Suppose A is any square invertible complex matrix. Then $$ C = \left[ \begin{array}{c|c} 0 & A \\ \hline A^{\dagger} & 0 \end{array} \right] $$ is a Hermitian matrix.

My Question: Is there a way to do something similar to get a unitary matrix from A? If necessary, suppose $detA = 1$.

Please let me know if it is not clear what I mean. The motivation comes from a quantum information theory question.

  • $\begingroup$ Not sure what you mean by "get a unitary matrix from A" $\endgroup$ – Steven Sagona Sep 7 '18 at 1:08
  • $\begingroup$ Yes. See this question, which is closely related. $\endgroup$ – John Polcari Sep 7 '18 at 1:13
  • $\begingroup$ @StevenSagona I mean something perhaps like how I "got the Hermitian matrix C from A". Essentially, I want to multiply a vector by the matrix A, but in quantum information theory, you only have unitary matrices to act on vectors. $\endgroup$ – wanderingmathematician Sep 7 '18 at 14:19

Let $H$ be any Hermitian matrix and $Id$ the identity matrix both of order $k$.

Thus, $Id\pm iH$ are invertible matrices. Define $U=(Id+iH)(Id-iH)^{-1}$.

Let $H=VDV^*$ be a spectral decomposition of $H$, i.e., $D$ is a real diagonal matrix and $V$ a unitary matrix.

Hence, $Id+iH=V(Id+iD)V^*$ and $(Id-iH)^{-1}=V(Id-iD)^{-1}V^*$. Thus, $U=V(Id+iD)(Id-iD)^{-1}V^*$.

Now you can easily check that $U$ is unitary, since its eigenvalues are $(1+i d_{ss})(1-i d_{ss})^{-1}$, $1\leq s\leq k$, (which have norm 1).

This idea comes from the Cayley parametrization (Check page 74 here).

Now you can construct any Hermitian matrix $H$ from $A$ and use the formula $U=(Id+iH)(Id-iH)^{-1}$ to define a unitary matrix $U$.


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