Properties of Complex Symmetric Matrices

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Properties of Complex Symmetric Matrices

Complex symmetry is a purely algebraic property, and it has no effect on the spectrum of the matrix. Indeed, for any given set of numbers,

$\begin{displaymath} \lambda_1,\lambda_2,\ldots,\lambda_n \in {\cal C}, \end{displaymath}$

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there exists a complex symmetric $n \times n$ matrix

whose eigenvalues are just the prescribed numbers (7.89); see, e.g., [233, Theorem 4.4.9].

A complex symmetric matrix may not even be diagonalizable. For example, consider the complex symmetric matrix

$\begin{displaymath} A = \left[ \begin{array}{cc} 2 {\tt i} & 1\\ 1 & 0 \end{array} \right],\quad \mbox{where}\quad {\tt i} = \sqrt{-1}. \end{displaymath}$

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The only eigenvalue of this matrix is $\lambda={\tt i}$ , with algebraic multiplicity

but geometric multiplicity

. In fact, the Jordan normal form of

is as follows:

$\begin{displaymath} Z^{-1} A Z = \left[ \begin{array}{cc} {\tt i} & 1\\ 0 & {... ... \begin{array}{cc} {\tt i} & 1\\ 1 & 0 \end{array} \right]. \end{displaymath}$

Thus,

is not diagonalizable.

If a complex symmetric matrix is diagonalizable, then it has an eigendecomposition that reflects the complex symmetry; see, e.g., [233, Theorem 4.4.13]. More precisely, a complex symmetric matrix is diagonalizable if and only if its eigenvector matrix, , can be chosen such that

$\begin{displaymath} Z^T A Z = \Lambda = {\mathop{\rm diag}\nolimits}\left(\lambd... ...da_2,\ldots,\lambda_n\right)\quad \mbox{and}\quad Z^T Z = I_n. \end{displaymath}$

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A matrix

with

columns that satisfies

is called complex orthogonal. The complex orthogonality of

in (7.91) reflects the complex symmetry of

We remark that the eigendecomposition (7.91) is the suitable adaptation of the corresponding decomposition for Hermitian matrices. Recall that for any matrix $A=A^{\ast}$ , the eigenvector matrix can always be chosen to be unitary:

$\begin{displaymath} Z^{\ast} A Z = \Lambda = {\mathop{\rm diag}\nolimits}\left(\... ...\ldots,\lambda_n\right)\quad \mbox{and}\quad Z^{\ast} Z = I_n. \end{displaymath}$

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The unitariness of

in (7.92) reflects the fact that

is Hermitian.

The reason why an eigendecomposition (7.91) does not always exist is that there are complex vectors with

$\begin{displaymath} z^T z = 0,\quad \mbox{but}\quad z\not=0. \end{displaymath}$

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Indeed, suppose

has an eigenvalue with a one-dimensional eigenspace and the vector

spanning that space satisfies (7.93). Then one of the columns of any eigenvector matrix

would be of the form $z_k = \gamma z$ , where $\gamma\not=0$ is a scalar. Then, by (7.93),

, while the complex orthogonality condition,

, in (7.91) would imply

. Note that for example (7.90), the vector $z=\left[\begin{array}{cc} {\tt i} & 1 \end{array}\right]^T$ spans the one-dimensional eigenspace associated with $\lambda={\tt i}$ and it satisfies (7.93).

Next: Properties of the Algorithm Up: Lanczos Method for Complex Previous: Lanczos Method for Complex Contents Index

Susan Blackford 2000-11-20