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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?
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?