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Diagonalization of Hermitian Matrices

The problem of diagonalization of Hermitian Matrices reduces trivially to the already solved problem of diagonalization of real symmetric matrices.

Consider the following equation:

\begin{displaymath}\left(\boldsymbol{A} + i \boldsymbol{B}\right) \cdot
\left(\b...
...ight) = \lambda
\left(\boldsymbol{u} + i \boldsymbol{v}\right)
\end{displaymath}

where A, B, u and v are all real and A = AT and B = -BT, so that C = A + i B is Hermitian. This can be done to any Hermitian matrix.

Separating real and imaginary parts the equation above reduces to:

\begin{displaymath}\left(
\begin{array}{cc}
\boldsymbol{A} & -\boldsymbol{B} \...
...ay}{c}
\boldsymbol{u} \\
\boldsymbol{v}
\end{array}\right)
\end{displaymath}

The matrix on the left is clearly symmetric because B = -BT.

Observe that if $\left(\begin{array}{c}\boldsymbol{u}\\ \boldsymbol{v}\end{array}\right)$is an eigenvector, then $\left(\begin{array}{c}-\boldsymbol{v}\\ \boldsymbol{u}\end{array}\right)$is also an eigenvector with the same $\lambda_i$. This means that the real matrix that corresponds to a Hermitian C has double the number of eigenvalues and they are all degenerate with multiplicity of 2, i.e., $\lambda_1$, $\lambda_1$, $\lambda_2$, $\lambda_2$, and so on, and the eigenvectors of C are u + i v and then $i\left(\boldsymbol{u} + i \boldsymbol{v}\right)$.



Zdzislaw Meglicki
2001-02-26