TITLE: Motivation of Adjoints and Normal Operators
QUESTION [8 upvotes]: 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.

REPLY [2 votes]: 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.