which the system will have a solution so I am not going to give the details I am sure you can see this happening in the previous problem this system will have a solution for all B which satisfy the extra condition can you tell me the extra condition that must be satisfied so I want the answer in terms of B I know that D 3 must be 0 okay you please I leave this as an exercise then I have something like let us say R okay beta times B 1 plus gamma times B 2 plus delta times B 3 equal to 0 my question is what are these values beta gamma delta and why is it that I have written a single equation not several equations with this I will stop.