I have a question related with singular values of matrix sums.

Let's assume I have matrices $A$, $B$, and $D$ (positive, semi-definite) where $D = A + B$. For singular values of $D$, I know that

$$ λ(D) \le ∑λ(A)+∑λ(B). $$

However, if I want to calculate the singular values of $D = A + cB$, where $c$ is a constant (real number) other than just $1$, does the assumption above still hold?

Thus, can I assume that for the singular values of $D = A + cB$ the following holds

$$ ∑λ(D) \le ∑λ(A) + c∑λ(B). $$

  • $\begingroup$ Could you define $\lambda(D)$ and possibly check whether the following MO posts mathoverflow.net/questions/4224/eigenvalues-of-matrix-sums and its follow-up mathoverflow.net/questions/31475/singular-values-of-matrix-sums may help prove or disprove your inequalities? $\endgroup$
    – Luc Guyot
    Commented Aug 3, 2016 at 21:44
  • 3
    $\begingroup$ This is true; it's the triangle inequality for the trace norm. $\endgroup$ Commented Aug 3, 2016 at 21:52
  • 1
    $\begingroup$ (assuming of course $c \ge 0$) $\endgroup$ Commented Aug 4, 2016 at 1:22
  • $\begingroup$ So, it won't be true for $c < 0$ @RobertIsrael? Because, I may also use negative real numbers. $\endgroup$
    – ciyo
    Commented Aug 4, 2016 at 14:13
  • $\begingroup$ If $A$, $B$ and $A + c B$ are all positive semidefinite, the singular values are just the eigenvalues and their sum is the trace. Then of course $\sum \lambda(A+cB) = \sum \lambda(A) + c \sum \lambda(B)$. You only need an inequality when $A+cB$ might not be positive semidefinite, and that inequality will require $c \ge 0$. $\endgroup$ Commented Aug 4, 2016 at 16:15


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.