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# 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.

## Want to join the conversation?

• The rref of any n by n square matrix is the nth identity matrix
as per

http://www.wolframalpha.com/input/?i=RowReduce&a=*F.ReducedRowEchelon.reducematrix-_%7B%7Ba%2C+b%2C+c%7D%2C+%7Bd%2C+e%2C+f%7D%2C+%7Bg%2Ch%2Ck%7D%7D

and

http://www.wolframalpha.com/input/?i=RowReduce&a=*F.ReducedRowEchelon.reducematrix-_%7B%7Ba%2C+b%2C+c%2C+D%7D%2C+%7Bd%2C+e%2C+f%2C+G%7D%2C+%7Bg%2Ch%2Ck%2C+l%7D%2C+%7Bm%2Cn%2Co%2Cp%7D%7D

So why emphasize that the rref of A should be I?
(1 vote)
• Incorrect. any nxn matrix is not automatically linearly independent.
http://www.wolframalpha.com/input/?i=RowReduce+%7B%7B1%2C+2%2C+3%7D%2C%7B+7%2C+8%2C+1%7D%2C%7B+2%2C+4%2C+6%7D%7D

These properties arise when it IS linearly independent (i.e. the rref will be the I matrix)
• There is something still missing in this proof for me. At ~, 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?
• 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.
• So does a linear operator T have to be bijective to be invertible and/or vice versa?
• 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.
• 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.
• 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.
• 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.

Some properties:
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).
• Is there a way to find the inverse of a function without using matrices?
• 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.
• @ , 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.