- Introduction to the inverse of a function
- Proof: Invertibility implies a unique solution to f(x)=y
- Surjective (onto) and injective (one-to-one) functions
- Relating invertibility to being onto and one-to-one
- Determining whether a transformation is onto
- Exploring the solution set of Ax = b
- Matrix condition for one-to-one transformation
- Simplifying conditions for invertibility
- Showing that inverses are linear
Simplifying conditions for invertibility
Showing that a transformation is invertible if and only if rref(A) is equal to the identity matrix. Created by Sal Khan.
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- The rref of any n by n square matrix is the nth identity matrix
So why emphasize that the rref of A should be I?(1 vote)
- Incorrect. any nxn matrix is not automatically linearly independent.
These properties arise when it IS linearly independent (i.e. the rref will be the I matrix)(16 votes)
- There is something still missing in this proof for me. At ~4:00, Sal says "rref(A) = an nXn Matrix where every column is a linearly independent pivot column." -- this is not the definition of the Identity matrix, because an rref(A) matrix does not necessarily have 1 along diagonals.
What about a rref(A) like this:
0 1 0 0
1 0 0 0
0 0 1 0
0 0 0 1
This is not the same as the Identity matrix! What am I missing?(2 votes)
- That matrix isn't reduced row echelon form. It's reduced, but it's not echelon. In rref, if you start from the left and move right, the first 1 will be in the first column, the second 1 will be in the second column, and so on. Thus, if you have an nXn matrix and every column is a pivot column, you get an identity matrix.(3 votes)
- So does a linear operator T have to be bijective to be invertible and/or vice versa?(2 votes)
- Yes, but this is true regardless of whether or not T is a linear operator. Any bijective function is invertible, and any invertible function is bijective.
An alternate way to state this is:
A function f is invertible if and only if f is bijective.(2 votes)
- Shouldn't it be possible for a function to be invertible without having its domain rank equal to its codomain rank? For example, imagine a function that maps from integers in one dimension to integers in two dimensions, which just traverses them in a spiraling motion, starting with 0 -> [0,0], 1 -> [0,1], 2 -> [1,0], 3 -> [0,-1], 4 -> [0,-1], 5 -> [0,2], 6 -> [1,1], etc. This function is invertible even though it has a rank mismatch. It's not representable as a matrix though.(2 votes)
- The problem in your example is in the very begining. First of all, in Linear Algebra we talk about transformations from R^n to R^m, and it is not mere by chance. It is because we work with transformations from a vector space to another vector space, both over a certain field.
In your example, the set of integer numbers is not even a field (such as the real numbers or the rational numbers), and this is because you cannot define a product with an inverse (because the inverse of 2 is 1/2, which is not integer). By the same logic the set of ordered pairs of integers is not a vector space, such as R^2.
So the theory it is seen in these videos cannot be applied to your example.(2 votes)
- I read about left and right inverses in which case m =/= n. What would it mean to have a left or right inverse.(1 vote)
- Suppose A is mxn and has a left inverse. Then there is an nxm matrix B such that BA=Inxn, where Inxn is the nxn identity matrix.
Now suppose instead A has a right inverse. Then there is an nxm matrix C such that AC=Imxm, where Imxm is the mxm identity matrix.
A matrix has a left inverse if and only if it is one-to-one (aka injective).
A matrix has a right inverse if and only if it is onto (aka surjective).(3 votes)
- Is there a way to find the inverse of a function without using matrices?(2 votes)
- We haven't found the inverse of a transformation yet here, or an inverse matrix.
When you have y=f(x) (y in terms of an independent variable x) solve it for x in terms of y to get the inverse function (e.g. the inverse of "y=3x" is "x=y/3", or you could just exchange x and y to get "y=x/3").(1 vote)
- I always thought that A is invertible if the determinant of A is not 0 ( A invertible <=> |A| =/= 0 ). Is it not true?(1 vote)
- Yes, that is true. That the determinant must be non-zero is a consequence of the matrix being onto and one-to-one.(2 votes)
- @1:15, it makes sense to me that "Ax is onto" means that rank(A) = m (if A is mxn). However, maybe I'm missing something, because I haven't figured out how to prove or even explain this. Is there something from previous videos I didn't get?(1 vote)
- The relevant video is: https://www.khanacademy.org/math/linear-algebra/matrix-transformations/inverse-transformations/v/determining-whether-a-transformation-is-onto
For A (not A x) to be surjective, we require that for every b in Rm we can find at least one x such that A x = b.
Equivalently, this means that this system never has no solutions for x, regardless of our choice of b.
This means that rref(A) can not have any rows which are entirely zero.
This means that rref(A) must have m pivot columns (since A has m rows), thus A has m linearly independent column vectors, so its column space C(A) spans all of Rm, and the dimension of C(A), or the rank, is also m.(2 votes)
- Can it be thought like that?: If we go from R^n space to R^m space (n > m) more than one point in R^n must be assigned to one point in R^m because R^n has more points (so such translation isn't one-to-one)
If we go from R^n to R^m (now m > n) there is either a point in R^m which isn't assigned to any point in R^n (so translation isn't "onto") or there is a point in R^m which is ascribed to at least two elements of R^n.(1 vote)
- His last if should be an iff, right ?
A is invertible iff rref(A) = In(1 vote)
Well the 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 Rm, --the whole question is, is T invertible. And we showed several videos ago that a function, and a transformation is really just a function, that a function is invertible if it meets two conditions. So invertible. So I don't have to keep writing the word over again. You have to have two conditions. It has to be onto or essentially it has to map to every member of your codomain and it also has to be one-to-one. Another way of saying one-to-one is that every member of your codomain is mapped to by at most one member of your domain. And we did several videos where we thought well 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 only met if the rank of A is equal to the number of rows in your transformation matrix is equal to m. And in the last video I just showed that this is only true if every one of your column vectors are linearly independent or that they all are basis vectors for your column space or that the rank of your matrix it has to be equal to n. Now 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. Your matrix has to be n-by-n. That's what this implies. If both of these are true, then m has to be equal n and you're dealing with a square matrix. Even more, you're dealing with the square matrix where every one of the columns are linearly independent, so this is our A. A looks like this. A1, A2, all the way to An. 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 its rank has to be equal to m, which is the 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 column vectors are bases for your column space, or that if you put them into reduced row echelon form, what are you going to get? Well all of these guys are 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, bunch of 0's, and then you're going to have a 0, 1, a bunch of 0's like this. They're all going to be 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 write this. So you have an n. So the reduced row echelon form of A has to be equal to an n-by-n matrix, cause A is n-by-n, where every column is a linearly independent pivot column. And I mean by definition of reduced row echelon form 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 where every column is a linearly independent pivot column? Well that is just a matrix that has 1's down the diagonal and everything else is a 0. Or, you'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 just 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 T that is a mapping from Rn to-- well it has to map to the same dimension space so from Rn to Rn. It's equal to some square matrix n-by-n, it has to be an n-by-n matrix, times our vectors in our domain. And it's only going to be invertible if 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 and I could've 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 m. But maybe I could leave them there. Let me leave those m's there because that's the big takeaway. The big takeaway is that in order for the transformation matrix to be invertible, 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.