Suppose, we have a matrix $A$ which is diagonalizable i.e. there exist an invertible matrix $B$ such that $B^{-1}AB = D$ where $D$ is a diagonal matrix. Is $B$ unique ? I don't think so.

  • 2
    $\begingroup$ No, it is not. For instance $2B$ will do as well. Given $B$, try to characterize all the matrices that diagonalize $A$ using $B$ and all the matrices that commute with $A$. $\endgroup$
    – Julien
    Oct 26, 2013 at 17:54
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    $\begingroup$ Consider what happens if $A$ is already diagonal. Then $B$ can be any diagonal matrix, and $B^{-1}AB=D$ will be diagonal. So in this case $B$ is not unique $\endgroup$
    – coffeemath
    Oct 26, 2013 at 17:55
  • $\begingroup$ The only thing which is unique are the eigenvalues and their associated eigenspaces. How you order them or how you choose their bases stuffed in $D$ and $B$ does not matter :) $\endgroup$ Oct 26, 2013 at 20:09
  • $\begingroup$ For the equality $PIP^{-1}=I$ to hold, is $P$ unique? $\endgroup$
    – user1551
    Oct 26, 2013 at 20:28

1 Answer 1


Of course it's not unique -- at the very least you can multiply it with from the right by any permutation matrix and/or diagonal matrix with nonzero entries on the diagonal. If there are any multiple eigenvalues you get even greater freedom.

(The only exception is if $A$ is a $1\times 1$ matrix over $\mathbb F_2$. Then $B$ is unique for the trivial reason that in that case $I$ is the only possible invertible matrix of the right size).


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