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The trace norm of an operator M ∈ L(H, H′) is equal to the sum of itssingular values.

Question

The trace norm of an operator M ∈ L(H, H′) is equal to the sum of its singular values.

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Solution

The trace norm of an operator M in the space of bounded linear maps from H to H' is indeed equal to the sum of its singular values. Here's a step-by-step explanation:

  1. First, we need to understand what the singular values of an operator are. For any operator M, the singular values are the square roots of the eigenvalues of the positive semi-definite matrix MM*. Here, M* denotes the adjoint of M.

  2. The singular values are always non-negative and can be ordered in a decreasing sequence. This sequence is denoted as {σi}, where i ranges over the dimensions of the Hilbert spaces H and H'.

  3. The trace norm (also known as the nuclear norm) of an operator M, denoted as ||M||1, is defined as the sum of its singular values. Mathematically, this can be written as ||M||1 = Σi σi.

  4. Therefore, by definition, the trace norm of an operator M is equal to the sum of its singular values.

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