Mathematics Asked by Antonio Maria Di Mauro on February 23, 2021

**Definition**

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

- $$

biglangle pmb u,f(pmb v)bigrangle=biglangle f^*(pmb u),pmb vbigrangle

$$

for all $pmb vin V$ and $pmb uin U$.

**Theorem**

For every transformation $f:Vrightarrow U$ 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 $mtimes n$ 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

$$

biglanglepmb u^i,f(pmb v_j)bigrangle=biggllanglepmb u^i,sum_{h=1}^ma_{h,j}pmb u_hbigglrangle=sum_{h=1}^ma_{h,j}langlepmb u^i,pmb u_hrangle=sum_{h=1}^ma_{h,j}delta_{i,h}=a_{i,j}

$$

and

$$

biglangle f^*(pmb u^i), pmb v_ibigrangle=biggllangle f^*biggl(sum_{h=1}^mb_{h,i}pmb u_hbiggl),pmb v_jbigglrangle=biggllanglesum_{h=1}^mb_{h,i}f^*(pmb u_h),pmb v_jbigglrangle=biggllanglesum_{k=1}^nsum_{h=1}^ka^*_{k,h}b_{h,i}pmb v_k,pmb v_jbigglrangle=sum_{k=1}^nsum_{h=1}^ma^*_{k,h}b_{h,i}langlepmb v_k,pmb v_jrangle=sum_{k=1}^nsum_{h=1}^mlanglepmb v_j,pmb v_krangle a^*_{k,h}b_{h,i}=sum_{k=1}^nsum_{h=1}^mc_{j,k} a^*_{k,h}b_{h,i}

$$

where $b_{h,i}=langle pmb u^h,pmb u^irangle$ 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 $sum_{k=1}^nsum_{h=1}^mc_{j,k} a^*_{k,h}b_{h,i}=(Ccdot A^*cdot B)_{j,i}$ so that I have to choice the elements of $A^*$ such that $(Ccdot A^*cdot B)_{j,i}=a_{i,j}$ 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?

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 $u in U$: $$u = sum_{i=1}^m langle u, u_irangle u^i.$$ In particular, if we consider $u = f^*(v^i)$, then $$f^*(v^i) = sum_{j=1}^m langle f^*(v^i), u_jrangle u^j$$ 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 $$f^*left(sum_{i=1}^n a_i v^iright) = sum_{i=1}^n sum_{j = 1}^m a_i langle v^i, f(u_j)rangle u^j.$$ 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 &= leftlangle 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^krangle u_krightrangle \ &= sum_{i=1}^n sum_{j=1}^m sum_{k=1}^m a_i langle v^i, f(u_j)rangle langle u, u^krangle langle u^j, u_k rangle \ &= sum_{j=1}^m sum_{k=1}^m leftlangle sum_{i=1}^n a_i v^i, f(u_j)rightrangle langle u, u^krangle delta_{j,k} \ &= sum_{j=1}^m langle v, f(u_j)rangle langle u, u^jrangle \ &= leftlangle v, fleft(sum_{j=1}^m langle u, u^jrangle u_jright)rightrangle \ &= langle v, f(u) rangle. end{align*} Thus, the proposed function is indeed an adjoint, establishing existence.

Correct answer by user837206 on February 23, 2021

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