Transpose of a matrix product

I've got a handful of matrices here I have the matrix a that's an M by n matrix you can see it has n columns and M rows and actually let me throw in one entry there it might be useful this is the this is the jth column so the M throw is going to look like this am J that's that entry right there and then I have matrix B defined similarly but instead of being an M by n matrix B is an N by M matrix and so this entry right here let me just and I'm I realize this might be useful this is going to be my nth row and it's going to be my jth column and then I also wrote out there transposes so if you look at the transpose of B B was an N by M matrix now the transpose is going to be an M by n matrix and each of its rows become its columns and the same thing I did for a its transpose is right there a was M by n the transpose is n by M and each of these rows become each of these columns now fair enough let's define two new matrices right now let's define the matrix C let's define the matrix the C we do it over here or we do it over here I think the real estate will be valuable in this video let's define my matrix C is being equal to the product of a and B so what's what are going to be the dimensions of C well M by n matrix times an N by M matrix these two have to be equal even for the matrix matrix product to be defined it's going to result with an M by M matrix so it's going to be an M by M matrix now let's define another matrix let's call it B D and it's equal to B transpose B transpose times a transpose and the dimensions are going to be the same because this is an M by n times and n by M so these are the same so which is a requirement for this product to be defined and so the dimensions of V are going to be M by M M by M so let's explore a little bit with the different entries see our going to look like so let me write my matrix C right here so it's just going to have a bunch of entries C 1 1 C 2 2 C 1 2 all the way to C 1 M you can imagine if it's an M by M matrix you have CMM over here you know how to you know how this drill goes but what I'm curious about is how do we figure out what C the general CIJ is how do we figure out what a particular entry is we know that C is the product of a and B so to get to a particular entry in C and we've seen this before so it's a particular entry in C so C IJ it's going to be you can view it as the dot product of the I throw in a the I throw in a with the jth column in B with the jth column in B just like that and what's that going to be equal to it's going to be equal to AI 1 AI 1 times b 1j b 1j plus AI to a I 2 times B 2 j b2 j and you're just going to keep going until you get to the last term here a I n AI n times the last term here BN j BN j fair enough now what about our matrix D what are its entry is going to look like so D similarly it's going to look like you know you're going to have D 1 1 D 1 2 sorry D 1 2 all the way to D 1 M you're gonna have D mm I could keep putting entries here but I'm curious about just that some general entry here let's say I want to find D D sub J I D sub J I that's what I want to find so I want to find a general way for any particular entry of D the jate row a nice column which is a little bit different than the convention we normally use these letters but it's fine the first one is DS row the second one is DS is this entries column so how do we figure that out so d sub J I it's going to be equal to D is the product of these two guys so to get D jethro and I column entry here we essentially take the dot product of the J throw here so we're gonna take the dot product of the J throw here which is that right there with the ith column of a with the ith column of a which is that right there so I'm going to take the dot product of that and you might already see something interesting here this thing right here is equivalent to that thing right there and this thing right here is equivalent to that thing right there because we took the transpose is but let's actually just write it out so what is this dot product going to be equal to its going to be bi J well let me write it this way it's going to be B IJ times a I 1 or we could write it as a I 1 times b 1j and it's going to be plus B 2 J times AI 2 which is the same thing as AI 2 times B 2 j AI 2 times B 2 J and you're going to keep going until you get BN j times AI n or you could write that as AI n times BN j b & j now notice something these two things are equivalent they're completely equivalent statement the d sub ji is equivalent to c sub IJ let me write that so D or I could write C sub IJ is equivalent to D sub J I or another way you could say it is anything that's at row all the entries that set row I column J in C is now in row J column I in D and this is true for all the entries true for all entries I stayed as general as possible so what does this mean this is the definition of a transpose so we now get that C C transpose is equal to D or you could say that C is equal to D transpose now this is pretty interesting because how did we define these two we said that our matrix C is equal to our matrices matrix the matrix product a and B and we said that D is equal to our matrix product B transpose times a transpose I did that those definitions right there here are the definitions now we just found out that D is equal to the transpose of C so we could write that C transpose which is the same thing as a times B transpose is equal to D so it is equal to D which is just B transpose a transpose and this is a pretty neat takeaway this is a pretty neat takeaway here if I take the product of two matrices and then transpose it it's equivalent to switching the order and or transposing them and then taking the product of the reversed order B transpose a transpose which is a pretty pretty neat takeaway and you can actually extend this to an arbitrary number of matrices that you're taking the product of if you're taking I'm not proving it here but it's actually a very simple extension from this right now if you take the matrix C's let's say a let me do a different different letters X Y Z take their product and then transpose it it's equal to Z transpose Y transpose X transpose I haven't proven this general case and you could keep doing it with 4 or 5 or n matrices multiplied with each other but it generally works and you could essentially prove it using what we proved in this video right here that you take the product of two matrices take the transpose it's equal to the product of the transpose is in reverse order