Short easy question, I just want someone to double check what I did.

We are given that $T$ is an invertible, normal matrix. We are asked to show that $T^{-1}$ is also normal, and find it's unitary diagonlization.

What I did:

$T$ is normal if and only if there is a unitary matrix $U$ and diagonal matrix $D$ such that $T=UDU^{-1}$

if $T=UDU^{-1}$ then $T^{-1}=(UDU^{-1})^{-1} = U^{-1}D^{-1}U$

$U$ and $U^{-1}$ are still the same (unitary) and $D^{-1}$ is still diagonal. So $T^{-1}$ is unitary diagonlizable, so it is normal, and as stated above, the unitary decomposition is $T^{-1}=U^{-1}D^{-1}U$

  • $\begingroup$ I think you are right on the solution. $\endgroup$ – randomname Jan 5 '14 at 20:17
  • 1
    $\begingroup$ don't forget to mention that $D$ is invertible (ie the diagonal coefficient are non-zero). $\endgroup$ – imj Jan 5 '14 at 20:25

Your answer is almost correct (though uses a nontrivial theorem about normal matrices). Two missing/incorrect points:

  • If $T=UDU^{-1}$ then $T^{-1}=(UDU^{-1})^{-1}=UD^{-1}U^{-1}$ (taking inverse changes order of multiplication).
  • As imj noted, we should also argue for the invertibility of $D$ before (or after) we write it down, but it is easy now: as the above equation implies, the matrix $U^{-1}T^{-1}U$ will be an inverse for $D=U^{-1}TU$.

You can go also directly from the definition.


Invert both sides.


Use that $(T^{*})^{-1}=(T^{-1})^*$, which also folows directly from the definition.


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