Suppose I have an arbitrary orthogonal projector $\Pi$ and two density operators $\rho, \sigma$. Is it true that:

$$ ||\Pi (\sigma - \rho) \Pi||_1 \le || \sigma - \rho ||_1 $$

where $||\cdot||_1$ denotes the trace norm?


2 Answers 2


Yes, you can use Cauchy interlacing theorem to prove that.

Let $M = \sigma - \rho$ and dimension of the space is $n$.

In an appropriate basis $\Pi = I_k \oplus 0_{n-k}$. Assume $k=n-1$ (in general situation we can use an induction).

Then $\Pi M \Pi$ is the principle submatrix of $M$ of size $n-1 \times n-1$ (plus a row and a column of $0$s, but we discard that). So that, its eigenvalues $\mu_i$ are interlaced by eigenvalues $\lambda_i$ of $M$, that is $$ \lambda_1 \le \mu_1 \le \lambda_2 \le \mu_2 \le \lambda_3 \le \dots \le \mu_{n-1} \le \lambda_n. $$

Now, if $\mu_i \ge 0$ then $|\mu_i| \le |\lambda_{i+1}|$. Otherwise, $|\mu_i| \le |\lambda_{i}|$. Hence $$ \sum_{i=1}^{n-1}|\mu_i| \le \sum_{i=1}^n|\lambda_i| $$

  • $\begingroup$ Cool proof, I didn't know about this result. $\endgroup$
    – Rammus
    Commented Feb 21, 2022 at 13:02
  • $\begingroup$ Nice, thanks! So this also holds for any σ and rho even if they have different traces, correct? $\endgroup$
    – NYG
    Commented Feb 21, 2022 at 13:44
  • $\begingroup$ Yes, they can be any Hermitian matrices. $\endgroup$
    – Danylo Y
    Commented Feb 21, 2022 at 13:47

Here is a slightly more general alternative to Danylo's answer. The trace norm satisfies the inequality $$ \| X Y Z \|_1 \leq \|X\| \|Y\|_1 \|Z\| $$ for all choices of operators $X$, $Y$, and $Z$ that can be composed as $XYZ$ (so nothing needs to be Hermitian). Note that the norms $\|X\|$ and $\|Z\|$ on the right-hand side refer to the spectral (or operator) norm.

Hence, in the case at hand, we have $$ \|\Pi(\sigma - \rho) \Pi\|_1 \leq \|\Pi\| \|\sigma - \rho\|_1 \|\Pi\| \leq \|\sigma - \rho\|_1 $$ because the spectral norm of any orthogonal projector is equal to 1 (unless it is zero).

The inequality $$ \| X Y Z \|_1 \leq \|X\| \|Y\|_1 \|Z\| $$ is actually true for any unitarily invariant norm in place of the trace norm. One way to prove this is to first use the fact that every contraction (i.e., operator having spectral norm at most 1) can be expressed as a convex combination of unitary operators, and then use the convexity and unitary invariance of the norm.

  • 1
    $\begingroup$ Cool! And in the case of $\Pi$ it's just an average of 2 unitaries. $\endgroup$
    – Danylo Y
    Commented Feb 21, 2022 at 21:01
  • 1
    $\begingroup$ Just a note: You can prove this for any Schatten norm using the generalized Hölder inequality which states in this case that $\| XYZ \|_t \leq \| X \|_{p} \| Y \|_{q} \| Z \|_r$ where $1/t = 1/p + 1/q + 1/r$ and $t,p,q,r \in (0,\infty]$. You get the desired inequality for $p=r=\infty$. $\endgroup$ Commented Feb 22, 2022 at 9:19

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