Matrix representation of a unitary operator, change of basis

[Kashmir]

If $U$ is an unitary operator written as the bra ket of two complete basis vectors :$U=\sum_{k}\left|b^{(k)}\right\rangle\left\langle a^{(k)}\right|$

$U^\dagger=\sum_{k}\left|a^{(k)}\right\rangle\left\langle b^{(k)}\right|$

And we've a general vector $|\alpha\rangle$ such that $|\alpha\rangle=\sum_{a^{\prime}}\left|a^{\prime}\right\rangle\left\langle a^{\prime} \mid \alpha\right\rangle$

Sakurai writes at pg 50 :
"how can we obtain $\left\langle b^{\prime} \mid \alpha\right\rangle$, the expansion coefficients in the new basis? answer is very simple: Just multiply (1.5.9) by $\left\langle b^{(k)}\right|$
$\left\langle b^{(k)} \mid \alpha\right\rangle=\sum_{l}\left\langle b^{(k)} \mid a^{(l)}\right\rangle\left\langle a^{(l)} \mid \alpha\right\rangle=\sum_{l}\left\langle a^{(k)}\left|U^{\dagger}\right| a^{(l)}\right\rangle\left\langle a^{(l)} \mid \alpha\right\rangle .$
$(1.5 .1$
In matrix notation, (1.5.10) states that the column matrix for $|\alpha\rangle$ in the new basis can be obtained just by applying the square matrix $U^{\dagger}$ to the colum matrix in the old basis:
$\quad(\mathrm{New})=\left(U^{\dagger}\right)($old $)$"So if the matrix representing $U^\dagger$ is applied on to the matrix representing $|\alpha\rangle$ ,it gives the vectors representation in the new basis. But when I apply $U^\dagger$ onto say an basis vector $\left|a_{1}\right\rangle$ ,it doesn't give me the vectors representation in new basis as shown below :

$\begin{aligned} U^{\dagger}\left|a_{1}\right\rangle &=\sum_{k}\left|a^{k}\right\rangle\left\langle b^{k} \mid a_{1}\right\rangle \\ &=\sum_{k}\left(\left\langle b^{k} \mid a_{1}\right\rangle\right) \cdot\left|a^{k}\right\rangle \end{aligned}$

 

[vanhees71]

Why should it? You rather have
$$\ket{b^k}=\hat{U} \ket{a^k}.$$
Sakurai in the quoted text uses the adjoint of this
$$\bra{b^k}=\bra{a^k} \hat{U}^{\dagger}.$$