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# Simplifying conditions for invertibility

## Video transcript

with a whole premise of the last series of videos was really just trying to get at trying to figure out whether some transformation T let's have some transformation that is a mapping from let's say it's a mapping from RN to R M the whole question is is T invertible is t invertible is t invertible and we showed several videos ago that a function and the transformation is really just a function that a function is invertible if it meets two conditions so invertible invertible i don't have to keep writing the word over then you have to have two conditions it has to be on two or essentially it has to map to every member of your codomain and it also has to be one to one one to one another way of saying one to one is that every member of your codomain is mapped to by it at most one member of your domain and we did several videos where we thought well we if we had a transformation a linear transformation that's defined by a matrix a where this would be an M by n matrix we said that this is going to be met if the rank this is met if the rank of and this is only met if the rank of a is equal to the number of rows in your transformation matrix is equal to M and the last video I just show that this is only true if every one of your column vectors are linearly independent or that they all our basis vectors for your calm space or that the rank the rank of your matrix it has to be equal to n now so in order for something to be invertible in order for the transformation to be invertible both of these things have to be true your rank of a has to be equal to m and your rank of a has to be equal to n so in order to be invertible a couple of things have to happen in order to be invertible your rank of your transformation matrix has to be equal to m which has to be equal to n so M has to be equal to n so we have an interesting condition you have to have a square matrix you have to have a square matrix your matrix has to be an N by n that's what this implies if B both of these are true then M has to be equal to n and you're dealing with the square matrix even more you're dealing with the square matrix where every one of the columns are linearly independent so this is going to be so this is our a a looks like this a 1 a 2 all the way to a n since the rank of a is equal to n and this is of course an N by n matrix we just said that this has to be the case because both its rank both M its rank has to be equal to M which is a number of rows and its rank has to be equal to n which is the number of columns so the rows and columns have to be the same but the fact that your rank is equal to the number of columns that means that all of your all of your column vectors our basis for your column space or that if you put them into reduced row echelon form so if you put this into reduced row echelon form what are you going to get well all of these guys our basis vectors so they're all going to be associated with pivot vectors or they're all going to be associated with pivot columns so this is going to be 1 0 a bunch of zeros and then you're going to have 0 1 a bunch of zeros like this they're all going to be associated with pivot columns so all associated with pivot columns when you go into reduced row echelon form so all of them are pivot columns it's an N by n matrix so what is an N by n matrix where every column is a pivot column what is an N by n matrix let me like this so you have an N so the reduced row echelon form of a the reduced row echelon form of a has to be equal to an N by n matrix goes because a is n by n n by n matrix where every column is a linearly independent pivot column and I mean by definition of reduced row echelon form they can't be you know you can't have the same pivot column twice where every column is a linearly independent pivot column it's a little bit redundant but I think you get the idea so what is an N by n matrix for every column is a linearly independent pivot column well that is just a matrix that has ones down the diagonal ones down the diagonal and everything else is a zero everything else is a zero or we've seen this matrix before this is just an N by n identity matrix or the identity matrix on N or on RN so if you multiply this matrix times any member of RN you're going to get that matrix again but this is interesting we now have a pretty usable condition for invertibility we can say that the transformation the transformation T that is a mapping from R n to well it has to map to the same dimension space so from RN to RN it's equal to it's equal to some square matrix n by n has to be an N by n matrix times our vectors in our domain and it's only going to be only invertible only invertible if the reduced row echelon form the reduced row echelon form of our transformation matrix is equal to the identity matrix on n I mean I could have written an M here I could have written an M here and I could have said this is an M by n matrix but the only way that this is going to be true is if this is also an N and this is also an end but maybe I could leave them there let me leave those M's there they because that's the big takeaway the big takeaway is that if in order for the transformation matrix to be invertible it's the only way it's invertible is if the reduced row echelon form of our transformation matrix is equal to an N by n identity matrix the identity matrix is always going to be n by n so that's a big takeaway now we'll use that in the future to actually solve for transformations or solve for inverses of transformations