# A Isomorphism between a linear space and its dual

1. Mar 27, 2016

### Geofleur

I have been trying to prove the following theorem, for a finite dimensional vector space $X$ and its dual $X^*$:

Let $f:X\rightarrow X^*$ be given by $f(x) = (x|\cdot)$, where $(x|\cdot)$ is linear in the first argument and conjugate linear in the second (so I am using the mathematicians' convention here). Then $f$ is bijective if and only if [$(y|x)=0$ for all $x \in X$ implies that $y = 0$].

I have been able to show the "only if" part - that $f$ being bijective implies the statement about the inner product. For the "if" part of the proof (where we assume that [$(y|x)=0$ for every $x \in X$ implies $y = 0$] and prove that $f$ is bijective), I am able to prove that $f$ is injective, but I am running into a problem in proving that $f$ is surjective. Here is my attempt:

Let $x^*$ be an arbitrary linear functional in $X^*$. Because the space $X$ is finite dimensional, we may choose an orthonormal basis $\{e_i\}_{i=1}^n$ and write an arbitrary vector in $X$ as $x = \sum_{i=1}^{n}x_i e_i$. Then we have $x^*(x) = x^*(\sum_{i=1}^{n} x_i e_i) = \sum_{i=1}^{n}x_i x^*(e_i)$. Now we need to find a $y \in X$ such that $f(y) = x^*$, i.e., such that $(y|x) = x^*(x)$ for all $x \in X$. For any $y \in X$ we can write $y = \sum_{j=1}^{n} y_j e_j$, so maybe choosing the $y_j$ appropriately will do the job. We have $(y|x) = (\sum_{j=1}^{n} y_j e_j | \sum_{i=1}^{n} x_i e_i) = \sum_{j} \sum_{i} y_j \overline{x}_i (e_j|e_i) = \sum_{i} y_i \overline{x}_i$. The problem is that I want this to equal $x^*(x) = \sum_{i=1}^{n}x_i x^*(e_i)$, but the former expression involves $\overline{x}_i$ while the latter involves $x_i$. If I had used the physicists' convention and made the inner product conjugate linear in the first argument instead, this problem would not have occurred. But it seems strange that a theorem like this would be dependent upon a convention like that. Any insights?

2. Mar 27, 2016

### Samy_A

Isn't the problem that in $(y|x) = x^*(x)$, the LHS is conjugate linear in x, while the RHS is linear in x?

In other words, shouldn't f be defined as $f(x) = (\cdot|x)$?

3. Mar 27, 2016

### micromass

Staff Emeritus
Another is by saying $X$ and $X^*$ have the same dimension.

4. Mar 27, 2016

### Geofleur

@Samy_A: That makes sense - it means I would need to make a correction in the book I am reading, though (Analysis, Manifolds, and Physics, Choquet-Bruhat).

@micromass: Do you mean I was wrongfully assuming that $X$ and $X^*$ have the same dimension? I thought that all I had assumed is that the action of $x^*$ is completely determined by its effect on the basis vectors of $X$. Did I sneak something else in without realizing it?

5. Mar 27, 2016

### micromass

Staff Emeritus
No, it's correct. In finite dimensions, $X$ and $X^*$ have the same dimension. But once you know that, the proof is done. See the alternative theorem. This says that a linear map between two spaces of the same finite dimension is an isomorphism iff it is surjective iff it is injective.

6. Mar 27, 2016

### Geofleur

Ah - I understand! Thanks :-D

7. Mar 27, 2016

### micromass

Staff Emeritus
If you're interested in quantum mechanics or functional analysis, you probably will want to think about how much this result generalizes to infinite dimensions...

8. Mar 28, 2016

### WWGD

Are you assuming a default inner product defined?

9. Mar 28, 2016

### Geofleur

I think so? I am defining a sequilinear mapping $X \times X \rightarrow \mathbb{C}$ such that $(x,y) \mapsto (x|y)$, where

$(x|y) = \overline{(y|x)}$, and
$(\alpha x + \beta y|z) = \alpha(x|z) + \beta(y|z)$.

I ended up just writing a note in the margin pointing out that $(x|\cdot) \in X^*$ if one uses the physics convention; otherwise, we have $(\cdot|x) \in X^*$ as Samy_A suggested.

Last edited: Mar 28, 2016