In many places, I see a form of representing a matrix with:

$A = \sum_\lambda \lambda \left | \lambda \right> \left < \lambda \right |$

Where $\lambda$ is an eigenvalue, and $\left | \lambda \right >$ is an eigenvector.

One that I find a pretty good example is here

enter image description here

$W$ is defined to be a unitary:

$W \left | \lambda \right > = e^{i\theta_\lambda} \left | \lambda \right >$

$V$ is the same as $W$, but has eigenvalues $e^{ih(\theta_\lambda)}$ instead of just $e^{i\theta_\lambda}$.

However, $V$ is constructed with $V = \sum_\lambda e^{ih(\theta_\lambda)} \left | \lambda \right > \left < \lambda \right |$ not $V\left | \lambda \right > = e^{ih(\theta_\lambda)} \left | \lambda \right >$.

Why is this the case, and what is the reasoning we can represent matrices this way? I also see this with singular value transforms, but I assume that the reasoning is similar.

  • $\begingroup$ Assuming $|\lambda\rangle$ are orthonormal, $A = \sum_\lambda \lambda|\lambda\rangle\langle\lambda|$ is simply saying $A$ can be diagonalized under the basis $|\lambda\rangle$. $\endgroup$
    – Guangliang
    Oct 8, 2022 at 2:15

1 Answer 1


This notation is representing the fact that a given $A$ is diagonal. In particular, the spectral theorem guarantees that not only we can diagonalize $A$, but we can use the spectrum and eigenprojectors to do so, when $A=A^*$ is self-adjoint (hermitian). This is the general answer for the title and the first part of the question.

For the second part you gave an example related to the operator $W$, and it was not clear to me what was the problem. The point of the algorithm is to input the phases $e^{ih(\theta_\lambda)}$ because note that for the states $\vert \lambda\rangle$ you may see that (and here I am guessing that this was the confusion, which might not be it) if we would have used the states $V\vert \lambda \rangle = e^{ih(\theta_\lambda)} \vert \lambda \rangle $ to construct the operator we would have $\sum_\lambda V\vert \lambda \rangle \langle \lambda \vert V^\dagger = \sum_\lambda e^{ih(\theta_\lambda)}\vert \lambda \rangle \langle \lambda \vert e^{-ih(\theta_\lambda)} = \sum_\lambda \vert \lambda \rangle \langle \lambda \vert = I$. This comes from the fact that one of the points of representing the operators instead of the vectors avoids the fact that every state is equivalent to the entire line defining the equivalence class $\vert \lambda \rangle \sim e^{i \alpha }\vert \lambda \rangle $ for any complex $\alpha$. Global phases are not physical.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.