Proving the triangle inequality for the trace norm

Question

I have been trying to show that $$\|M+N\|_1 \le \|M\|_1 + \|N\|_1$$ However, I seem to be missing some fundamental property of either how the trace or the square root acts on these sums of matrices, or how the Hilbert-Schmidt inner product can be used. I can expand it easily, getting \begin{align*} \|M+N\|_1 &= {\rm tr}|M+N|\\ &={\rm tr}\Big[\sqrt{(M+N)(M+N)^\dagger}\Big]\\ &={\rm tr}\Big[\sqrt{MM^{\dagger}+MN^{\dagger}+NM^{\dagger}+NN^{\dagger}}\Big] \end{align*}

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 inner product. So what am I missing here?

Answer 1

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.

Answer 2

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 :/