Let A, B be vector spaces of traceless matrices such that A is orthogonal to B (in Frobenius inner product sense) and $S = \{ a + b\ |\ a \in A, b \in B \}$ is an internal direct sum.
Is there relationship between matrix 2-norm (biggest singular value) of $S$ and $A, B$? I have "intution" that following relationship might hold, but I do not know how to approach it: for all $s\in S$ (with unique decomposition $s=a+b$) $$ \lVert s \rVert_2^2 \leq \lVert a \rVert_2^2 + \lVert b \rVert_2^2 $$
I'd say your guess that $\|s\|^2\leq\|a\|^2+\|b\|^2$ was an educated one given---from what I could find---it seems to hold true for $2\times 2$ matrices that fit your scenario. However, consider the following subspaces of $\mathbb C^{3\times 3}$: $$ A:=\Big\{\begin{pmatrix}c&0&0\\0&-c&0\\0&0&0\end{pmatrix}:c\in\mathbb C\Big\}\qquad B:=\Big\{\begin{pmatrix}0&0&0\\0&0&c\\0&c^*&0\end{pmatrix}:c\in\mathbb C\Big\} $$ Certainly $A,B$ are vector spaces of traceless matrices which are (Frobenius) orthogonal: for all $c_1,c_2\in\mathbb C$ $$ {\rm tr}\Big(\begin{pmatrix}c_1&0&0\\0&-c_1&0\\0&0&0\end{pmatrix}^* \begin{pmatrix}0&0&0\\0&0&c_2\\0&c_2^*&0\end{pmatrix} \Big)={\rm tr}\begin{pmatrix}0&0&0\\0&0&-c_1^*c_2\\0&0&0\end{pmatrix}=0 $$ However, choosing $c_1=c_2=1$ we find \begin{align*} \Big\|\begin{pmatrix}1&0&0\\0&-1&0\\0&0&0\end{pmatrix}+\begin{pmatrix}0&0&0\\0&0&1\\0&1&0\end{pmatrix}\Big\|_2^2=\Big\|\begin{pmatrix}1&0&0\\0&-1&1\\0&1&0\end{pmatrix}\Big\|_2^2=\frac{3+\sqrt5}2\approx 2.618\ldots \end{align*} which is larger than $$ \Big\|\begin{pmatrix}1&0&0\\0&-1&0\\0&0&0\end{pmatrix}\Big\|_2^2+\Big\|\begin{pmatrix}0&0&0\\0&0&1\\0&1&0\end{pmatrix}\Big\|_2^2=1+1=2\,. $$ The reason the inequality $\|a+b\|^2\leq\|a\|^2+\|b\|^2$ fails in this case is the following: evaluating $\|a+b\|^2$ results in \begin{align*} \|a+b\|^2&=\sup_{\|x\|=1}\big\|(a+b)x\|^2=\sup_{\|x\|=1}\big\langle(a+b)x,(a+b)x\rangle\\ &=\sup_{\|x\|=1}\big\langle x,(a^*a+a^*b+b^*a+b^*b)x\rangle\\ &=\sup_{\|x\|=1}\|ax\|^2+2{\rm Re}(\langle x,a^*bx\rangle)+\|bx\|^2 \end{align*} Now if the maximizing $x$ in this supremum were always such that ${\rm Re}(\langle x,a^*bx\rangle)=0$, then we would obtain \begin{align*} \sup_{\|x\|=1}\|ax\|^2+2{\rm Re}(\langle x,a^*bx\rangle)+\|bx\|^2&=\sup_{\|x\|=1}\|ax\|^2+\|bx\|^2\\ &\leq\sup_{\|x\|=1}\|ax\|^2+\sup_{\|y\|=1}\|by\|^2=\|a\|^2+\|b\|^2\,. \end{align*} One special case where this is true is one where $A,B$ are orthogonal in the sense that $a^*b=0$ for all $a\in A$, $b\in B$. However, as the above example shows this need not be the case and the term $\langle x,a^*bx\rangle$ can very well increase the norm of $\|a+b\|^2$ beyond the value of $\|a\|^2+\|b\|^2$.