If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:11:13

Video transcript

a couple of videos ago I made the statement that the rank the rank of a matrix a is equal to the rank of its transpose and I made a bit of a hand wavy argument it was at the end of the video and I was tired it was actually the end of the day and I thought it was it'd be worthwhile to I maybe flush this out a little bit because it's an important take away it'll help us improve understand everything we've learned a little bit better so let's just understand what the actually I want to start with the rank of a transpose the rank of a transpose the rank the rank of a transpose is equal to the dimension of the column space of a transpose that's the definition of the rank and what is it the dimension of the column space of a transpose is the number of basis vectors basis vectors for the column space of a transpose that's what dimension is for any subspace you figure out how many basis vectors you need in that subspace and you count them and that's your dimension so it's the number of basis vectors for the column space of a transpose which is of course the same thing this thing we've seen multiple times is the same thing as the row space of a row space of a right the columns of a transpose are the same thing as the rows of a is because you switch the rows and the columns now how can we figure out the number of basis vectors we need for the column space of a transpose or the row space of a so let's just think about what the column space of a transpose is telling us so it's equivalent to so let's say let me draw a like this let me draw a of some matrix a let's say it's an M by n matrix and let me just write it as a bunch of row vectors I could also write it as a bunch of column vectors but right now let's just stick to the row vectors so we'd have row 1 then the transpose of column vectors we can just write that's Row 1 and we're going to have Row 2 and I have Row 2 and we're going to go all the way down to row M right it's an M by n matrix each of these vectors are members of RN because they're going to have n entries in them because we have end so that's what a is going to look like a is going to look like that and then a transpose a transpose all of these rows are going to become columns a transpose is going to look like this r1 r2 all the way to RM and this is of course going to be an N by M matrix you swap these out so all these rows are going to be columns right and obviously the column space or maybe not so obviously the column space of a transpose the column space of a transpose is equal to the span the span of r1 r2 all the way to RM right it's equal to the span of these things or you could equivalently call it it's equal to the span of the rows of a that's why it's called so called the row space so this is equal to the span of the rows rows of a these two things are equivalent now these are the span that means this is some subspace it's all of the linear combinations of these columns or all the linear combinations of these rows if we want the basis for it we want to find a minimum set of linearly independent vectors that we could use to construct any of these columns or that we could use to construct any of these rows right here now what happens when we put a into reduced row echelon form so let's just we do a bunch of row operations row operations to put it into reduced row echelon form right you do a bunch of row operations and you eventually you'll get something like this you'll get the reduced row echelon form of a the reduced row echelon form of a is going to look something like this you're going to have some pivot rows some rows that have pivot entries let's say that's one of them let's say the second let's say that's one of them this will all have zeros all the way down this one will have zeros your pivot entry has to be the only nonzero entry in its column and everything to the left of it also has to be 0 let's say that this one isn't these are some nonzero values these are 0 so we have another pivot entry over here everything else is 0 and let's say everything else our non-pivot entries so you come here and you add a certain number of pivot rows or a certain number of pivot entries right and you got there by performing linear row operations on these guys so those linear row operations you know I take three times Row 2 and I add it to Row 1 and I write you know then that's going to become my new Row 2 and you keep doing that and you get these things here so these things here are linear combinations of those guys or another way to do it you could reverse those row operations I could start with these guys right here and I could just as easily perform the reverse row operations you know you can any linear operation you can perform the reverse of it we've seen that multiple times you could perform row operations you could perform row operations with these guys to get all of these guys or another way to view it is these vectors here these row vectors right here they span all of these all of these or all of these row vectors can be represented as linear combinations of your pivot rows right here obviously you're going to have your non pivot rows are going to be all zeros are going to be all zeros and those are useless those are useless but your pivot rows if you take linear combinations of them you can clearly put you can clearly kind of do reverse row echelon form and get back to your matrix so all of these guys can be represented as linear combinations of them and all of these pivot entries are by definition our our Bibles almost by definition they're linearly independent right because I've got a one here no one else has a one there so no one else can be so this guy can definitely not be represented as a linear combination of the other guy so why am I going through this whole exercise well we started off saying we wanted a basis we wanted a basis for the row space we wanted some minimum set of linearly independent vectors that spans everything that these guys can span well if all of these guys can be represented as linear combinations of these row vectors in reduced row echelon form or these pivot rows in reduced in reduced row echelon form and these guys are all linearly independent then they are a reasonable basis so these pivot rows right here that's one of them this is the second one this is the third one maybe they're the only three this is just my particular example that would be a suitable basis for the row space so let me write this down the pivot rows pivot rows in reduced row-echelon form of a are a basis are a basis for the row space row space of a and the row space of a is the same thing you know or the column space of a transpose row space of a is the same thing as a column space of a transpose we've seen that multiple times now so if we want to know the dimension of your column space where you just count the number of pivot rows you have so you just count the number of pivot rows so the dimension of your row space which is the same thing as a column space of a transpose is going to be the number of pivot rows you have in reduced row echelon form or even simpler the number of pivot entries you have because every pivot entry as a pivot row so we can say we can write that the rank the rank of a transpose is equal to the number of pivot entries pivot entries in reduced row echelon form of a right because every pivot entry corresponds to a pivot row those pivot rows are a suitable basis I'll compete for the entire row space because every row can be read with a linear combination of these guys and since that all of these can be then anything that these guys can construct these guys can construct fair enough now what is the rank of a this is the rank of a transpose that we've been dealing with so far the rank of a the rank the rank of a is equal to the dimension the dimension of the column space of a or you could say it's the number of vectors in the basis for the column space of a so if we take that same that same matrix a that we used above and we instead we write it as a bunch of column vectors so c1 c2 all the way to CN we have end columns right there the column space is essentially the subspace that's spanned by all of these characters right here right spanned by each of these column vectors so the column space of a is equal to the span of c1 c2 all the way to CN that's the definition of it but we want to know the number of basis vectors and we've seen before we've done this multiple times the basis vectors or a suitable basis vectors could be if you put this into reduced row echelon form and you have some pivot entries and their corresponding pivot columns so some pivot entries with their corresponding pivot columns just like that maybe that's like that and then maybe this one isn't one and then this one is so you have a certain number of pivot columns you have a certain number of pivot columns let me do them in another color right here when you put a into reduced row echelon form we learned that the basis vectors or the basis columns that form a basis for your column space are the columns that correspond to the pivot columns so this the first column here is a pivot column so this guy could be a basis vector the second column is so this guy could be a pivot vector and maybe the fourth one right here so this guy could be a pivot vector so in general you just say hey how many if you want to if you want to count the number of basis vectors because we only have to know what they are to figure out the rank we just have to know the number they are well you say well for every pivot column here we have a basis vector over there so we could just count the number of pivot columns but the number of pivot columns is equivalent to just a number of pivot entries we have because every pivot entry gets its own column so we could say that the rank of a the rank of a is equal to the number number of pivot entries of pivot entries in the reduced row echelon form of a and as you can see very clearly that's the exact same thing that we deduced was equivalent to the rank of a transpose so the dimension of the column space of a transpose or the dimension of the row space of a so we can now write our conclusion the rank of a is definitely the same thing as the rank of a transpose