Is transition matrix between two orthonormal bases orthogonal? Is it true that the transition matrix between two orthonormal bases is orthogonal?
If true, why? I do really not understand. How can I prove that (some ideas)? Thank you all!
 A: As user1551 points out, it's false in general, but true when talking about orthonormal bases with respect to the real Euclidean inner product and real entries. 
So for easier understanding, let's suppose this is given and the vector space is $(\mathbb{R}^n, \langle . \rangle)$ with standard euclidean product, let $e = (e_1, \dots, e_n)$ be standard basis and have two orthonormal bases $b = (b_1, \dots, b_n)$ and $c = (c_1, \dots, c_n)$.
Have two orthogonal matrices $B$ and $C$, where the columns of $B$ (resp. $C$) are exactly the basis vectors of $b$ (resp. $c$) expressed in coordinates of the standard basis. (standard situation in e.g. an exercise, where the matrix columns are just the vectors)
Let $M(id, b, c)$ denote the change of base matrix from $b$ to $c$. 
It's always helpful to remind yourself of what an entry in a change of base matrix stands for. The $j$-th column if $M(id, b, c)$ is the vector $b_j$ expressed in coordinates of $c_1, \dots, c_n$
As you probably know, when changing basis from $b$ to $c$, you can also first go from $b$ to $e$ and then from $e$ to $c$, i.e.
$$M(id, b,c) = M(id, e, c) \cdot M(id, b, e)$$
What is $M(id, b, e)$? As said, the $j$-th column of this matrix is $b_j$ expressed in coordinates of the standard basis. So $M(id, b, e) = B$. 


*

*Can you say which matrix $M(id, e, c)$ is equal to?

*Then, with the knowledge of $B$ and $C$ being orthogonal, it will be easy for you to show that $M(id, b, c)$ is orthogonal as well.

