Motivation of Adjoints and Normal Operators

Question

What is the motivation of adjoints and normal operators. By "motivation," I mean an example, such as a proof, where it is natural to use them.

Answer 1

In some applications, the adjoint operator has a meaning. For instance, in signal processing, there are linear operators $T$ that map a signal (think: a sample of human speech) into a different representation (think: a digital representation), and then the adjoint $T^{*}$ maps the corresponding representations back into signals (e.g. digital signals into audio). A simple example is the Fourier transform for periodic signals.

If you're familiar with Fourier series, then an example of the previous type is the Fourier transform: if $T : L^{2}(\mathbb{T}) \to \ell^{2}(\mathbb{Z})$ is given by $Tf = (\hat{f}(n))_{n \in \mathbb{Z}}$, where $\hat{f}(n) = \int_{\mathbb{T}} f(x) e^{-i 2 \pi x} \, dx$, then $T^{*}a = \sum_{j \in \mathbb{Z}} a_{j} e_{j}$, where $e_{j}(x) = e^{i 2 \pi j x}$, is the adjoint. The interpretation is periodic "speech" signals get mapped to their "digital" Fourier coefficients by $T$, and $T^{*}$ takes sequences of coefficients to periodic signals.

Another nice example of this is in the theory of diffusion processes. Roughly speaking, a diffusion process is determined by the partial differential equation $\frac{\partial u}{\partial t} - L u = 0$. If $L^{*}$ is the adjoint operator of $L$, then the partial differential equation $\frac{\partial u}{\partial t} - L^{*}u = 0$ governs the time reversal of the diffusion. (The time reversal of a diffusion is another diffusion that looks like the original going backwards.) In other words, in this case, taking adjoints is interpreted as reversing time.

Answer 2

The idea of adjoint came from Lagrange who used integration by parts to differentiations and multiplications from one function to another in the integral. $$ \int (Lf)gdx-\int f(L^*g)dx = \mathscr{L}(f,g). $$ $L^*$ was the Lagrange adjoint obtained by integrating by parts. Lagrange used his formula to come up with variation of parameters in order to reduce an ODE to lower order, and his formula was later used to study the first symmetric differential equations arising from Fourier's separation of variables technique for solving his Heat equation. Fourier's application, however, naturally involved cases where $L=L^*$, which singled this case out for further study.

Sturm, along with Liouville initiated a study of "symmetric" ODEs in this context of Fourier and Lagrange, and they studied the associated orthogonal eigenfunction expansions as well. Endpoint conditions were imposed that would force the evaluation terms $\mathscr{L}(f,g)$ to vanish. Such conditions arose naturally in the context of Fourier's study of the Heat Equation. This led to operators that were symmetric on the domain of functions that were sufficiently differentiable and satisfied the endpoint conditions:

$$ \int_a^b (Lf)g dx = \int_a^b f(Lg)dx. $$ They realized that, just as Fourier had found, there were resulting eigenfunctions, discrete eigenvalues, and functions could be expanded in these orthogonal eigenfunctions. It was quite a remarkable thing considering that linear space had not been defined yet, and they were working in an infinite-dimensional space. It wasn't until decades later that symmetry was used to study matrices, and to find similar orthogonal expansions in eigenvectors of symmetric matrices.

So it all seems a little unnatural because infinite-dimensional analysis of symmetry and eigenfunctions came well before the analysis of finite-dimensional cases, which makes the natural applications inaccessible in the study of finite-dimensional Linear Algebra. The most abstract came first, which is also rather unusual in Mathematics.

By the way, I'm not sure where the study of normal operators started, but a normal $N$ can be written as $N=A+iB$ where $A$, $B$ are selfadjoint and commute with each other.

Answer 3

In order to keep things elementary as possible, let's work in $\mathbb{C}^n$.

Get a linear map $A:\mathbb{C}^n\rightarrow\mathbb{C}^n$. Ask yourself if it is possible to decompose $\mathbb{C}^n$ as a direct sum of $n$ one-dimensional mutually orthogonal $A$-invariant spaces, or, said in another way, ask yourself when your linear map $A$ is unitarily equivalent to a multiplication operator. This is equivalent to find an orthonormal base of $\mathbb{C}^n$ made of eigenvectors of $A$, i.e. you have to find an orthonormal base $\{e_1,...,e_n\}$ of $\mathbb{C}^n$ and $\lambda_1,...,\lambda_n\in\mathbb{C}^n$ such that $$Ae_1=\lambda_1e_1,...,Ae_n=\lambda e_n.$$ Suppose it is possibile to perform such a decomposition. Then, if $v\in\mathbb{C}^n$ and $c_1,...,c_n\in\mathbb{C}^n$ are such that $$v=\sum_{k=1}^nc_ke_k$$ you obtain: $$\langle A^*Av,v\rangle=\langle Av,Av\rangle=\sum_{k=1}^n|\lambda_k|^2|c_k|^2=\sum_{j=1}^n\sum_{k=1}^n\sum_{h=1}^n|\lambda_h|^2c_k\bar{c_j}\delta_{h,k}\delta_{h,j}=\\=\sum_{j=1}^n\sum_{k=1}^n\sum_{h=1}^n|\lambda_h|^2c_k\bar{c_j}\langle e_k,e_h\rangle \langle e_h,e_j\rangle=\sum_{j=1}^n\sum_{k=1}^n\sum_{h=1}^nc_k\bar{c_j}\langle e_k,\lambda_he_h\rangle \langle \lambda_he_h,e_j\rangle=\\=\sum_{j=1}^n\sum_{k=1}^n\sum_{h=1}^nc_k\bar{c_j}\langle e_k,Ae_h\rangle \langle Ae_h,e_j\rangle=\sum_{j=1}^n\sum_{k=1}^n\sum_{h=1}^nc_k\bar{c_j}\langle A^*e_k,e_h\rangle \langle Ae_h,e_j\rangle\\=\sum_{k=1}^n\sum_{h=1}^nc_k\langle A^*e_k,e_h\rangle \langle Ae_h,\sum_{j=1}^nc_je_j\rangle=\sum_{k=1}^n\sum_{h=1}^n \langle c_k\langle A^*e_k,e_h\rangle Ae_h,\sum_{j=1}^nc_je_j\rangle=\\=\sum_{h=1}^n \langle \langle A^*\sum_{k=1}^nc_ke_k,e_h\rangle Ae_h,\sum_{j=1}^nc_je_j\rangle=\\=\sum_{h=1}^n \langle \langle A^*v,e_h\rangle Ae_h,v\rangle=\sum_{h=1}^n \langle A\left(\langle A^*v,e_h\rangle e_h\right),v\rangle=\\= \langle A\left(\sum_{h=1}^n\langle A^*v,e_h\rangle e_h\right),v\rangle=\langle AA^*v,v\rangle.$$ Then $$\forall v\in\mathbb{C}^n, \langle (A^*A-AA^*)v,v\rangle=0.$$ Then, if $B:=A^*A-AA^*$, you have: $$\forall v,w \in\mathbb{C}^n, 0=\langle B(v+w),(v+w)\rangle=\\=\langle Bv,v\rangle+\langle Bv,w\rangle+\langle Bw,v\rangle+\langle Bw,w\rangle=\langle Bv,w\rangle+\langle Bw,v\rangle,$$ and $$\forall v,w \in\mathbb{C}^n, 0=\langle B(v+iw),(v+iw)\rangle=\\=\langle Bv,v\rangle-i\langle Bv,w\rangle+i\langle Bw,v\rangle+\langle Bw,w\rangle=-i\langle Bv,w\rangle+i\langle Bw,v\rangle.$$ Then, multiplying the second equation by $i$, adding the result to the first equation and dividing by 2, you get: $$\forall v,w\in\mathbb{C}^n, \langle Bv,w\rangle=0,$$ and so $\forall v\in\mathbb{C}^n, Bv=0$ and so $A^*A-AA^*=B=0$, or: $$A^*A=AA^*,$$ i.e. $A$ is normal.

Vice versa, the spectral theorem for normal operators states that if a linear map $A:\mathbb{C}^n\rightarrow\mathbb{C}^n$ is normal, then you can decompose $\mathbb{C}^n$ as a direct sum of $n$ one-dimensional mutually orthogonal $A$-invariant spaces.

So here is the crux of the matter: a linear map $A:\mathbb{C}^n\rightarrow\mathbb{C}^n$ is unitarily equivalent to a multiplication operator if and only if $A$ is normal.