I am investigating the biggest singular value of Kronecker Sum. I'd like to understand if my argument is correct.
Let $A$, $B$ be square matrices of size $n$. Let $I$ be an identity matrix of size $n$. Let $S = A \otimes I + I \otimes B$. Now let $v_S$ be a right singular vector of $S$ (we will analogously define $v_A$ and $v_B$) associated with the largest singular value. Matrix norm $\lVert \cdot \rVert$ will always refer to a matrix 2-norm (or euclidian vector norm)
Now from SVD we have $\lVert S v_s\rVert = \lVert\sigma_1u_s\rVert = \sigma_1 \lVert u_s \rVert = \sigma_1$.
We can decompose $v_s$ using Kronecker Product $v_s = v_l \otimes v_r$. To obtain the following: $$ Sv_s = (A \otimes I + I \otimes B)(v_l \otimes v_r) \\ = (A \otimes I)(v_l \otimes v_r) + (I \otimes B)(v_l \otimes v_r) \\ = (Av_l \otimes v_r)+ (v_l \otimes Bv_r) $$
Let's calculate the norm with help of the law of cosines $$ \lVert (Av_l \otimes v_r) + (v_l \otimes Bv_r) \rVert^2 \\ = \lVert (Av_l \otimes v_r) \rVert^2 + \lVert (v_l \otimes Bv_r) \rVert^2 + 2 \lVert (Av_l \otimes v_r) \rVert\lVert (v_l \otimes Bv_r) \rVert \cos(Av_l \otimes v_r, v_l \otimes Bv_r) $$
Let's fisrt focus on $\lVert (Av_l \otimes v_r) \rVert^2$ $$ \lVert Av_l \otimes v_r \rVert^2 = \lVert Av_l\rVert^2\lVert v_r\rVert^2 \\ \leq \lVert Av_l \rVert^2 \textit{ (from fact that } v_s \text{has norm of } 1 \text{)}\\ \leq \sigma_A^2 $$
We can analogously consider $v_l \otimes Bv_r$.
Now let's focus on $\cos(Av_l \otimes v_r, v_l \otimes Bv_r)$. That's the part I am least sure about.
$$ \cos(Av_l \otimes v_r, v_l \otimes Bv_r) = \\ = Tr((Av_l \otimes v_r)^H(v_l \otimes Bv_r)) \\ = Tr((v_l^HA^H\otimes v_r^H)(v_l \otimes Bv_r)) \\ = Tr(v_l^HA^Hv_l\otimes v^H_rBv_r) \\ = Tr(v_l^HA^Hv_l)Tr(v^H_rBv_r) \\ = Tr(v_lv_l^HA^H)Tr(v_rv_r^HB) $$
We know that $v_lv_l^H$ is equal to a $c_l = \lVert v_l \rVert^2$. We can analogously consider $v_r$. Based on that we have:
$$ Tr(v_lv_l^HA^H)Tr(v_rv_r^HB) = \\ c_lc_rTr(A^H)Tr(B) = \\ c_lc_rTr(A)Tr(B) $$
We know that $\lVert v_s \rVert = 1 = c_lc_r$ (from properties of Kronecker Product). Summing all of that up we have:
$$ \sigma_S^2 \leq \sigma_A^2 + \sigma_B^2 + Tr(A)Tr(B) $$
Please criticise my reasoning.