Prove the existence and uniqueness of the adjoint function of a linear function

Question

Definition

Let be $V$ and $U$ two vector spaces equipped with an inner product. So given a linear transformation $f:V\rightarrow W$ a function $f^*:U\rightarrow V$ is called adjoint of $f$ if

1. $$ \big\langle \pmb u,f(\pmb v)\big\rangle=\big\langle f^*(\pmb u),\pmb v\big\rangle $$

for all $\pmb v\in V$ and $\pmb u\in U$.

Theorem

For every transformation $f:V\rightarrow U$ there exist a unique adjoint $f^*$ satisfying the condition of the above definition.

So to follow my proof attempt.

• Chose a basis $\mathcal{V}:=\{\pmb v_1,...,\pmb v_n\}$ for $V$ and a basis $\mathcal{U}:=\{\pmb u_1,...,\pmb u_m\}$ of $U$. Then $f$ can be characterized by the $m\times n$ matrix $A$ such that $$ f(v_j)=\sum_{i=1}^ma_{i,j}\pmb u_i $$ for any $j=1,...,n$. Now let $\mathcal{V}':=\{\pmb v^1,...,\pmb v^n\}$ and $\mathcal{U}':=\{\pmb u^1,...,\pmb u^m\}$ the reciprocal bases of $\mathcal{V}$ and $\mathcal{U}$ respectively. So since $f$ and $f^*$ are both linear transformation it is suffices to check $(1)$ for $u$ equal to an arbitrary element of the basis $\mathcal{U}'$ and for $v$ equal to an arbitrary element of the basis $\mathcal{V}$. So we observe that $$ \big\langle\pmb u^i,f(\pmb v_j)\big\rangle=\biggl\langle\pmb u^i,\sum_{h=1}^ma_{h,j}\pmb u_h\biggl\rangle=\sum_{h=1}^ma_{h,j}\langle\pmb u^i,\pmb u_h\rangle=\sum_{h=1}^ma_{h,j}\delta_{i,h}=a_{i,j} $$ and $$ \big\langle f^*(\pmb u^i), \pmb v_i\big\rangle=\biggl\langle f^*\biggl(\sum_{h=1}^mb_{h,i}\pmb u_h\biggl),\pmb v_j\biggl\rangle=\biggl\langle\sum_{h=1}^mb_{h,i}f^*(\pmb u_h),\pmb v_j\biggl\rangle=\biggl\langle\sum_{k=1}^n\sum_{h=1}^ka^*_{k,h}b_{h,i}\pmb v_k,\pmb v_j\biggl\rangle=\sum_{k=1}^n\sum_{h=1}^ma^*_{k,h}b_{h,i}\langle\pmb v_k,\pmb v_j\rangle=\sum_{k=1}^n\sum_{h=1}^m\langle\pmb v_j,\pmb v_k\rangle a^*_{k,h}b_{h,i}=\sum_{k=1}^n\sum_{h=1}^mc_{j,k} a^*_{k,h}b_{h,i} $$ where $b_{h,i}=\langle \pmb u^h,\pmb u^i\rangle$ for each $h=1,...,m$. So if $B$ and $C$ are the matrix of change of basis from $\mathcal U'$ to $\mathcal U$ and from $\mathcal V$ to $\mathcal V'$ respectively then it seems to me (IS THIS TRUE?) that $\sum_{k=1}^n\sum_{h=1}^mc_{j,k} a^*_{k,h}b_{h,i}=(C\cdot A^*\cdot B)_{j,i}$ so that I have to choice the elements of $A^*$ such that $(C\cdot A^*\cdot B)_{j,i}=a_{i,j}$ but unfortunately I don't be able to do this: it seems that the above equality imply that $A^*=C^{-1}\cdot A^{tr}\cdot B^{-1}$ but I am not sure about this.

I point out that my proof attempt is the same that is given by Ray M. Bowen and C. C. Wang in Introduction to vectors and tensors and they use the above theorem to prove though the canonical isomorphism the existent of an adjoint function between the dual spaces. So could someone help me, please?

Answer 1

I must admit, I trailed off at the end there, but I think it started well. I think the missing ingredient here is that, just as $\mathcal{U}'$ is dual (or reciprocal) to $\mathcal{U}$, so too is $\mathcal{U}$ dual to $\mathcal{U}'$. As such, the following holds for all $u \in U$: $$u = \sum_{i=1}^m \langle u, u_i\rangle u^i.$$ In particular, if we consider $u = f^*(v^i)$, then $$f^*(v^i) = \sum_{j=1}^m \langle f^*(v^i), u_j\rangle u^j$$ and hence $$f^*(v^i) = \sum_{j=1}^m \langle v^i, f(u_j)\rangle u^j. \tag{2}$$ Importantly, this gives us $f^*(v^i)$ purely in terms of known vector quantities, and indeed as a linear combination of $\mathcal{U}'$. In this way, we can build a matrix $A$ for $f^*$ from $\mathcal{V}'$ to $\mathcal{U}'$ by setting $(A)_{ij} = \langle v^i, f(u_j)\rangle$.

There are some extra steps here that the above argument glossed over. My working has assumed the existence of (at least one) function $f^* : V \to U$ satisfying the definition of an adjoint. I've shown the action of $f^*$ on the basis $\mathcal{V}'$ is unique (i.e. every adjoint maps the basis vectors in $\mathcal{V}'$ to the same vectors). This almost proves uniqueness; we would need to show that $f^*$ is linear, then use the fact that linear maps are defined uniquely on bases.

Conversely, in order to establish the existence of a dual, we could start by using $(2)$ as a definition, and extend $f^*$ linearly to all of $V$. That is, we would define $$f^*\left(\sum_{i=1}^n a_i v^i\right) = \sum_{i=1}^n \sum_{j = 1}^m a_i \langle v^i, f(u_j)\rangle u^j.$$ We would then have to verify that $f^*$ satisfies the definition of adjoint. That is, we need to show, for all $u \in U$, $$\langle f^*(v), u \rangle = \langle v, u \rangle.$$ We have, \begin{align*} \langle f^*(v), u \rangle &= \left\langle \sum_{i=1}^n \sum_{j = 1}^m a_i \langle v^i, f(u_j)\rangle u^j, \sum_{k = 1}^m \langle u, u^k\rangle u_k\right\rangle \\ &= \sum_{i=1}^n \sum_{j=1}^m \sum_{k=1}^m a_i \langle v^i, f(u_j)\rangle \langle u, u^k\rangle \langle u^j, u_k \rangle \\ &= \sum_{j=1}^m \sum_{k=1}^m \left\langle \sum_{i=1}^n a_i v^i, f(u_j)\right\rangle \langle u, u^k\rangle \delta_{j,k} \\ &= \sum_{j=1}^m \langle v, f(u_j)\rangle \langle u, u^j\rangle \\ &= \left\langle v, f\left(\sum_{j=1}^m \langle u, u^j\rangle u_j\right)\right\rangle \\ &= \langle v, f(u) \rangle. \end{align*} Thus, the proposed function is indeed an adjoint, establishing existence.