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UIn page 6 of the following paper: https://arxiv.org/pdf/0904.2557.pdf, in the proof of theorem 3:"Suppose equation (30) holds. We can diagonalize $C_{ab}$. This involves choosing new basis $\{F_a\}$ for $\mathcal{E}$...".

Does this mean the matrix representation of the bilinear form $b(E)=\langle\psi|E^{\dagger}E|\psi\rangle$ with respect to the basis that diagonalizes it? (Here, $\psi$ is a fixed codeword.)

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    $\begingroup$ It's unclear to me what your first question means. Could you try rephrasing it? $\endgroup$
    – Peter-Jan
    Aug 24, 2023 at 9:39
  • $\begingroup$ How can we formulate diagonalizability of $[C_{ab}]$ in terms of new basis? $\endgroup$
    – Star21
    Aug 24, 2023 at 10:59

1 Answer 1

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How can we formulate diagonalizability of [𝐶𝑎𝑏] in terms of new basis?

$[C_{ab}] = \langle \psi | F_a^{\dagger} F_b | \psi \rangle$

How will equations (26), (27), (28) be written?

$ \langle \overline{0} | F_a^{\dagger} F_b | \overline{1}\rangle = 0$. Similairly in Equations 27 and 28 the error terms are replaced by elements of the new basis $\{ F_a \}$

Why equation (31) holds?

If we assume the outcome of $\langle \psi | E^\dagger E |\psi \rangle$ only depends on the error and not on the state, (i.e the outcome C is only a function of $E$ not $\psi$), this means that $\langle \psi | E^\dagger E |\psi \rangle = \langle \phi | E^\dagger E |\phi \rangle$.

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