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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

$$ ∑λ(D) \le ∑λ(A) + c∑λ(B). $$

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  • $\begingroup$ 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
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    $\begingroup$ This is true; it's the triangle inequality for the trace norm. $\endgroup$
    – Christian Remling
    Commented Aug 3, 2016 at 21:52
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    $\begingroup$ (assuming of course $c \ge 0$) $\endgroup$
    – Robert Israel
    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 $c \ge 0$. $\endgroup$
    – Robert Israel
    Commented Aug 4, 2016 at 16:15

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