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.

  • 1
    $\begingroup$ 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

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 $b(x)=a(x)$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.