# matrix elements of unitary operator

I have read that the matrix elements of a unitary operator are the same in a pair of orthonormal basis such that {${|a_i\rangle}$} and {${|b_i\rangle}$} are orthonormal bases and ${|b_i\rangle} = U{|a_i\rangle}$:

$$\langle{a_i}|U|a_j\rangle = \langle{b_i}|U|b_j\rangle$$

I have 2 questions:

$$\langle{a_i}|U|a_j\rangle = \langle{a_i}|U1|a_j\rangle$$

and using ${1} =UU^{\dagger} = U^{\dagger}U:$

$$\langle{a_i}|U|a_j\rangle = \langle{a_i}|UU^{\dagger}U|a_j\rangle$$

$$\langle{a_i}|U|a_j\rangle =\langle{a_i}U|U^{\dagger}|Ua_j\rangle$$

$$\langle{a_i}|U|a_j\rangle = \langle{b_i}|U^{\dagger}|{b_j}\rangle$$

Is this true? And if not, what did I do wrong?

Also, intuitively, why must it be that the matrix representation of a unitary operator looks the same in such bases?

Edit

The original question mistakenly started off: "I have read that the matrix elements of a unitary operator are the same in any pair of orthonormal basis" which as shown by celtschk in his answer is clearly false.

Counterexample: Be $$U=\pmatrix{1&0\\0&-1}$$ This is easily checked to be an unitary matrix. In the orthonormal basis $$\left|a_0\right>=\pmatrix{1\\0}, \left|a_1\right>=\pmatrix{0\\1}$$ we have $\left<a_i\middle|U\middle|a_j\right> = (-1)^i\delta_{ij}$. The basis $$\left|b_i\right> = \frac{1}{\sqrt{2}}\left(\left|a_0\right> + (-1)^i\left|a_1\right>\right)$$ can easily be checked to be an orthonormal basis as well, but \begin{aligned} \left<b_i\middle|U\middle|b_j\right> &= \frac{1}{2}\left(\left<a_0\right| + (-1)^i\left<a_1\right>\right)U\left(\left|a_0\right> + (-1)^j\left|a_1\right>\right)\\ &= \frac{1}{2}\left(\left<a_0\middle|U\middle|a_0\right> + (-1)^i\left<a_1\middle|U\middle|a_0\right> + (-1)^j\left<a_0\middle|U\middle|a_1\right> + (-1)^{i+j}\left<a_1\middle|U\middle|a_1\right>\right)\\ &=\frac{1}{2}\left(1-(-1)^{i+j}\right)=1-\delta_{ij} \end{aligned} Clearly $(-1)^i\delta_{ij} \ne 1-\delta_{ij}$.
A simpler counterexample would be the above matrix with $\left|b_0\right> = \left|a_1\right>$ and $\left|b_1\right> = \left|a_0\right>$.