 I had written down this theorem yesterday let us look at a proof of that okay and also discuss other consequences how one could use elementary matrices and then as I mentioned yesterday a summary of how elementary matrices are related to invertible matrices then homogeneous equations non-homogeneous okay that is what we will discuss today. So let me write down the theorem once again I have a square matrix with real entries then the following statements are equivalent for a square matrix the following statements are equivalent. First statement is A is invertible second statement A is row equivalent to the n by n identity matrix that is the second statement. Third statement is A is a product of elementary matrix A is a product of elementary matrix okay so we want to see how these statements are equivalent okay let me prove so the usual schema will be to prove A implies B, B implies C and then C implies A okay I will follow that idea then you could mean that these three statements are equivalent. Proof first A implies B okay I want to show that if A is invertible then A is row equivalent to the n by n identity matrix remember the problem that I gave you yesterday if R is a row reduced echelon matrix which I know is invertible then R is equal to identity okay suppose A is invertible let me just write down the statement A suppose that A is invertible we are required to prove that A is row equivalent to identity I will write that statement also to show that A is row equivalent to the n by n identity matrix. Let us take R as the row reduced echelon form of A so let R be the row reduced echelon matrix row equivalent to A see we know that every matrix can be reduced to a row reduced echelon form that form I am calling as R as before then we had seen yesterday that okay if R is a row reduced echelon matrix row equivalent to A it means A is equivalent to R okay this is the notation for that we had seen yesterday that this means I can write R as P times A where P is a product of elementary matrix if B is row equivalent to A then B is P times A where P is a product of elementary matrix that is what we proved yesterday okay so R is equal to P times A where P is a finite product of elementary matrix this was demonstrated yesterday I am making use of that here now what I know is that a product of each elementary matrix is invertible product of invertible matrix is invertible that was proved yesterday so P is invertible I am given that A is invertible I know then that the product P is invertible so R is invertible so R is identity since P is invertible and A is invertible we conclude that R is invertible but we know that if this happens R is a row reduced echelon matrix if it is invertible then it must be equal to identity okay and so R is equal to I so I go back and substitute in this equation okay from this can we directly say A is row equivalent to the identity matrix A is row equivalent to R that is what I have written down here R is identity so O is row equivalent to L that is statement B A is row equivalent to the identity matrix okay so that proves A implies P okay B implies C so you will see that the results today for instance will be kind of a summary of all that we have discussed till now okay B implies C let us prove this next A is row equivalent to given that A is row equivalent to the N by N identity matrix I must show that A is a product of elementary matrices I will appeal to the same result that we proved yesterday if B is row equivalent to A then B is equal to P times A where P is a product of elementary matrices then I equals P times A where P is a product of elementary matrices you remember I want to show that A is a product of elementary matrices if A is is that okay we are proving B implies C I want to show that A is a product of elementary matrices okay so what I will now do is list identity is EL minus 1 etc E 2 E 1 times A that is the I know that P is a product of elementary matrices I have written down this product EL EL minus 1 etc E 1 times A okay please remember that this is P I know that I can write like this now what I know is that each elementary matrix is invertible so I will pre-multiply by first EL inverse then EL minus 1 inverse etc so let me just do the first step also write what I am doing pre-multiply by EL inverse sorry EL inverse each elementary matrix is invertible I know each elementary matrix is invertible so I take this equation pre-multiply by EL inverse for instance I get okay I will do fully EL inverse EL minus 1 inverse etc E 1 inverse pre-multiply by this I get on the left EL inverse EL minus 1 inverse etc E 2 inverse E 1 inverse times identity is equal to A E 1 E 2 EL minus 1 yes that is perfect so this is what I have this means what A is equal to E 1 inverse etc EL inverse which is a product of elementary matrices this is a product of elementary matrices because the inverse of each elementary matrix is another elementary matrix it is an elementary matrix of the same type does not matter it is an elementary matrix so I have written A as a product of elementary matrices what is the reason since the inverse of an elementary matrix is again an elementary matrix this last step because the inverse of any elementary matrix is again an elementary matrix so I have written E as a product of elementary matrix okay so that is that statement B implies C okay finally C implies A, C is A is the statement C is the matrix A is a product of elementary matrices statement A is capital A is invertible but this is something we have seen yesterday okay C implies A we have already seen C implies A let A be a product E 1, E 2, etc E s where each E i is an elementary matrix I must conclude that A is invertible okay but since each E i is invertible the product on the right is invertible that is A is invertible right away there is nothing C implies A follows from the fact that each elementary matrix is invertible and the fact that a product of invertible matrices is invertible okay so this is one summary this is one summary connecting row equivalence elementary matrices invertible matrices one of the corollaries of this result is the following one of the consequences I will state that as a corollary let A be invertible I have an invertible matrix A okay then I know that A is row equivalent to identity what this result says is you do the same sequence of elementary row operations that you do on A to get I on I you will get A inverse from A to I you perform a sequence of elementary row operations this corollary says you perform this sequence of elementary row operations on the N by an identity matrix you will get the inverse matrix A inverse the same sequence of elementary row operations on A yielding I the same sequence of elementary row operations performed on A yielding I when applied to identity yield A inverse the same sequence yields A inverse okay that is a statement proof is the statement clear so this gives a method to construct the inverse of a matrix if you know it is invertible by a sequence of elementary row operations proof there is a sequence of elementary row operations that I do on the matrix A to get the matrix identity which means I can write identity as let us say EL EL minus 1 etc E 2 E 1 A the first operation is E 1 second operation E 2 etc last one is EL I do these operations in this sequence on the matrix A then I get the identity matrix I know that this happens because A is invertible I am using the previous result if A is invertible then A is row equivalent to the identity matrix I want to show that when I do this same sequence of operations on I I get A inverse okay what is given is that A is invertible okay so I post multiply by A inverse since A inverse exists by post multiplying by A inverse this equation I post multiply by A inverse I get the following I get A inverse on the left on the right EL EL minus 1 etc E 1 E 2 E 1 A into A inverse matrix multiplication is associative so I insert a bracket in the end like this but this you see is precisely EL EL minus 1 etc E 2 E 1 being operated on I times I I am emphasizing that this is done on I I could have just left this as it is but this does not tell you that you are performing these operations on I so that is why the last I I have retained here it is not necessary to include this right identity is the multiplicative identity matrix is the multiplicative identity of the multiplication operation so I am making it a point to retain this I to emphasize the last part that you are doing this sequence in the same order on the identity matrix to get A inverse okay let us consolidate do one numerical example and proceed with the theory example where we must know that the matrix is invertible okay so let me give a numerical example find the inverse of the following 3 by 3 matrix 1 1 0 0 1 1 1 0 1 see when I do the elementary row operations I will eventually know if this matrix is invertible okay but I have chosen this example in such a way that it has an inverse that is I have verified that this matrix is row equivalent to identity I am going to use the previous result I will do the same sequence of elementary row operations on I that I do for A to get A inverse okay for that it is convenient to append the identity matrix to the matrix A and do the operations for this A and identity together so I will append this matrix together with the identity 1 1 0 0 1 1 1 0 1 put a vertical line and then write identity of order 3 I must reduce this part to the identity keep doing the same sequence of operations on the second part if this first part reduces the identity the second part should reduce to A inverse this time I am not going to write down the row operations tell me if the steps are correct that is the first step I am keeping the first row as the pivot row the operation elementary row operations are performed keeping the first row the next step is I will keep the second row as it is and then I add second and the third rows so to get 0 0 2 I add the second and the third rows minus 1 1 1 I add the second and the third rows next step is divide by 2 I will do that here it is okay just let me write down once again 1 0 minus 1 1 minus 1 0 0 1 1 0 1 0 0 1 minus 1 by 2 1 by 2 1 by 2 I want to make these two entries 0 so the next operation I will keep the third row as it is I am keeping this as it is just add 1 0 0 second row minus this plus this okay so I see that at the last step I get identity this is equivalent to A sorry this is identity and so I know that the second part must be A inverse so A inverse I take 1 by 2 outside 1 minus 1 1 1 1 minus 1 minus 1 1 1 this is the inverse of A it is always a good practice to check that this satisfies the equation A B equals identity one must actually verify A B that is you can call this one must verify that A B equal to B A equal to identity okay but towards end of today's lecture we will see that it is enough if you verify one of them it is enough if you verify that either A B is identity or B A is identity we will show that a matrix which has a left inverse or a right inverse if the matrix is a square matrix then it will be invertible okay. So it is always a good practice to verify that you have got the inverse correctly some other consequences typically one that would relate to homogeneous equations non-homogeneous equations is the following theorem for a square matrix A the following statements are equivalent first statement as before A is invertible A is invertible second statement the homogeneous system A x equal to 0 has only the trivial solution x is equal to 0 that is the second statement third statement is now about non-homogeneous systems the non-homogeneous system A x equal to B has a solution for all B that is the third statement the non-homogeneous equation has a solution for all B these three statements are equivalent if A is invertible then any non-homogeneous system will have a solution B element of R n okay again we can prove A implies B B implies C C implies A I would follow a slightly different approach I will prove that A and B are equivalent then prove that A and C are equivalent then it would follow that these statements are equivalent okay this is not a very efficient way of doing it but this is rather easier. So I will prove A if and only if B first okay prove file as I mentioned I want to show that A and B are equivalent I want to show that A and B are equivalent okay first A implies B consider so statement A is with me A is in capital A is invertible consider the system A x equal to 0 I know that A inverse exist I will pre-multiply by A inverse A inverse into the 0 vector that is a 0 vector A inverse into A is identity by definition matrix multiplication is associative so this is identity x right hand side is 0 that is x equals 0 so if A is invertible then the homogeneous system has only the 0 solution I must prove the converse B implies A B implies A is something that we have already seen in a slightly different language B implies A the homogeneous system has 0 as the only solution then we know that A is row equivalent to identity okay this was proved sometime ago A is row equivalent okay even the previous no this was proved sometime ago if A x equal to 0 has 0 as only solution then A is row equivalent to the identity so let me write this in this case A is row equivalent to the identity appeal to the previous theorem appeal to the previous theorem B implies A previous theorem statement A is A is invertible statement B is A is row equivalent to Y we have proved that these two are equivalent so I will simply appeal to that in this case A is row equivalent to Y by the previous theorem previous result A holds by the previous result A holds okay so A and B are equivalent that has been demonstrated as I mentioned I will next prove that A and C are equivalent I will first prove A implies C that is again as easy as A implies B earlier okay the first part A implies B if A is invertible we have shown that homogeneous equation has 0 as the only solution so let us prove A implies C consider X as A inverse B given any B I know that A is invertible that statement A I look at the vector X defined as A inverse B B is a given right hand side vector then AX is A inverse B is B that is X solves AX equal to B. Now this I can do for any right hand side vector B all that I need to do is pre-multiply the right hand side vector B by the inverse of A which I know because statement A holds statement A holds A inverse exists pre-multiply the right hand side vector by A inverse I get a solution so that is A implies C I need to show C implies A this is probably the toughest okay so I need to show C implies A that would complete the proof of the theorem I want to show that A is invertible so as before I look at the row reduced echelon form of A let R be the row reduced echelon matrix row equivalent to A I will show that R has the last row non-zero we show that the last row of R is non-zero is it clear that this would mean that R is identity if R is identity then it means A is row equivalent to Y appeal to the previous theorem A is invertible so we are proving C implies A. So if I show that the last row of R is non-zero then it follows that A is invertible to show that the last row of R is non-zero I need to show that this system R X equals I look at a specific right hand side vector all entries 0 except the last entry which is 1 if I show that this system has a solution then it follows the last row of R cannot be 0 because if the last row of R is 0 then the last row of R into X will give me 0 on the right hand side I have 1 so that is not possible. So if I show that this system has a solution X then it follows that the last row of R is not 0 okay so to show that R X equal to 0 has a solution X so this is what we will show I suppose this is clear I have taken this specific entry to be 1 this could actually be any non-zero entry okay but R is row reduced echelon form of A so R is PA where P is a product of elementary matrices if B is row equivalent to A then B is P times A where P is a product of elementary matrices so what happens to this system let us call this as B prime I will call this B star I am rewriting this system R X equal to B star is PA X equal to B star let me call this as system 1 this is system 1 system 1 has been rewritten in this form okay. If R X equal to B star I want to show that this has a solution going back if I show that this system has a solution I should call the system 2 rather this is system 2 I am claiming that these two systems are equivalent I am claiming that if 1 is if X is a solution of system 1 then X is a solution of system 2 and conversely that is clear because R is a row reduced echelon form of A okay. I want to show that this has a solution I will show that system 2 has a solution it then follows that this has a solution if system 1 has a solution then R has to be invertible so A is invertible okay but look at system 2 what is the property of P that I have not used P is a product of elementary matrices elementary matrices are invertible product of invertible matrices is invertible so P is invertible post multiply by P inverse I get A X equals P inverse B star I will call this B double star so A X equals B double star does this system have a solution I have till now not used the fact that system condition C holds I will use that here condition C holds so whatever be the right hand side vector see this is condition C condition C whatever be the right hand side the system A X equal to B has a solution whatever be the right hand side the system A X equal to B this has a solution A is equal to B double star this time this has a solution X so system 2 has a solution so system 1 has a solution so R the last row of R has to be non-zero so A is invertible that is the last line of the proof so R is equal to I and so A is row equivalent to I so that by the previous theorem A is invertible that is the proof of this theorem okay. You have any questions let us move on we need to see to conclude this topic we need 2 more results let me complete them in today's lecture 1 corollary of this result it is actually a corollary of the previous result let me write down this result I am not going to prove this if B is row equivalent to A then B equal to P A where P is invertible if B is row equivalent to A then B can be written as P times A where P is invertible what we have proved earlier is that if B is row equivalent to A then B equal to P times A where P is a product of elementary matrices we know that if you have a product of elementary matrices then that must be invertible conversely if B equal to P times A P is invertible then we know by the previous theorem that P can be written as a product of elementary matrices and so B is something like EL EL minus 1 etc E 1 times A so B is row equivalent to A that is a proof okay so please fill up the gaps write down the complete proof yourselves that is one corollary another consequence is the following this is something that I mentioned in that numerical example if a square matrix has a right inverse or a left inverse if a square matrix is either right invertible or left invertible then it is invertible if A is left invertible okay let me write this as if A has a left inverse or a right inverse then A is invertible so for square for a square matrix it is enough to verify one equation either AB equals identity or BA equals identity in order to show that it is invertible okay quick proof let us first check the left inverse case suppose that A has a left inverse I will call it B suppose A has a left inverse I am calling that as B then BA equals identity left inverse so when I post multiply pre multiply A by B I get identity I want to show that A is invertible I want to appeal to the previous theorem which connects invertibility with homogeneous systems I will show that the system A x equal to 0 has only x equal to 0 as a solution so consider A x equal to B A x equal to 0 consider A x equal to 0 I want to show that the system has 0 as the only solution can you guess the next step pre multiply by B BA x is B 0 B into 0 is 0 BA is identity x is equal to 0 so what I have shown is that A x equal to 0 implies x equal to 0 that is 0 is the only solution of the homogeneous system appeal to the previous theorem I know that A is invertible that is a left inverse case if it has a left inverse then we have shown that it is invertible we must deal with the right inverse case but in mathematics there is always this reduction I would like to use what I have proved just now can I reduce this step to the previous step can I make use of the fact that if A has a right inverse then some matrix has a left inverse second part let this is the second part suppose that A has a right inverse there exists C such that AC equals identity that is A has a right inverse then C has a left inverse A by appealing to the first part I know that C is invertible I know that C is invertible by the first part go back to this equation post multiply by C inverse go back to this equation AC equal to identity post multiplying post multiplying by C inverse we get AC C inverse equals C inverse that is A equals C inverse but I know that inverse of the inverse is the original matrix so A inverse is C that is C is the inverse of A that is A is invertible A is invertible and its inverse is C okay so this is another consequence once I did invertibility for a square matrix implies invertibility finally we have the last result I have A as a product of K matrices where each AI is square each AI is square so capital A is a square matrix then A is invertible if and only if each AI is invertible okay this is a this is the last result in this topic this again follows from what we have discussed earlier one way is easy anyway if each AI is invertible then the product is invertible so A is invertible so let me just write in short hand notation AI invertible implies A is invertible this is done each component if each factor is invertible then entire product is invertible conversely we must show that if A is invertible then each factor is invertible this we have not seen before suppose that A is invertible we would like to show that each factor is invertible okay I will proceed by showing a single factor is invertible at a time I will consider the system AKX equal to 0 consider the system coming from the last factor homogeneous coming from the last factor consider AKX equal to 0 I pre-multiply by AK minus 1 etc A1 then I have A1 A2 etc AK minus 1 AKX equals 0 I pre-multiply the right hand side also by this but I am having the right hand side vector as 0 so the resultant vector is also 0 but this entire product is A so I have AX equal to 0 I have AX equal to 0 but I know A is invertible so by the previous theorem X equal to 0 is the only solution since A is invertible it follows that X is equal to 0 that is what we have shown is that AKX equal to 0 has X equal to 0 as the only solution AKX equal to 0 has X equal to 0 as the only solution that is what we have shown remember we have started with AKX equal to 0 that implies AX equal to 0 since A is invertible X is equal to 0 so this implies AK is invertible it is square so AK is invertible I will go back to this equation this formula for A this formula for A becomes A into AK inverse I will call it A prime A into AK inverse this time it is A1 A2 etc AK minus 1 I call it A prime this time it is A1 A2 etc AK minus 1 I am post multiplying by AK inverse which I know exists now yes A is invertible AK is invertible inverse is invertible so A prime is invertible so AK minus 1 must be invertible by the same argument as we have seen just now it follows that each factor is invertible proceeding similarly it follows that each AI is invertible okay so you see that again it is idea of homogeneous equations having 0 as the only solution that we have made use of improving that a certain matrix is invertible okay so that completes our discussion on matrices elementary row operations formalizing Gaussian elimination then discussing the notion of elementary matrix discussing notion of an invertible matrix finally the summary connecting invertibility with product of elementary matrices A the invertible matrix being necessarily row equivalent to identity matrix homogeneous equation necessarily having 0 as only solution non-homogeneous equation always having a solution whatever be the right hand side vector okay so that is the summary we would from the next lecture onwards we would discuss the notion of vector spaces examples of vector spaces linear transformations matrix representations properties of linear transformations and other things okay I will stop here.