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1. $s_n(A+iB) \not\le s_n(2A+iB)$, for example by setting \begin{equation*} A = \begin{bmatrix} -2 & -6\\ -6 & -2 \end{bmatrix}\qquad B = \begin{bmatrix} 10 & 4\\ 4 & 16 \end{bmatrix}. \end{equation*} In this case, $s_2(A+iB) = 7.7192$ and $s_2(2A+iB) = 4.9433$.

2. $\prod_{k=1}^n s_k(A+iB) \not\le \prod_{k=1}^n s_k(2A+iB)$ for the same example as above. The lhs is 156, while the rhs is 129.2440.

3. The operator norm version also does not hold (as shown in the nice counterexample by Gerald Edgar)

EDIT

If, however, $A$ and $B$ are assumed to be positive definite, then these inequalities probably hold. As a hint why they might hold (I have not had the time to check any of the other cases), consider $C=A+iB$ and $D=2A+iB$ with $A,B \ge 0$. Then,

\begin{equation*} \begin{split} \prod_{j=1}^n s_j(C) = |\det C| &= \det(A)\prod_{j=1}^n[1+ s_j(A^{-1/2}BA^{-1/2})^2]^{1/2}\\ \prod_{j=1}^n s_j(D) = |\det D| &= \det(A)\prod_{j=1}^n2\left[1+ \frac{s_j(A^{-1/2}BA^{-1/2})^2}{4}\right]^{1/2}\\ &= \det(A)\prod_{j=1}^n[4+ s_j(A^{-1/2}BA^{-1/2})^2]^{1/2} \ge |\det(C)|. \end{split} \end{equation*}

2 last point

Only the weakest version of your inequality seems to hold (haven't had time to think about it yet); the other two are easily invalidated by the counterexample below.

1. $s_n(A+iB) \not\le s_n(2A+iB)$, for example by setting \begin{equation*} A = \begin{bmatrix} -2 & -6\\ -6 & -2 \end{bmatrix}\qquad B = \begin{bmatrix} 10 & 4\\ 4 & 16 \end{bmatrix}. \end{equation*} In this case, $s_2(A+iB) = 7.7192$ and $s_2(2A+iB) = 4.9433$.

2. $\prod_{k=1}^n s_k(A+iB) \not\le \prod_{k=1}^n s_k(2A+iB)$ for the same example as above. The lhs is 156, while the rhs is 129.2440.

3. The operator norm version also does not hold (as shown in the nice counterexample by Gerald Edgar)

1

Only the weakest version of your inequality seems to hold (haven't had time to think about it yet); the other two are easily invalidated by the counterexample below.

1. $s_n(A+iB) \not\le s_n(2A+iB)$, for example by setting \begin{equation*} A = \begin{bmatrix} -2 & -6\\ -6 & -2 \end{bmatrix}\qquad B = \begin{bmatrix} 10 & 4\\ 4 & 16 \end{bmatrix}. \end{equation*} In this case, $s_2(A+iB) = 7.7192$ and $s_2(2A+iB) = 4.9433$.

2. $\prod_{k=1}^n s_k(A+iB) \not\le \prod_{k=1}^n s_k(2A+iB)$ for the same example as above. The lhs is 156, while the rhs is 129.2440.