Graduate Isomorphism between a linear space and its dual

[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?

 

[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 $f(x) = (\cdot|x)$?

 

[micromass]

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

 

[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?

 

[micromass]

@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?

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.

 

[Geofleur]

Ah - I understand! Thanks :-D

 

[micromass]

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...

 

[WWGD]

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?

Are you assuming a default inner product defined?

 

[Geofleur]

Are you assuming a default inner product defined?

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.