The complex numbers and the algebra of all $n \times n$ complex matrices are particular examples of (complex) Banach algebras with involution. The involution is somewhat (though not always perfectly) analogous to complex conjugation. The algebra of bounded linear operators on a Hilbert space is one example of such an algebra (mentioned in the reference of Halmos in Theo's comment), which includes both these cases as subcase. In that case, the involution sends an operator to its adjoint, and the analogy with complex conjugation is quite strong. The behaviour of an operator with respect to this involution is strongly reflected in the behaviour of its spectrum (in the case of finite dimensional vector spaces, spectrum of a matrix equates to the set of eigenvalues of that matrix).
However, the complex numbers is unique among (complex) Banach algebras, in that the only Banach division algebra (that is, Banach algebra in which every non-zero element has
a multiplicative inverse) up to isomorphism is the complex numbers itself (this is the Banach-Mazur theorem).

I think you mix everything. When we speak about congruence, the matrices do not act on vectors and, consequently, the angles cannot be preserved. The "similar" in "similar triangles" has nothing to do with the "similar" of "similar matrices". When you write "affine transformations (i.e., uniform scaling, rotation, reflection, translation[?])", you are far from the definition of an affine function.

An affine function is in the form $f:x\in\mathbb{R}^n\rightarrow Ax+b\in\mathbb{R}^p$ where $A\in M_{p,n},b\in\mathbb{R}^p$ are fixed.

Similarity of triangles in the plane. It is associated (up to a translation) to a transformation in the form $z\in\mathbb{C}\rightarrow az$ (for direct similarity) or $z\in\mathbb{C}\rightarrow a\overline{z}$ (for inverse similarity) where $a=u+iv$ is a fixed complex. It's a composition of homothety, rotation and, eventually, symmetry. The associated linear application has the form $\begin{pmatrix}u&-v\\v&u\end{pmatrix}$ (in a vector subspace of $M_2$ that is isomorphic to $\mathbb{C}$).

When we speak about similarity of matrices, the matrix $A$ is considered as a linear function and acts on vectors:
$y=Ax$. By a change of basis $y=Py',x=Px'$ and $y'=P^{-1}APx$, that is, the new matrix is $P^{-1}AP$.

When we speak about congruence of matrices, the matrix $A$ is considered as a bilinear form and acts on a couple of vectors $(x,y)$: $\phi(x,y)=x^TAy$. By a change of basis $y=Py',x=Px'$ and $x^TAy=x'^TP^TAPy'$, that is, the new matrix is $P^TAP$.

Note that, if we use the standard inner product $<.>$, $\phi(x,y)=<Ay,x>=<y,A^Tx>$, that is the definition of the adjoint $A^T$.

- The orthogonal matrices $U$ bridge the two previous notions because $U^T=U^{-1}$: by an orthonormal change of basis, a matrix $A$ can be considered as the matrix of a bilinear form or as the matrix of a linear function.

EDIT. Answer to jjjjjj . Let $T_1,T_2$ be two triangles. According to the standard definitions,

$T_1,T_2$ are similar iff $f(T_1)=T_2$ where $f $ is affine (cf. 1.) with $A=\lambda U$ where $\lambda>0$ and $U$ is orthogonal.

$T_1,T_2$ are congruent iff $f(T_1)=T_2$ where $f $ is affine (cf. 1.) with $A=U$ orthogonal.

Note that if $\det(U)=1$, then the angles are preserved and if $\det(U)=-1$, then the angles are transformed in their opposite.

Assume that $T_1,T_2$ are congruent and placed on a sheet. If $\det(U)=1$, then we can drag $T_1$ onto $T_2$. If $\det(U)=-1$, then we flip $T_1$, after that, we drag it onto $T_2$.

## Best Answer

Let's recap some prelims.

By the spectral theorem, any Hermitian matrix is unitarily diagonalizable with real eigenvalues (see here). Further, a Hermitian matrix is positive definite if and only if its eigenvalues are all positive (see here).

The product of two Hermitian matrices $X$ and $Y$ is Hermitian if and only if they commute i.e. $XY=YX$ (see here).

A set of diagonalizable matrices commutes if and only if the set is simultaneously diagonalizable i.e. there is a single invertible matrix $V$ s.t. $V^{-1}XV$ is diagonal for all $X$ in the set (see here).

Applying these to your setup, if $P$ and $B-A$ are Hermitian and positive definite, their product is Hermitian only if they commute, in which case they must be simultaneously diagonalizable (and hence have the same eigenvectors) with each eigenvalue of $P(B-A)$ given by the product an eigenvalue of $P$ and an eigenvalue of $B-A$, implying $P(B-A)$ is positive definite.