# Ensuring injectivity of an operator

1. Sep 26, 2007

The following puzzles me, and help is highly appreciated, as always:

I am to ensure the injectivity of a linear operator A, which is the unique operator defined by the bilinear form a(. , .). So, the book says that a simple and natural condition which guarantees the desired is: $\alpha ||v||^2 \leq a(v, v)$, for all v, and, after a few steps, it follows that $\alpha ||v|| \leq ||Av||$ (for all v), and the book now says that injectivity follows easily from this inequality, but I can't see how.

2. Sep 26, 2007

### matt grime

You realize that the inequality

\alpha||v||^2 < a(v,v)

has absoultely no relation to A as stated right? How are you using a(-,-) to define A?

Anyway, that is completely immaterial. If Av=0, then you can reach an obvious conclusion that v=0.

3. Sep 26, 2007

Well, the book says that the a(-,-) defines a unique linear operator A with <Au, v> = a(u, v), for all u, v.

It is tragic how one can oversee such trivial things. Thanks.

4. Sep 27, 2007

### matt grime

The book might say that, but you didn't.

5. Sep 27, 2007

I didn't because I thought this was the only way to define this unique linear operator, which is, apparently, not correct, since you asked. I was told there is a theorem about this, but I wasn't able to find it. I'd appreciate some enlightment related to this matter, i.e. how exactly does a bilinear form a(-,-) : V x V --> F define a unique linear operator A : V --> V' , where V' is the dual space to V?

The definition above ( <Au, v> = a(u, v), for all u,v) means that A is the linear operator which maps every vector u to the functional F from V' such that, for this very u, F(v) = a(u, v), for all v.

Is there another way to look at this?

6. Sep 27, 2007

### matt grime

The point is that you didn't say what the linaer operator A had to do with a(-,-) at all! An inner product (x,x) is exactly the same as specifying a matrix so that that (x,x)=x^tAx, but you didn't say anything about that at all.

7. Sep 27, 2007

### morphism

Actually, he did say that A was "the unique operator defined by the bilinear form a(. , .)."

8. Sep 27, 2007

### matt grime

And that means what, precisely? Nothing. Only if we are to _presume_ information not specified does that mean anything at all. Saying 'a(-,-) specifies a unique linear operator' is completely untrue, if that is all the information supplied. However, as I pointed out that is immaterial.

9. Sep 28, 2007

matt, perhaps I misunderstood what I was said (I'll ask the person once again) about the "bilinear form vs. unique linear operator" issue. Perhaps, if we define that operator as above, i.e. <Au, v> = a(u, v), for all u, v, then this operator A is unique and can be identified with the bilinear form a(-,-)?

I realize this is completely immaterial now (since my primary question was already answered), but I'm still interested in demistifying this.

10. Sep 28, 2007

### matt grime

The point was you didn't say what conditions A had to satisfy! Look at what you wrote:

"a linear operator A, which is the unique operator defined by the bilinear form a(. , .)"

in post 1 (Notice you talk of 'a linear operator', by the way, then say it is unique...) and

"a(-,-) defines a unique linear operator A with <Au, v> = a(u, v), for all u, v."

Call me dumb, but having to read between the lines to guess what you meant to write is a bit tricky sometimes - it is better to be safe than sorry, so include *how* a(-,-) defines a linear operator rather than presuming that we will guess.

Bear in mind that I could choose to use a(-,-) to define A satisfying a(x,y)=<Ax,y> or I could use it to define B to satisfy a(x,y)=<x,By>, where B is of course A transpose, so no, the linear operator is not unique anyway.

Last edited: Sep 28, 2007
11. Sep 28, 2007