how can I obtain the lower bound for this case? For an arbitrary pair of orthogonal bases $\phi$, $\varphi$ that construct the matrix A and which of them are $n \times n$. mutual coherence for matrix $A$ is defined as te maximal inner product between columns from these two bases:
$$\mu(A)=\displaystyle\max_{1 \leq i,j \leq n} |\varphi_{i}^{T}\phi_{j}|.$$
The mutual coherence of such two-ortho matrices satisfies $1/\sqrt{n} \leq \mu(A) \leq 1$. How can I obtain this lower bound of $\mu(A)(i.e. 1/\sqrt{n} )?$
Can anyone help me in this case?
Thank you!
 A: Since the matrix $(\varphi_i^T\phi_j)=\varphi^T\phi$ is orthogonal$^{\color{red}{*}}$, the question is equivalent to:

Given a square orthogonal matrix $U=(u_{ij})\in\mathbb{R}^{n\times n}$, we have that
  $$
\frac{1}{\sqrt{n}}\leq\max_{1\leq i,j\leq n}|u_{ij}|\leq 1.
$$

Since the spectral norm of $U$ is equal to one, the upper bound is obvious (as the 2-norm is an upper bound for the absolute value of any entry of $U$). The lower bound can be obtained as follows. Since $U$ is orthogonal, its rows/columns have the Euclidean norm equal to one. So, e.g.,
$$
\sum_{i=1}^n u_{ij}^2=1, \quad j=1,\ldots,n,
$$
and we have
$$
1=\sum_{i=1}^n u_{ij}^2\leq n \left(\max_{1\leq i\leq n}|u_{ij}|\right)^2
\quad\Rightarrow\quad
\max_{1\leq i\leq n}|u_{ij}|\geq\frac{1}{\sqrt{n}}, \quad j=1,\ldots,n.
$$
Consequently,
$$
\max_{1\leq i,j\leq n}|u_{ij}|
=
\max_{1\leq j\leq n}\left(\max_{1\leq i\leq n}|u_{ij}|\right)
\geq 
\max_{1\leq j\leq n}\left(\frac{1}{\sqrt{n}}\right)=\frac{1}{\sqrt{n}}.
$$

Note that this cannot be obtained by a direct application of the equivalence between the matrix 2-norm and the so called max-norm, as this leads to the lower bound $1/n$ instead of $1/\sqrt{n}$. However, it is more or less a direct application of the equivalence between the vector 2- and $\infty$-norms.

$^{\color{red}{*}}$ $(\varphi^T\phi)^T(\varphi^T\phi)=\phi^T\underbrace{\color{blue}{(\varphi\varphi^T)}}_{=I}\phi=\phi^T\phi=I$.
