can someone please explain self-adjoint operator intuition to me.
And why when $T^* = T^{-1}$, $T$ preserves the inner product and therefore preserves the the orthonormal basis and the length and distance?
thank you
Self-adjoint operator intuition
adjoint-operatorsinner-productsunitary-matrices
Related Solutions
Let me first recall the definition of the transpose. Let $T: \ \ X \longrightarrow Y$ be a linear operator from $X$ to $Y$. The dual $X^*$ is the space of (continuous) linear functional on $X$, i.e., linear operators from $X \longrightarrow \mathbb{C}$. So if $x\in X^*, \ v \in X$, $x(v) \in \mathbb{C}$. We may even denote the composition of $x$ with $v$ as
$$ x(v):= (x,v). \ \ \quad \ \ (0) $$
Now the transpose of $T$ is the linear operator $ T^t : Y^* \longrightarrow X^*$ defined as
$$ (y, Tv) = ( T^t y, v) \ \ \quad \ \ (1) $$
for all $ v\in X, \ y \in Y^*$. Take now bases of $X,Y$ (say $e_i, \ f_j$) and natural bases of $X^*,Y^*$ (satisfying $ (\eta_i , e_j) = \delta_{i,j}$ and so on). Assume $\dim X = n$ and $\dim Y = m$. So for example $(0)$ becomes
$$ (x,v) = \sum_{i} x_i v_i \ \ \quad \ \ (2) $$
The matrix of $T$ in these bases is defined as
$$ Te_i = \sum_{j=1}^m T_{j,i} f_j $$
(for $i=1,2,\ldots, n$). Using the convention above we have
$$ Tv = T\sum_i v_i e_i = \sum_{i,j} v_i T_{j,i} f_j $$
so that the components of $(Tv)_k$ are given by $ (Tv)_k = \sum_i T_{k,i} v_i$. These are the usual matrix-vector multiplication rules.
You can see that the transpose is associated to the matrix $(T^t)_{i,j}=T_{j,i}$. Note that $T$ corresponds to a $m\times n$ matrix.
However what ultimately allows this to work, or alternatively, what makes definition $(1)$ works, is the following simple fact.
Once you have a matrix of $m\times n$ numbers $T_{i,j}$, you can form $n$-dimensional vectors via $\sum_{j=1}^m T_{k,j} v_j$ ($k=1,\ldots n$) but also $m$-dimensional vectors via $\sum_{i=1}^n T_{i,k} w_i$ ($k=1,\ldots m$).
The latter operation corresponds (using the usual matrix-vector multiplications rules) to
$$ (w_1,w_2, \ldots w_m) \left(\begin{array}{cccc} T_{1,1} & T_{1,2} & \cdots & T_{1,n}\\ T_{2,1}\\ \vdots\\ T_{m,1} & & \cdots & T_{m,n} \end{array}\right) $$
according to the usual rules (row-vector times matrix). To form the adjoint we proceed in a very similar way. However now we use the scalar product to identify linear functionals. That is, instead of $(2)$ we use the scalar product $\langle \bullet, \bullet \rangle$:
$$ \langle x, v \rangle = \sum_{j=1}^n x_j^* v_j $$
Then the construction of the adjoint is the same as that for the transpose, but you get the additional complex conjugate terms.
In component of course, the matrix associated to the adjoint satisfies:
$$ (T^*)_{i,j} = T_{j,i}^* $$
First let's start with the transpose operator. The transpose can be defined without reference to an inner product and only requires the vector space structure. The way to think about the tranpose is that it turns a linear map $f$ from $V$ to $W$ into a linear map $f^T$ from $W^*$ to $V^*$, and it does this in the most "natural" way possible.
If words, if we are trying to define a map from $W^*$ to $V^*$ we start with a function $g \in W^*$ and we want to turn this into a function that acts on vectors in $V$ and returns a real number. Well we know that $g$ can act on vectors in $W$ and we know that using $f$ we can turn vectors in $V$ into vectors in $W$ and chaining these together gives us a way to let $g$ act on vectors in $V$!
In symbols, what we just did above is define $f^T(g)(v) = g(f(v))$. We first use $f$ to map $v$ from a vector in $V$ to a vector in $W$ and then we use our element $g \in W^*$ to map it to a real number.
Now the Hermitian adjoint operator is a way of turning a function $f: V \to W$ into a function $f^*: W \to V$. The best way of thinking about this operator is through the transpose. The transpose gives us an operator $f^T: W^* \to V^*$. To turn this into a map from $W$ to $V$, we need to fix isomorphisms between $W$ and $W^*$ and $V$ and $V^*$. To map a vector $w$ from $W$ to $V$, you use the isomorphism between $W$ and $W^*$ to map $w$ to $w' \in W^*$, then you use the tranpose to map it to a vector $v' \in V^*$ and then you use the isomorphism between $V$ and $V'$ to map it to a vector $v \in V$.
You could describe the above construction with the following diagram: $\require{AMScd}$ \begin{CD} W^* @>{f^T}>> V^*\\ @VV{\cong}V @VV{\cong}V\\ W @>{f^*}>> V \end{CD}
Now the connection with inner product spaces and the Riesz Representation Theorem is that an inner product on a vector space $V$ gives you a "natural" way to define an isomorphism between $V$ and $V^*$.
For every vector $v \in V$, we can define the map $\varphi_v: V \to K$ (where $K$ is the base field) by $\varphi_v(\cdot) = \langle \cdot,v \rangle$. By the linearity of the inner product, $\varphi_v$ is an element of the dual space $V^*$. In addition the map from $V$ to $V^*$ given by $v \to \varphi_v$ is (anti)-linear and by the non-degeneracy of the inner product, it is injective. The Riesz Representation Theorem says that this injection is also a surjection and therefore a bijection. This means that whenever we have an inner product we can define $\varphi_v$ for every $v \in V$ and the map sending $v$ to $\varphi_v$ is a bijection between $V$ and $V^*$.
Now putting these two concepts together (the tranpose and the connection between inner product spaces and dual spaces), we can define the transpose $f^T$ without the inner product structure as a map from $W^* \to V^*$. The inner product structure gives us isomorphisms between $V$ and $V^*$ and $W$ and $W^*$, and chaining these isomorphisms together with the tranpose we can get a Hermitian adjoint map $f^*$ which maps $W$ to $V$.
EDIT: Added connection with $\mathbb{C}$.
Whenever we start writing linear transformations as matrices we first have to fix a basis. This also gives an alternate way of defining an isomorphism between $V$ and $V^*$ (search for "dual basis") and it is a useful exercise to verify that these two isomorphisms between $V$ and $V^*$ agree if and only if the basis that you started with is orthonormal when the base field is $\mathbb{R}$. In this case, let $v_1,\ldots,v_n$ be an orthonormal basis for $V$ and let $v^1,\ldots,v^n$ be its dual basis. Similarly, let $w_1,\ldots,w_m$ be an orthonormal basis for $W$ and let $w^1,\ldots,w^n$ be its dual basis.
Now if we let the matrix representation of $f$ be
$$\begin{bmatrix} f_{11} & \ldots & f_{1n} \\ \vdots & \ddots & \vdots \\ f_{m1} & \ldots & f_{mn} \end{bmatrix}$$
then we have $f(v_i) = \sum_j f_{ji} w_j$. Now if we consider the transpose we have
$$f^T(w^i)(v_k) = w^i(f(v_k)) = w^i\left( \sum_j f_{jk} w^j \right) = f_{ik} = \left( \sum_j f_{ij} v^j \right)(v_k)$$
so
$$f^T(w^i) = \sum_j f_{ij} v^j$$
so the matrix representation of $f^T$ is exactly the traditional transpose of the matrix representation of $f$.
Now let us consider the adjoint in the case of $K = \mathbb{C}$. In this case we have that the map from $v$ to $\varphi_v$ is anti-linear.
Let $w = \sum_i c_i w_i$. Then $\varphi_w = \sum_i \overline{c_i} w^i$. Now if we apply $f^T$ to this we get
$$f^T(\varphi_w) = f^T\left(\sum_i \overline{c_i} w^i\right) = \sum_j \left(\sum_i \overline{c_i}f_{ij}\right) v^j$$
Finally, when we apply the anti-linear isomorphism between $V$ and $V^*$ we get that
$$ f^*(w) = \left(\sum_i c_i\overline{f_{ij}}\right) v^j $$
so the matrix representation of $f^*$ is
$$\begin{bmatrix} \overline{f_{11}} & \ldots & \overline{f_{1m}} \\ \vdots & \ddots & \vdots \\ \overline{f_{n1}} & \ldots & \overline{f_{nm}} \end{bmatrix}$$
which is the conjugate transpose. The basic intuition is that the isomorphism between $W$ and $W^*$ conjugates all of the coefficients of $w$, then this is fed through the transposed matrix $f^T$ and then all of the coefficients are conjugated once again which flips the coefficients of $w$ back and conjugates the coordinates of $f^T$.
EDIT 2: For completeness, I figured I would add why the adjoint as I defined it above is equivalent to the traditional definition of the adjoint as the unique operator $f^*$ such that for all $v,w$ we have
$$\langle f(v), w \rangle = \langle v, f^*(w) \rangle.$$
If you look at the commutative diagram above, we will start with an element $w \in W$ in the bottom left and trace it using the two paths to the top right. First, we can go right to $f^*(w) \in V$. Then we can use the isomorphism between $V$ and $V^*$ to go up to $\varphi_{f^*(w)}$.
Alternatively, we could first go up using the isomorphism between $W$ and $W^*$ to get to $\varphi_w$ and then move right to get to $f^T(\varphi_w)$ so we have that
$$ f^T(\varphi_w) = \varphi_{f^*(w)} $$
as elements of $V^*$. Now evaluating both sides on $v \in V$ gives
$$ \langle f(v), w \rangle = \varphi_w(f(v)) = f^T(\varphi_w)(v) = \varphi_{f^*(w)}(v) = \langle v, f^*(w) \rangle $$
as desired.
Best Answer
Think about a symmetric matrix, it’s a simpler case. I think that the term arise from an observation: matrices that are diagonal if we describe them with orthonormal bases have a symmetric structure if we change coordinates with an orthogonal transformation. This is a different way to interpret the def immediatley using the specrtal theorem.