Let $T_{1}$,....,$T_{n}$ be bounded linear operators on a complex Hilbert space $H$ and $\alpha_{1}$,....,$\alpha_{n}$ be complex numbers. Let
$M_\alpha=\sum_{i=1}^n\alpha_{i}T_{i}$.
If $n=2$.
Let $N_{\alpha}= \overline{\alpha_{2}}T_{1}- \overline{\alpha_{1}}T_{2}$. Then
$$M_\alpha^* M_\alpha+N_\alpha^* N_\alpha = (|\alpha_{1}|^{2}+|\alpha_{2}|^{2}) (T_{1}^{*}T_{1}+T_{2}^{*}T_{2}). $$
Assume that $n>2$. Is it possible to construct an operator $N_\alpha$ such that $$M_\alpha^* M_\alpha+N_\alpha^* N_\alpha=(\sum_{i=1}^n|\alpha_{i}|^2)(\sum_{i=1}^nT_{i}^*T_{i})\;?$$
To make my point clearer, let us take $n$ to be $3$. Note that $M_{\alpha} = \sum_{i = 1}^3 \alpha_i T_i$, assume without loss of generality that $\overrightarrow{\alpha} = (\alpha_1, \alpha_2, \alpha_3)$ is a unit vector in $\mathbb{C}^3$. By Gram Schmidt, we can find $\overrightarrow{\beta} = (\beta_1, \beta_2, \beta_3)$ and $\overrightarrow{\gamma} = (\gamma_1, \gamma_2, \gamma_3)$ such that $\overrightarrow{\alpha}, \overrightarrow{\beta}, \overrightarrow{\gamma}$ form an orthonormal basis of $\mathbb{C}^3$. Define two operators $$E_\beta = \sum_{i = 1}^3 \beta_i T_i, \qquad E_\gamma = \sum_{i = 1}^3 \gamma_i T_i,$$ then $$M^∗_\alpha M_α+M^∗_\beta M_\beta + M^*_\gamma M_\gamma = T_1^* T_1 + T_2^* T_2 + T_3^* T_3.$$ When $\overrightarrow{\alpha}$ is not a unit vector, you need to multiply the right hand side by the square of length of $\overrightarrow{\alpha}$. This construction also works for general $n$, you need to construct $n-1$ more operator instead of $2$ here.