# Uniqueness of a (generalized) (orthonormal) eigenbasis when it exists

1. In Jordan decomposition of a complex square matrix $M = P J P^{-1}$, the Jordan canonical form $J$ is unique up to permutation of the diagonal Jordan blocks $J_i$'s along the diagonal. $P$ consists of a generalized eigenbasis as its columns.

2. If a complex or real square matrix can be similar to a diagonal matrix $M = P D P^{-1}$, the diagonal matrix $D$ is unique up to permutation of the diagonal entries along the diagonal line. $P$ consists of an eigenbasis as its columns.

3. If a complex or real square matrix can be unitarily/orthogonally similar to a diagonal matrix $M = P D P^{H}$, the diagonal matrix $D$ is unique up to permutation of the diagonal entries along the diagonal line. $P$ consists of an unitarynormal/orthonormal eigenbasis as its columns.

I wonder what we can say about the uniqueness of $P$ in each case? Can it be unique up to some matrix transformation?

Thanks!

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Note that, in the three cases, if $M=I$ then $D=I$ and $P$ can be any invertible matrix. So you cannot expect much uniqueness in general.
Thanks! in that particular example, $M=I$ has single eigen value $1$ and any nonzero vector is its eigenvector, and accordingly $P$ can be any invertible matrix. Maybe what I asked should be how much uniqueness $P$ can have in terms of the eigenvalues and eigenvectors of $M$? –  Tim Nov 25 '12 at 22:22
Well, you can always permute the Jordan blocks, so if you multiply $P$ by a proper permutation (in any of the three examples) you still get another $P$. And even if you prescribe a canonical order for $D$ (say eigenvalues in non-increasing order and block sizes in non-increasing order) you can still have repetitions so permutations still play a role. –  Martin Argerami Nov 25 '12 at 22:55
It is trivial to permute the columns of $P$ in the same way of permuting $D$ or $J$. If the $D$ or $J$ is fixed, how much flexibility/uniqueness can $P$ enjoys in the three cases? –  Tim Nov 25 '12 at 22:59
It still depends a lot. Take $D=I$, and still $P$ can be any invertible matrix. Take $D$ diagonal with all diagonal entries different, and then if you fix order, $P$ is unique. Take $J$ to be a full Jordan block, and you can still multiply $P$ by any invertible upper-triangular Toeplitz matrix. –  Martin Argerami Nov 25 '12 at 23:12