In Nielsen & Chuang's book equation 2.172 says $$A=\sum_{i}|\widetilde{\psi_i}\rangle \langle \widetilde{\psi_i}| = \sum_j |\widetilde{\phi_j}\rangle \langle \widetilde{\phi_j}|.$$

Then it makes the assumption that A can be written as $$A=\sum_{k} \lambda_{k}|k\rangle\langle k|$$ where the kets are orthonormal vectors and lambdas are strictly positive. On what basis are these two assumptions made? Namely the positivity and orthonormality.

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    $\begingroup$ That is the spectral theorem for positive semidefinite matrices. It says that for a postive-semidefinite matrix we can diagonalize it and it will have nonnegative eigenvalues. I expect Nielsen and Chuang should have something dedicated to the spectral theorem as it is quite an important tool. $\endgroup$
    – Rammus
    Mar 10, 2021 at 15:58

1 Answer 1


A linear operator $A$ is said to be normal if $AA^\dagger = A^\dagger A$. The significance of the class of normal operators is that these are precisely the operators that have an orthonormal basis of eigenvectors. Consequently, if $A$ is normal then

$$ A = \sum_k \lambda_k |\psi_k\rangle\langle\psi_k|\tag1 $$

for some complex numbers $\lambda_k$ and orthonormal basis $|\psi_k\rangle$. It is easy to see that $\lambda_k$ are $A$'s eigenvalues and $|\psi_k\rangle$ are $A$'s eigenvectors associated with the respective eigenvalues $\lambda_k$. Nielsen & Chuang introduces normal operators on page 70 and proves the above theorem in box 2.2 on page 72.

All Hermitian and unitary operators are normal. Moreover, it is easy to check that the eigenvalues of a Hermitian operator are real. If in addition the operator is positive (i.e. positive semi-definite) then its eigenvalues are non-negative real numbers.

Now, the operator $A$ in equation $(2.172)$ is easily seen to be positive (and therefore Hermitian by definition). Thus, $A$ is normal and can be written as in $(1)$. Moreover, since $A$ is positive then $\lambda_k$ are non-negative real numbers.

  • $\begingroup$ Thanks! Why does it say that the lambdas are strictly positive though? It is apparent to me now why they are non-negative but not why they are strictly positivie. $\endgroup$ Mar 11, 2021 at 7:40
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    $\begingroup$ We can drop the zero terms. The remaining ones have strictly positive $\lambda_k$. The sums may then be shorter than space dimension $n$, but that's perfectly acceptable here. $\endgroup$ Mar 11, 2021 at 7:42
  • $\begingroup$ ohh yeah that makes sense :) Thank you $\endgroup$ Mar 11, 2021 at 7:45

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