Question

Show that a unitary matrix is a normal matrix, that is, that it commutes with its transpose conjugate [see (9.2)]. Also show that orthogonal, symmetric, antisymmetric, Hermitian, and anti-Hermitian matrices are normal.

Short answer

A unitary matrix is a normal matrix, that is, it commutes with its transpose conjugate. Also, orthogonal, symmetric, antisymmetric, Hermitian, and anti-Hermitian matrices are normal.

Step-by-step solution

Given information.

Consider A as a normal matrix.

Normal matrix.

The matrices that are unitarily diagonalizable are known as normal matrices. A generic normal matrix has no constraints on its eigenvalues, whereas all Hermitian matrices are normal but have real eigenvalues.

Show that a unitary matrix is a normal matrix, and hence show that orthogonal, symmetric, anti-symmetric, Hermitian and anti-Hermitian matrices are normal. 

When a normal matrix A operates with its Hermitian conjugate $A^{†}$, then $A{A}^{†}=A^{†}A.$

Now, for a unitary matrix, $A^{†}=A^{-1}$, which means $A{A}^{†}=A\left(A^{-1}\right)=I=\left(A^{-1}\right)A=A^{†}A.$

Similarly, for an orthogonal matrix, $A^T=A^{-1}$, for which A is real.

$A{A}^{†}=A{A}^T=A{A}^{-1}=I=A^{-1}A=A^{T}A=A^{†}A$

For a symmetric matrix, $A^T=A$, for which A is real.

$A{A}^{†}=A{A}^T=AA=A^{T}A=A^{†}A$.

For an antisymmetric matrix, $A^T=-A$, for which A is real.

_{$A{A}^{†}=A{A}^T=A(-A)=(-A)A=A^{T}A=A^{†}A$}

For a Hermitian matrix, $A^{†}=A$, for which A is real.

_{$A{A}^{†}=A\left(A\right)=A^{†}A$}.

Finally, for an anti-Hermitian matrix, $A^{†}=-A$for which A is real.

_{$A{A}^{†}=A(-A)=(-A)A=A^{†}A$}.

Hence, proved.