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Not sure how to phrase my title correctly. I set myself the following problem:

Suppose I have a vector space $V$ with basis $\{a_n\}$, and a map $a: V \rightarrow V^*: a_i \mapsto a_i^*$ (where $a_i^*$ is defined as usual: $a_i^*(a_j)=\delta_{ij}$). This map is defined in terms of a basis, but how basis-dependent is it, really? We know there is no canonical map, but how close can we get in some sense?

To formalize this, I consider another basis $\{b_n\}$ and map $b: b_i \mapsto b_i^*$ and ask: when does $a=b$? My answer is: when $a$ and $b$ are orthogonal transformations of each other.

We require that $\forall k: a(b_k) = b(b_k)$. First we note that:

$$b(b_k) = b_k^* = \sum_{i}c_i b_i \mapsto c_k$$

What does $a(b_k)$ do to the same vector? First we need to describe the ${b_n}$ in terms of the ${a_n}$. So let $$b_i = \sum_jd_{ij}a_j$$

Now: $$a(b_k) = a(\sum_{j}d_{kj}a_j) = \sum_{j}d_{kj}a(a_j) = \sum_jd_{kj}a_j^*$$

So: $$a(b_k)(\sum_{i}c_i b_i) $$ $$= (\sum_jd_{kj}a_j^*)(\sum_{i}c_i b_i)$$ $$= (\sum_jd_{kj}a_j^*)(\sum_{i}c_i \sum_l d_{il}a_l)$$ $$= (\sum_jd_{kj}a_j^*)(\sum_{i,l}c_i d_{il}a_l)$$ $$= \sum_jd_{kj}a_j^*(\sum_{i,l}c_i d_{il}a_l)$$ $$= \sum_jd_{kj}a_j^*(\sum_{i}c_i d_{ij}a_j)$$ $$= \sum_jd_{kj}\sum_{i}c_i d_{ij}$$ $$= \sum_i c_i \sum_{j}d_{kj} d_{ij}$$

And now for the punchline: for this to equal $c_k$, it is sufficient that

$$\sum_{j}d_{kj} d_{ij} = \delta_{ik}$$

In other words, when $d$ is written out as a matrix, each row must be normalized, and orthogonal to all other rows. So two dual maps defined on bases are equal when those bases are orthogonal transformations of each other. I may have to think a little more about necessity.

Is that correct? Is there a simpler way to see all this? For my next trick, I hope to investigate why the tensor product (defined in terms of basis vectors) is unique.

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    $\begingroup$ I suppose this is basically equivalent to showing that orthogonal transformations (i.e., those that map orthonormal bases to each other) don't change the inner product or something? $\endgroup$ – A_P Jul 24 at 23:19
  • $\begingroup$ Exactly: see my answer. $\endgroup$ – Berci Jul 24 at 23:22
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Your proof is correct.

For another treatment, define an inner product $\langle,\rangle$ such that the given basis $a_1,\dots,a_n$ becomes orthonormal.
Specifically, simply define $\ \langle a_i,a_j\rangle:=\delta_{ij}\ $ and extend it linearly in both variables.
This implies that $a(a_i)=\langle a_i,\_\rangle$ for each $i$, hence $a(x)=\langle x,\_\rangle$ for all vectors $x$.

Since $a_i$ is orthonormal, they behave exactly like the standard basis with respect to the standard inner product.
In particular, any other basis $b_i$ is orthonormal w.r.t $\langle,\rangle$ iff the transition matrix (whose entries are $a_j$-coordinates of $b_i$, that is $\,a_j^*(b_i)$) is orthogonal.

Now, if $b_i$ is orthonormal, then we conclude $b(x)=\langle x,\_\rangle$ as above, hence in this case $b(x)=a(x)$.

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  • $\begingroup$ Thanks! Somehow the necessity of my condition isn't jumping out at me. What am I missing? $\endgroup$ – A_P Jul 24 at 23:44
  • $\begingroup$ Two bases generate the same inner product iff they are orthonormal w.r.t. each other, that is, iff their transition matrix is orthogonal. $\endgroup$ – Berci Jul 25 at 0:18
  • $\begingroup$ In my proof I show sufficiency but not necessity that the matrix is orthogonal. I'm sure there's something simple I'm not seeing. $\endgroup$ – A_P Jul 26 at 4:19
  • $\begingroup$ I didn't aim to prove the other direction (though it may be true). That would be $b(x)=\langle x, _\rangle$ implies that the transition matrix is orthogonal. $\endgroup$ – Berci Jul 26 at 8:34

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