- #1

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I have thought about this, but am unsure if my answer is correct. My claim is that ##B## is the only diagonal matrix to which ##A## is similar, up to a rearrangement (or permutation) of the eigenvectors of ##A## in an ordered eigenbasis for ##ℝ^{n}##. Thus, if you rearrange the vectors in your eigenbasis, you will obtain a different diagonal matrix for ##[L_{A}]_{β}## but this will be merely a rearrangement of the diagonal entries of B.

Is this true? If so, how might I go about proving this conjecture of mine?

BiP