A bra vector is not a column vector but a vector in an abstract Hilbert space. Usually you organize the components of a vector with respect to a basis as a column vector and the matrix elements of an operator as a matrix. Of course, it's much more convenient to use the Dirac formalism since then you don't need to think about this organization much. For the operation of an operator on a vector you have
$$\hat{A} |\psi \rangle=\sum_{j,k=1}^{\infty} |j \rangle \langle j|\hat{A}|k \rangle \langle k|\psi \rangle=\sum_{j,k=1}^{\infty} |j \rangle A_{jk} \psi_k.$$
Now you organize the vector components as a column and the matrix elements as a matrix,
$$\tilde{A}=(A_{jk})=\begin{pmatrix} A_{11} & A_{12} &\ldots \\ A_{21} & A_{22} & \ldots \\ \vdots & \vdots & \vdots \end{pmatrix},$$
$$\tilde{\psi}=(\psi_k)=\begin{pmatrix} \psi_1 \\ \psi_2 \\ \vdots \end{pmatrix}.$$
Then you can have for the components of ##|\phi \rangle=\hat{A} |\psi \rangle##
$$\tilde{\phi}=\tilde{A} \tilde{\psi}$$
in the sense of the usual matrix-vector product. Note, however, that you have vectors and matrices with infinitely many components.