# I Aren't all linear operators one-to-one and onto?

1. Mar 12, 2016

### pellman

Let W be a vector space and let A be a linear operator W --> W. Isn't it the case that for any such A, the kernel of A is the zero vector and the range is all of W? And that it is one-to-one from linearity? I ask because an author I am reading goes through a lot of steps to show that a certain operator is one-to-one and onto yet I thought it was a given for any linear operator.

2. Mar 12, 2016

### BiGyElLoWhAt

What if $A = \frac{d}{dx}$? One to One is bijective, right?
I don't believe that that operator is one to one, if you consider it acting on ln(x) in W. If you differentiate, you get 1/x, and the domain of those two functions are different (assuming no absolute value sign on ln(x)). ln(x): (0, inf) and 1/x: (-inf,0) U (0,inf). This is an injective mapping, which is not 1 to 1.

3. Mar 12, 2016

### BiGyElLoWhAt

So I think it depends on W, is what I'm gathering. If W is a vector space that includes all real functions, then no, it's not 1 to 1. I'm not 100% sure, though. Someone will likely step in and correct me on this =/

4. Mar 12, 2016

### WWGD

T(a):=0 is linear: T(a+b)=0=T(a)+T(b); T(ca)=0=cT(a). Projection operators are not 1-1: P(a,b,c)=P(a,b,d) ; $c \neq d$ For P(x,y,z):=P(x,y,0). Or, take a matrix with linearly-dependent rows/columns. EDITIt is not injective (independently of the choice of basis).

Last edited: Mar 12, 2016
5. Mar 12, 2016

### pellman

Projection operators! Yes. thank you for that counter example.

6. Mar 12, 2016

### PeroK

A linear operator (from finite dimensional W to W) is 1-1 iff it is onto. That may be what you were thinking.

7. Mar 12, 2016

### pellman

Yes. I just didn't remember that not all linear operators are one-to-one.

8. Mar 12, 2016

### Staff: Mentor

"finite dimensional W" is important here. It is not true if you drop that requirement.

9. Mar 12, 2016

### mathwonk

the most important linear operator is D = differentiation, on the (infinite dimensional) linear space W of smooth functions. Then notice that D of a constant function is zero, so it is not 1-1. (it is onto however.)

10. Mar 12, 2016

### Staff: Mentor

Add integration with the requirement f(0)=0 and you have the example for 1-1 but not onto.

11. Mar 12, 2016

### mathwonk

nice remark.

here we have the subspace Z of those smooth functions that equal zero at 0, and the complementary (real) line R of constant functions. then the space of all smooth functions is a direct product of Z and R, and D is 1-1 on Z and onto all smooth functions. the inverse of this restriction therefore, i.e. integration with f(0) = 0, is an injection from all smooth functions onto the subspace Z.

So differentiation D is essentially projection of all smooth functions onto the subspace Z, followed by an isomorphism from Z to all smooth functions. And the special integration above is essentially the inverse isomorphism of all smooth functions onto Z, followed by inclusion of Z into all smooth functions.

In general any linear surjection which has a kernel, restricts to an isomorphism on any subspace complementary to that kernel. Then the inmverse isomorphism is injective and not surjective.