MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Assume $A, B$ are self-ajoint compact operators. Is it true that $\|A+iB\|\le \|2A+iB\|$? Do we have a stronger inequality $\prod_{k=1}^ns_k(A+iB)\le \prod_{k=1}^ns_k(2A+iB)$ or even stronger one $s_n(A+iB)\le s_n(2A+iB)$, $n=1, 2, \ldots$, where $s_n$ are s-numbers?

share|cite|improve this question
Even the first inequality doesn't seem to hold - take $2A + iB = 0$ for non-zero $A$. – robibok Dec 27 '12 at 15:40
they are self-ajoint – Betrand Dec 27 '12 at 15:48
could you add some motivation behind the first inequality? – Suvrit Dec 27 '12 at 19:54
A motivation is from the scalar case, as in this article – Betrand Dec 27 '12 at 22:17
up vote 4 down vote accepted

I just tried a few random matrices... $$ \begin{align} &A=\begin{bmatrix} -0.1 & -0.4\cr -0.4&0\end{bmatrix}, \qquad B=\begin{bmatrix} 1.5 & -0.5 + i\cr -0.5 - i& 3.5\end{bmatrix}, \cr &\|A+iB\| = \frac{7}{2}+\frac{\sqrt{61}}{10}\approx 4.28 \cr &\|2A+iB\| = \frac{7}{2}+\frac{\sqrt{29}}{10}\approx 4.04 \end{align} $$

share|cite|improve this answer
  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)


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*}

share|cite|improve this answer
What if $A, B$ are positive definite? – Betrand Dec 27 '12 at 22:14
Thanks, but how do you define determinant? If $A, B$ are positive definite matrices, we have $|\det(2A+iB)|\ge |\det(A+iB)|$; see Lemma 5 of Kh. D. Ikramov, Determinantal inequalities for accretive-dissipative matrices, J. Math. Sci. (N. Y.), 121(2004) 2458-2464. – Betrand Dec 28 '12 at 19:29
I have shown above the determinant that you claim in your comment. The absolute value of the determinant of a complex matrix $X$ is the product of its singular values. Using that, my edit shows that $|\det(2A+iB)| \ge |\det(A+iB)|$, which of course, is probably the most trivial result of its type :-) – Suvrit Dec 28 '12 at 20:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.