 So, in this lecture we shall we shall sketch a proof of the fact that G l n r plus. So, this is those n cross n matrices. So, is that determinant of A is positive is path connected. So, we shall we shall see this as an application of whatever we have done so far and ok. So, let us denote. So, denote G by this G l n r plus ok. So, we will do this. So, first of all it suffices to show to connect to show that any matrix A in G can be connected to the identity by a path in G right. So, we have our G over here. So, given any matrix A we can connected to identity right using a path. So, that is what we are going to prove and we will do it in several steps. So, let us begin. So, let A be in G right. So, then the first step is then we can join A to B in G. So, when we say join A to B in G we always mean join by a path which is completely contained inside G right. So, where B is such that B 1 1 the first entry is non-zero yeah. So, this B 1 1 that should be non-zero ok. So, how do we do this? So, if A 1 1 is non-zero then we just take to be the constant path right. So, gamma of t is equal to A. So, the constant path is continuous and therefore, we can just take B to be equal to A right. So, on the other hand if A 1 1 is equal to 0. So, then since G l n since G is an open subset of M n r right and M n r has the product apology there exists an epsilon positive such that the set there is those B in M n r with the absolute value of A i j minus the absolute minus B i j is strictly less than epsilon for all i comma j right. So, this is going to be contained in G l n this G simply because G l n r plus is open inside M n r yeah and ok. So, we can for instance and M n r has a standard apology or product apology yeah. So, we can see this in many ways ok. So, we just take any B in U with B 1 1 not equal to 0 then take the path gamma from 0 1 to U given by gamma of t is equal to t times A plus 1 minus t times ok. So, this path is going to be completely contained inside U because each coordinate of gamma t i j right. So, this is A i j and this B i j. So, the distance between them is less than epsilon. So, for any t gamma t i j is going to be over here somewhere in between A i j and B i j. So, the distance of that from A i j is going to be less than epsilon. So, it is going to be contained inside U which is contained inside G right. So, this implies that. So, this gives a required right. So, we have joined A using a path to this matrix B and B 1 1 is non 0 ok. So, let us go to step 2 if B is in G and B 1 1 is non 0. So, we let. So, we call this E i j is these are the elementary matrices. So, these are matrices of the type ok. So, instead of E i j right. So, right. So, then there exists a matrix E 1 of the type 1 these entries are something and on the diagonal we have 1 and all these are 0. So, there is a matrix of this type says that E 1 when we multiply it on the left by B all these except for the first one all the other entries in the first column becomes 0 ok. So, this first one is going to be lambda. So, let us assume that B 1 1 is lambda ok. So, similarly there exists a matrix E 2 right of the type 1. So, the first row is of this type and in the diagonal we have 1's and all the other entries are 0 right such that E 1 B E 2 is of this type. So, lambda all the other entries in the first row are 0 all the other entries in the first column are 0 and here we have a matrix. So, we consider this map 0 1 this is G L and R plus this is given as follows. So, T goes to 1 minus T times identity plus T times E 1 into B into 1 minus T times identity plus T times E 2 ok. So, we need to check that first as a map of sets the image actually lands inside G L and R plus, but when we take determinant for each T let us look at this matrix 1 minus T times identity plus T times E 1 1 E 1 right. So, E 1 is a matrix of this type. So, one easily checks that once again this is going to be a matrix of this type. So, this implies determinant of this matrix is equal to 1 ok. So, similarly determinant of 1 minus T plus T E 2 is going to be 1. So, this matrix is now going to be of this type here 0 and the first is going to be of the same type as E 2 ok. So, this implies that determinant of gamma of T is actually equal to determinant of B which we know is positive. So, therefore, as a map of sets indeed the image of gamma lies inside G L and R plus next we want to check a discontinuous here, but gamma to check gamma is continuous we can just since G L and R plus since G L and R plus is contained in M and R and has a subspace topology it is it suffices to check that the coordinates of gamma are continuous right, but the coordinates of gamma are polynomials, but these coordinates are polynomials and so are continuous. That is easily checked when we multiply out these matrices these three matrices right it is clear that the coordinates will be polynomials in T and so gamma is continuous since each coordinate function is continuous. So, we have this path from 0 1 to G L and R plus now gamma of 0 is equal to we can check it is equal to easily it is equal to B and gamma of 1 is equal to E 1 B E 2 which we know is a matrix of this type ok. So, this completes step 2. So, this completes step 2. So, if B is a matrix such that B 1 1 is not 0 then B can be connected to a matrix of this type ok. So, let us go to step 3. So, if lambda is positive then this matrix lambda 0s D let us look at this path. So, so consider C to be equal to lambda all these 0s right. So, then and let lambda be positive right. So, then C can be joined to 1 matrix D prime ok. So, obviously this is in G yeah. So, we are starting with some matrix C in G of this type then we can join it to some D prime like this here where D prime belongs to G L n minus 1 R plus. So, how do we do that? So, we look at this path. So, we look at this matrix. So, T times lambda inverse plus 1 minus T. So, this is the diagonal matrix. So, in A 1 1 it has this entry and in all the other diagonal entries it has 1 right. So, what is this what happens over here at this entry? So, this is joining. So, lambda is positive. So, lambda inverse lies over here and let us say 1 is over here. So, then this is the straight line joining 1 with lambda inverse. So, and we just multiply this with C ok. So, when we take determinant. So, let us just once again we check that when we take this is equal to gamma of T. So, determinant of gamma of T is equal to T times lambda inverse plus 1 minus T into determinant of C which is therefore positive. So, therefore the image of gamma actually lands inside G L n R plus and once again this is continuous because each of the coordinate functions are polynomials in T in the variable T right. So, and notice that gamma of 0 is equal to C and gamma of 1 is. So, lambda inverse 1 times C which is lambda T. So, this is equal to 1 0 less than G L ok. So, if so similarly if lambda is strictly less than 0 then this path T times mod lambda inverse plus 1 minus T 1 1 times C connects C with minus 1 a matrix of this type right. So, let us say some D prime. So, now we want to show this matrix D prime can be connected to a matrix of the form 1 D prime prime. So, this is in G and we want to connect this to in G right via a path in G. So, for this let us check that. So, consider this matrix T times minus 1 this 2 cross 2 matrix this is equal to minus T and this has determinant T square plus 1 minus T whole square which is positive right and similarly T times this matrix plus 1 minus T times 0 1 has determinant positive. So, what this means is that this minus 1 minus 1 can be joined to 1 comma 1 in G L 2 R plus right. So, in fact, this first thing that is a path from 0 1 ok that is a path joining minus 1 minus 1 to 0 1 minus 1 0 in this path is completely contained inside G L 2 R plus right and similarly there is a path completely contained inside G L 2 R plus which joins this matrix to 1 1 0 0 the identity right. So, therefore, this minus identity can be joined to identity in G L 2 using a path inside G L 2 R plus right. So, we will use this by a path. So, there exists such a path because we can just come put these two paths together as we had seen when we proved that being when we defined path equivalence classes ok. So, now we define a map from gamma to 0 from 0 1 using this gamma ok let us say gamma tilde to G L and R plus by T and here all 0s here identity and 0 over here. So, clearly this path joins minus 1 minus 1 to 1 1 to the identity in G L and R plus. So, then the path from 0 1 to G L and R plus given by T goes to ok. So, this path is what we are calling gamma tilde T. So, gamma tilde T times this minus 1 d prime right. So, it joins minus 1 d prime to 1 prime prime let us say ok. So, thus we conclude that if C in G L and R plus is of the type it is a block matrix is of this type yeah. So, then C can be joined by a path in G L and R plus to a matrix 2 of the type right and then d prime prime is forced to be in G L and minus 1 R. So, now, so what are we done so far we start with a matrix A we connected it by a path to a matrix B such that B 1 1 is not equal to 0 then we connected it to a matrix C such that C is of this type. So, then we connected this to a matrix of the type 1 right. So, by induction. So, by induction on n we may assume that d prime prime can be connected identity n minus 1 in G L and R plus in G L using a path right. So, then the path 0. So, gamma of T in G L and R plus connects prime to the identity. So, we have this path is connected right. Thus we have proved there is only a sketch and I will leave it as an exercise to fill in the details and convince yourself that all the arguments are correct every element A in G L and R plus can be connected using a continuous path gamma from 0 1 to G L and R plus to the identity. So, this proves that G L and R plus is connected is path connected and in fact, connected because we know that part connected spaces are connected. So, we will end this lecture here.