Question

How do you determine whether the operator is onto?

Answer 1

w_1=-x+3x_2+2x_3 w_2 = 2x_1 + 4x_3 w_3=x_1 + 3x_2 + 6x_3

Answer 2

I get its one to one if the kernel only has one point

Answer 3

you want to create the matrix associated with the linear transformation first. Then find what its rank is. if the matrix is an m x n matrix, and the rank is equal to m, then the transformation is onto (surjective).

In other words, you want to see if the rows of the matrix are linearly independent.

Answer 4

Okay so for this matrix it would be [-1 3 2; 2 0 4; 1 3 6] right

Answer 5

right, that is the matrix, so now you need to row reduce it.

Answer 6

I got [1 0 2 ; 0 1 1.3333; 0 0 0 ]

Answer 7

ok, so there are 2 pivots and an all 0 row. That means that the rank of the matrix is 2. Therefore because the rank isnt 3, this linear transformation is not onto.

Answer 8

Oh I see that make sense, how about for one to one? My TA said something about the kernel having only one point

Answer 9

I dont quite get that is the kernel for this not [-2 -1.3333 1]?

Answer 10

so for one to one (injective), that is one way to do it. You can look at the equation:

Ax = 0

and see if x=0 is the only solution (in which case you have a one-to-one function) or if there is more in the Null Space (or Kernel, so much terminology lol).

Answer 11

Another way is to do what i did earlier. Row reduce the matrix, but this time, see if the rank matches n instead of m. If you have a m x n matrix, and its rank is equal to n, then the transformation in one-to-one

Answer 12

I like the second way better haha does that always work so in this case it's not one to one either

Answer 13

right, this one is not one to one either, because n = 3, and the rank is 2.

Answer 14

So basically these are the tests. Let A be a m x n matrix with rank r.

If r = m, the matrix is onto.

If r = n, the matrix is one-to-one.