# Dual basis map invariant under orthogonal transformation?

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.

• 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? – A_P Jul 24 at 23:19
• Exactly: see my answer. – Berci Jul 24 at 23:22

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

• Thanks! Somehow the necessity of my condition isn't jumping out at me. What am I missing? – A_P Jul 24 at 23:44
• Two bases generate the same inner product iff they are orthonormal w.r.t. each other, that is, iff their transition matrix is orthogonal. – Berci Jul 25 at 0:18
• 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. – A_P Jul 26 at 4:19
• 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. – Berci Jul 26 at 8:34