Given $RHR^{-1}=D$ where $H$ is Hermitian and $D$ is Diagonal. Show that $R$ is Unitary.

where Unitary Matrix: $U^\dagger U=UU^\dagger =I$ and Hermitian Matrix: $H=H^\dagger$

My try to this question

$RHR^{-1}=D \iff RH=DR$ and then applying dagger on both sides I get $HR^\dagger=R^\dagger D^\dagger$

I know $D$ consists of eigenvalues of $H$ and since $H$ is Hermitian implies eigen values are real.

Thus, I get $HR^\dagger=R^\dagger D$.

And the above equation implies $R^\dagger R H=H R^\dagger R $

  • $\begingroup$ In particular, note that you should provide context with your question. What have you tried? $\endgroup$ – Shaun Mar 19 '18 at 1:44
  • $\begingroup$ Perhaps define unitary. What do you want to show for R? $\endgroup$ – Michael Mar 19 '18 at 1:48
  • $\begingroup$ I have added what I tried for this question @Shaun $\endgroup$ – Ramandeep Singh Arora Mar 19 '18 at 2:00
  • $\begingroup$ Good! My downvote is now an upvote and I've retracted my vote to close. $\endgroup$ – Shaun Mar 19 '18 at 2:02
  • $\begingroup$ I have added the definitions of Hermitian and Unitary to the context. @Michael $\endgroup$ – Ramandeep Singh Arora Mar 19 '18 at 2:08

Given $RHR^{-1}=D$ where $H$ is Hermitian and $D$ is Diagonal. Show that $R$ is Unitary.

Here is a counter example which shows that $R$ is not unitary but $RHR^{-1}=D$.

$$ RHR^{-1} = \begin{pmatrix} \frac{1}{2} & -\frac{i}{2} \\ \frac{1}{2} & \frac{i}{2} \end{pmatrix} \begin{pmatrix} 1 & i \\ -i & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ i & -i \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 2 \end{pmatrix} = D $$ Note that $R$ is not unitary but $\sqrt{2}R$ is unitary.

  • $\begingroup$ Is there a condition I can put on R (or any other matrix) such that it becomes unitary. $\endgroup$ – Ramandeep Singh Arora Mar 19 '18 at 3:30
  • $\begingroup$ @RamandeepSinghArora Unfortunately, I have no idea $\endgroup$ – ChoF Mar 19 '18 at 3:36
  • $\begingroup$ thanks for the clarification above. $\endgroup$ – Ramandeep Singh Arora Mar 19 '18 at 4:36
  • $\begingroup$ See below for a condition. $\endgroup$ – Pietro Paparella Mar 19 '18 at 16:20

There is a fix to your statement.

Suppose that $A = SDS^{-1}$. If $\hat{D}$ is an invertible diagonal matrix and $\hat{S} := S\hat{D}$, then $A = \hat{S} D \hat{S}^{-1}$ since $\hat{D} D \hat{D}^{-1} = D$.

There is a partial converse that holds with an additional assumption; suppose that $A = SDS^{-1} = \hat{S} D \hat{S}^{-1}$, and suppose that $A$ has distinct eigenvalues. Let $D = \text{diag}(\lambda_1,\dots, \lambda_n)$. Let $s_i$ and $\hat{s}_i$ denote the $i^\text{th}$ column of $S$ and $\hat{S}$, respectively. Then, since $As_i = \lambda_i s_i$, $A \hat{s}_i = \lambda_i \hat{s}_i$, and each eigenspace is one-dimensional (the eigenvalues are distinct), it follows that $s_i = \alpha_i \hat{s}_i$, in which $\alpha_i \ne 0$. If $\hat{D} := \text{diag}(\alpha_1,\dots,\alpha_n)$, then $\hat{S} = S \hat{D}$ (right-multiplication by a diagonal matrix scales columns).

If $H$ is a Hermitian matrix, then the spectral theorem says that there is a unitary matrix $U$ such that $H=UDU^*$ (it is preferable to use an asterisk to denote the conjugate-transpose as the dagger symbol is typically used to denote the pseudo-inverse). If $H$ has distinct eigenvalues and $H=RDR^{-1}$, then there is an invertible diagonal matrix $\hat{D}$ such that $U=R\hat{D}$.

The situation does not generalize if the diagonalizable matrix has repeated eigenvalues. For instance, if $I$ denotes the $n$-by-$n$ identity matrix, then $I = S I S^{-1} = T I T^{-1}$, in which $S$ and $T$ are arbitrary invertible matrices.

  • $\begingroup$ Could you tell me what your additional condition of $R$ being unitary is? $\endgroup$ – ChoF Mar 19 '18 at 22:11

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