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# Determining whether a transformation is onto

Determining whether a transformation is onto. Created by Sal Khan.

## Want to join the conversation?

• Hello Sal. Thank you for such thorough videos. The example you give towards the end of the video (around 25 into video) seems to indicate that a transformation from Rn to Rm where n > m can never be surjective. The rank of the transformation matrix can never > n, which will always be less than the number of rows, m. Am I right?
• I agree...and I understand it in this way: let's say n=2 and m=3, since 3D space has more points than 2D, there is no way that a function could map all points in 3D by using points from 2D.
• Sal says T is Onto iff C(A) = Rm. But the definition of "onto" is that every point in Rm is mapped to from one or more points in Rn. So surely Rm just needs to be a subspace of C(A)? For example, if C(A) = Rk and Rm is a subspace of Rk, then the condition for "onto" would still be satisfied since every point in Rm is still mapped to by C(A). In that case, it appears wrong to say C(A) equals Rm. Am I misunderstanding?
• Good question James,

What your saying makes sense but consider the following. We think of `T` as a map from `R^n` to `R^m`. That means we input vectors of length `n` and we output vectors of length `m`. The "image" of `T` is the space `C(A)` and so it lives inside of `R^m` naturally. Thus, if `C(A)=R^k` then `k` must be less than or equal to `m`. So if the image contains all of `R^m` we must have `k=m`.
• Could someone please help me understand what is the homogeneous and non homogeneous that Sal is talking about in this and the next two videos about invertibility of transformations. Thanks a lot in advance.
• Homogeneous means that the one side is zero, i.e. Ax = 0.
• In the end at around 25.00 sal sir says it is not onto as rank is 2. But i didnt get the reason.. Anyone cud please explain.

"rank" is number of basis vectors
"number of basis vectors" is number of dimensions
"R^n" is a space of n dimensions

you want to be able to reach any (every) point in R^n, and those can be reached by a combination of at least "n" number of basis vectors, you need to have at least that many basis vectors in your matrix to have the "onto" condition

if you have too few basis vectors (can't reach every point of R^n), then the "onto" condition does not apply
• What if the last row of the rref(A) does not contain a pivot entry & was [00...00|0], i.e. a row of zeros? would this transformation still be surjective?
• I think this is not possible to happen. You are implying that a combination of the elements of b vector (from Ax=b) will always be zero. Meaning a1*b1+a2*b2+..an*bn, where 'a' terms are coefficients and constant, will always be 0 for every possible b in R^n. Which is not possible.

But it is possible for some b in R^n. And that means its not surjective. Sal also explains it on minutes of the video.
(1 vote)
• If I'm proving that a transformation is onto, does it suffice to show that "A" has a nonzero determinant (if the matrix is square)? Because aren't a reduced row echelon form of no zero rows, and a nonzero determinant, equivalent statements?
• Yes, for square matrices, it suffices to show the determinant is nonzero. This will ensure that A is onto and one to one.
• I understand that if last row of rref(A) is "all-zeros", then there will only be a few cases ("a few vector b's") that satisfy the equation and hence it will NOT be ONTO.

I also understand that if each row will have a pivot entry including the last row then it will be ONTO

But what if i have a rref(A) where my last row does not have "all zeros" and also does not have a pivot entry, but instead has a free variable, will then the tranformation be ONTO?
eg: rref(A) =
[1 0 0 0 0]
[0 1 0 0 0]
[0 0 3 0 0]
[0 0 0 1 4]
if this is my rref then?
• Your last row does have a pivot entry, the 1 in the fourth column. And this is not rref, but it will be after you divide your third row by 3.

But yes, this transformation is onto. You're mapping a five-dimensional space into a four-dimensional space, and you don't have any zero rows or columns.
• In the example that Sal provides around , is it obvious from the start that the rank of the 3x2 matrix can be at most 2 and therefore less than 3, and therefore not "onto" R3?