Consider I have
$$k = \| \mathbf{A} - \mathbf{S} \|_F^2 + \| \mathbf{A} - \mathbf{B} \|_F^2$$
where $\mathbf{A}$, $\mathbf{S}$, $\mathbf{B}$ are matrices, and $\| \cdot \|_F$ is the Frobenius norm. I just want to know if $k$ can be re-written with just one squared Frobenius norm, say, $k = \| \mathbf{A} - \mathbf{X}' \|_F^2$. Is it possible?
In other words, can we form the squared Frobenius norm $\|\mathbf{A} - \mathbf{X}'\|_F^2$ from $\|\mathbf{A}-\mathbf{S}\|_F^2 + \|\mathbf{A}-\mathbf{B} \|_F^2$? Do you know some helpful properties? If so, so what can be the value of $\mathbf{X}'$?
As the others have answered in the comments, what you ask is not possible. However you can obtain something that may be equally useful in some cases. The following holds: $$ \Vert\mathbf{A}-\mathbf{S}\Vert_F^2 + \Vert\mathbf{A}-\mathbf{B}\Vert_F^2 = 2\,\left\Vert\mathbf{A}-\frac12(\mathbf{S}+\mathbf{B})\right\Vert_F^2 + \frac12\, \Vert\mathbf{S}-\mathbf{B}\Vert_F^2 $$ (it can be easily proven by using $\Vert\mathbf{M}\Vert_F^2 = \mathrm{trace}(\mathbf{M}^\top\mathbf{M})$, expanding all terms on both sides and comparing). While this is not what you asked for, in case you are minimising $k$ w.r.t. $\mathbf{A}$ this is equally useful since the second term is constant and can be omitted from the optimisation problem. I think this is what @MichaelGrant hinted at when he asked which are variables and which are constants.