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Current time:0:00Total duration:14:10

Video transcript

let's see if taking the transpose of a matrix does anything to its determinant so a good place to start is just with the two-by-two scenario so let's do the two-by-two scenario so if I start with some matrix here let me just take its determinant so a b c d and then let me take its determinant so this is going to be equal to a d minus bc now let me take the transpose of this and then take its determinant so that would be the determinant of a see the columns turn into the rows and then BD and the rows turn into the columns now what is this going to be equal to this is going to be equal to ad minus BC again the only thing that happened is these two guys got swapped and they multiplied time to each other anyway so these two things are equivalent so at least for the 2 by 2 case for the 2 by 2 case the determinant of some matrix is equal to the the determinant of the transpose of that matrix now that's just the 2 by 2 case now I'm going to make an inductive argument trying to show you an argument by induction an argument by induction to show that this works for all n by n for all cases and the way you construct an argument by induction is you assume you assume that it's true for the N by n case assume true for the N by n case so let's assume so we assume that for if for n by n that it let's say I have some matrix I'm going to say I have some matrix let me call it matrix B let's say it's an N by n matrix we assume that the determinant of any matrix B that's n by n is equal to the determinant of B's transpose that's where we start off with our inductive argument and then we see if given this if given that is it also true is it true of n plus 1 by n plus 1 matrix because if we can if we can say look given that it's true for the N by n case it's going to be true for the n plus 1 by n plus 1 case then we're done because we know it's true for the base case the 2 by 2 case which you could say well that's your first n by n so if it's true for the 2 by 2 case it'll be 2 for the 3 by 3 case because that's just one increment but then if it's true for the 3 by 3 case it'll be true for the 4 by 4 case but if it's true for the 4 by 4 case it'll be true for the 5 by 5 case and you just keep going up like that so when you do a proof by induction you prove a base case and then you prove that if it's true for n 4 or in this case an N by n determinant if you can prove that given it's true for an N by internment it's going to be true for an n plus 1 by n plus 1 determinant or an n plus 1 by n plus 1 matrix then you have completed your proof so let's see if this is the case so let me construct let me construct an n plus 1 by n plus 1 matrix so let's say I have my matrix a my favorite letter to use for matrices I think the entire linear algebra is favorite letter to use for matrices and let's say it's an n plus 1 by n plus 1 matrix and just to simplify my notation let's just say that M let's say that M is equal to n plus 1 so we could call it an M by M matrix and what is it going to look like let's draw its entries right here so it's going to have I'm going to draw more than the normal amount of entries a1 1 this is X first row a 1 2 all the way to a 1 M we have M columns or actually n plus which is the same thing as n plus 1 that's not n times 1 columns this is n plus 1 as well and then we do our second row right here a 2 1 a 2 2 a 2 3 all the way to a to M then you have your third row right here a 3 1 on a32 a33 all the way to a 3m and then you go all the way down here at the end the you have your amp throw which is you could also say it's your n plus one throw just a so it's your M throw first column and then you have a sub M 2 and then a sub M 3 all the way to a sub mm fair enough now let me draw the transpose of a so a transpose is also going to be an N plus 1 by n plus 1 matrix which you could also write as an M by M matrix I'm just going to take the transpose of this C so the transpose of that this row becomes a column so it becomes a 1 1 then this entry right here is a 1 2 it's this entry right there then you have and you can go all the way down all you could have all the way down to a 1 M then this pink row becomes a pink column here a 2 1 you have a I wanted to do it in pink you have a 2 1 a 2 2 if you have an a 2 3 it goes all the way down to a to M if your green row right here it's your third one so it's a31 a32 a33 all the way down to a 3 m and then you can just skip a bunch of rows in this case but it's columns in this case so you just draw some dots and you have a m one-a m - I'm just going down this guy but this row is now going to become which was the last row is not going to become the last column am 3 all the way down to a mm and I have my transpose now let's take the determinant of a let's take the determinant of a now let me do it in purple so the determinant the determinant of my matrix a we could just go to go down this first row up here it's going to be equal to its going to be equal to a11 times the determinant of its sub-matrix so it's the determinant of this sub matrix right here the determinant of that sub matrix right there we could call that the sub matrix a sub 1 1 we've seen this notation before so it's the determinant of a sub 1 1 and then it is minus a 1/2 times the determinant times the determinant of its sub-matrix so you cross out that row in that column and so that is going to be a sub 1/2 and you're going to go all the way to and I don't know the sign on that is we could call it negative 1 to the to the 1 plus M that'll give us the right sign for the checkerboard pattern times the determinant of the sub-matrix for this guy so we call a 1 M where you cross out that guy's row that guy's column and you're just left with all of this stuff over here fair enough now let's look at the determinant of a transpose the determinant of a transpose we learned earlier you don't have to go down the first row or you don't have to even go down a row you could go down a column so let me be clear so for our determinant of a we went down this row and our sub matrices this was my first sub matrix my second sub matrix you know what it looks like you would cross out the second column and that row and whatever is left over would be the second sub matrix and so on and so forth but for our determinant of a transpose let's just go down this first column and get the sub matrices like that so this is going to be equal to let's get our first term right here a sub 1 1 times the determinant of its sub-matrix so what's the determinant of its sub-matrix it's going to be it's going to be its sub matrix you cross out its row and it's column and you're going to be left with you're going to be left with this thing right here now an interesting question is how does this thing that I've just squared off this guy's sub matrix we're - this guy's sub-matrix well if you look at it carefully this row from 8 to 2 to a 2m has now become a column from a 2 2 - a - M this row which is the next one from a 3 2 to a 3m has now become a column from a 3 2 to 8 3m you keep going down this last row has now become this column so this guy's sub matrix or the determinant or the thing we're going to take the determinant of right here is equal to the transpose of this guy so this is equal to a sub 1 1 a sub 1 1 transpose a sub 1 1 transpose and if you go through it so and then we go - this guy - a 1/2 times the determinant of his sub matrix and if we cross out this guy's row and that guy's column what is that going to look like his sub matrix is going to look like this it's going to have that there and it's going to have that right there now how does that compare to a 1/2 so a 1/2 is if you crossed out this and this and you're left with all of this right here all of this right here so once again you see that this Row is the same thing as this column that this Row is the same thing as this column that that Row is the same thing as that column so once again this the sub matrix we have to take the determinant of is equal to the transpose of this thing over here so it's equal to a 1 2 transpose this thing I could draw shaded in is equal to the transpose of this thing is equal to the transpose of that thing right there so in general each of these sub matrices when we go down this row is equal to the transpose of each of these so you keep going so then you're going to go all the way to plus minus 1 we're going to go all the way down to the minus 1 1 plus M times the determinant of it's going to be this guy it's going to be this guy transpose you can even do it if you go all the way to you cross that guy out and that guy out you're left with everything else on this on this matrix and that's equal to the transpose of if you cross this guy out and that guy out this row turns into that column that row turns into that column I think you see the point I don't want to beat a dead horse so that's going to be equal to a sub 1 M transpose now remember going into this inductive proof or proof by induction I assume that for remember this is an n plus 1 by n plus 1 matrix but going into it I assumed that for an N by n matrix the determinant of B is equal to the determinant of B transpose well these guys right here these are n by n matrices right this guy right here is an n plus 1 by n plus 1 by n plus 1 same thing for this guy right here but these guys right here are n by n n by n so if we assume for the N by n case that the determinant of a matrix is equal to the determinant of a transpose right this is a determinant of matrix this is the determinant of a trend of its transpose these two things have to be equal so we can then say that the determinant of a transpose is equal to this term a sub 1 1 times this but this is equal to this for the N by n case remember we're doing the n plus 1 plus by n plus 1 case but these sub matrices are one dimension smaller in each direction has one less row and one less column so these two things are equal so instead of writing this I can just write this so times the determinant of a sub 1 1 and then you keep going minus a sub 1 2 times the determinant instead of writing this I could write that because they're equal determinant a sub 1 2 all the way to plus minus 1 to the 1 plus M times the determinant of this these two things are equal that is equal to that that was our assumption and this inductive proof a sub 1 M and then you see that of course this line this blue line here is equivalent to this blue line there so we get that the determinant of a which is an n plus 1 by n plus 1 so this is the n plus 1 by n plus 1 case we get the determinant of a is equal to the determinant of a transpose and we got this assuming assuming that it is true is let me write assuming that it's true for n by n case and then we're done we've now proved that this is true in general because we proven the base case we've proven it for the two by two case and then we show that if it's true for the N case it's true for the n plus one case so if it's true for the two case it's going to be true for the three by three case it's true for the three by three case it's true for the four by four case so on and so forth but the takeaway is pretty neat you can take the transpose the determinant doesn't change