v_2^2), \notag \\
M_{33} &=&  -\frac{1}{v_L} (M_2 v_1 v_R + M_1 v_2 v_R) +  \a_3 (v_2^2 - v_1^2), \notag \\
M_{44} &=& -\frac{1}{v_R} (M_2 v_1 v_L + M_1 v_2 v_L) +  \a_6 (v_2^2 - v_1^2), \notag \\
M_{55} &=&  \gamma_1 (v_1^2 + v_2^2)+ 4 \gamma_2  v_1 v_2 + \gamma_3 v_L^2 + \gamma_4 v_R^2 - \mu_5^2.\end{aligned}$$ This $5\times 5$ charged Higgs mass-squared matrix can be diagonalized to obtain their mass eigenvalues as $$M^2_{Diag} = V^{\dagger} M_{H^\pm}^2 V,
\label{eq:hdiag}$$ where $M^2_{Diag}$ is the diagonalized charged Higgs boson mass-squared matrix and $V$ is the corresponding diagonalizing matrix. There are two zero eigenvalues corresponding to the two Goldstone bosons absorbed by the $W_R^\pm$ and $W^\pm$ bosons to give them mass. The Goldstone bosons primarily consist of $H_R^\pm$ and $\phi_1^\pm$ states respectively as their corresponding doublet neutral fields get the large non-zero VEVs. The other three eigenstates give the three physical charged Higgses and are linear combinations of $\phi_2^\pm$, $H_L^\pm$ and $\d^\pm$. Flavor constraints, such as, $K^0-\bar{K}^0$ and $B^0-\bar{B}^0$ mixings require the neutral component of the bidoublet field $\phi_2^0$ mass to be heavier than 15 TeV [@Zhang:2007da], forcing its charged counterpart to be very massive as well. So $\d^\pm$ can primarily mix only with $H_L^\pm$ 