I have been trying to show that $$||M+N|| \le ||M|| + ||N||$$ However, I seem to be missing some fundamental property of either how the trace or square root acts on these sums of matrices, or how the Hilbert-Schmidt I.N can be used. I can expand it easily, getting $$||M+N|| = Tr|M+N|=Tr[\sqrt{(M+N)(M+N)^\dagger}]$$which evaluates to $$Tr[\sqrt{MM^{\dagger}+MN^{\dagger}+NM^{\dagger}+NN^{\dagger}}]$$

Now obviously you can't just square both sides, as the square operations doesn't distribute over the trace function, nor does the square root over the sum, or this would be trivial using Hilbert-Schmidt I.N.

So what am I missing here?


2 Answers 2


For square matrices, $$\Vert A \Vert_1 = \max_U \vert \text{Tr}(UA)\vert$$ over all unitaries acting on the matrix space related.

$$\Vert A+B\Vert_1 = \max_U \vert \text{Tr}(U(A+B))\vert = \max_U \vert \text{Tr}(UA+UB)\vert = \max_U \vert \text{Tr}(UA) + \text{Tr}(UB)\vert$$

For the absolute value we have for any real number the inequality,

$$\vert \text{Tr}(UA) + \text{Tr}(UB)\vert \leq \vert \text{Tr}(UA)\vert + \vert \text{Tr}(UB)\vert$$

the (sup) maximum preserves this relation and you'll get the wanted result.

  • 1
    $\begingroup$ For a more general proof one strategy is to use the fact that the trace norm is related to singular values and then show that singular values of a sum of matrices is less than or equal to the sum of singular values for each matrix. These results can be found in Math SE. $\endgroup$
    – R.W
    Aug 27, 2022 at 7:29
  • $\begingroup$ Isn't the above exactly how you would prove said relation for singular values? $\endgroup$ Aug 27, 2022 at 18:18
  • $\begingroup$ To the best of my knowledge for more general p-norms this path would not be sufficient and one would need min max arguments; but I am not so sure. @NorbertSchuch $\endgroup$
    – R.W
    Aug 28, 2022 at 7:22
  • $\begingroup$ For the last inequality there, does the jump to $\le ||M||+||N||$ come from $U$ not necessarily maximising the individual norms of the matrices? $\endgroup$ Aug 29, 2022 at 10:51

Sleep deprivation has gotten to me

You use $$||M+N||^{2}=|Tr[(M+N)U]|^{2}=|Tr[MU]+Tr[NU]|^{2}$$ which is just the square of the modulus of the sum of two complex numbers and then use the regular triangle inequality.

$$|Tr[MU]+Tr[NU]|^{2}=Tr[MU]^2+Tr[NU]^2+Tr[MU]Tr[NU]^{*}+Tr[NU]Tr[MU]^{*}\le \\ ||M||^2+||N||^2+2||M||||N||=(||M||+||N||)^{2}$$

Unless sleep deprivation has also given me a wrong answer :/

  • $\begingroup$ Is there a maximization over projectors U which is missing, plus the fact that separating the two maximizations gives something larger? $\endgroup$ Aug 25, 2022 at 16:47
  • $\begingroup$ @NorbertSchuch I omitted the maximisation, you are correct. However, given the action of the unitary expands to the first inequality, it shouldn't give something larger, no? I mean it is just a complex number, the same bounds on triangle inequalities still apply, right? $\endgroup$ Aug 26, 2022 at 11:04
  • $\begingroup$ @NorbertSchuch Sorry I mean the first equality on the second line. $\endgroup$ Aug 26, 2022 at 11:15
  • $\begingroup$ @NorbertSchuch does my edit address the issue. I realize what you mean't, the unitary max not maximise the individual traces, but they themselves are still upper bound by the schatten-1 norms. $\endgroup$ Aug 26, 2022 at 15:13

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