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 $n$ numbers,

$$\lambda_1,\lambda_2,\ldots,\lambda_n \in {\cal C},$$ (195)

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

$$A = \left[ \begin{array}{cc} 2 {\tt i} & 1\\ 1 & 0 \end{array} \right],\quad \mbox{where}\quad {\tt i} = \sqrt{-1}.$$ (196)

The only eigenvalue of this matrix is $\lambda={\tt i}$, with algebraic multiplicity $2$ but geometric multiplicity $1$. In fact, the Jordan normal form of $A$ is as follows:

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

Thus, $A$ is not diagonalizable.

If a complex symmetric matrix $A$ 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 $A$ is diagonalizable if and only if its eigenvector matrix, $Z$, can be chosen such that

$$Z^T A Z = \Lambda = {\mathop{\rm diag}\nolimits}\left(\lambd... ...da_2,\ldots,\lambda_n\right)\quad \mbox{and}\quad Z^T Z = I_n.$$ (197)

A matrix $Z$ with $n$ columns that satisfies $Z^T Z = I_n$ is called complex orthogonal. The complex orthogonality of $Z$ in (7.91) reflects the complex symmetry of $A$.

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 $Z$ can always be chosen to be unitary:

$$Z^{\ast} A Z = \Lambda = {\mathop{\rm diag}\nolimits}\left(\... ...\ldots,\lambda_n\right)\quad \mbox{and}\quad Z^{\ast} Z = I_n.$$ (198)

The unitariness of $Z$ in (7.92) reflects the fact that $A$ is Hermitian.

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

$$z^T z = 0,\quad \mbox{but}\quad z\not=0.$$ (199)

Indeed, suppose $A$ has an eigenvalue with a one-dimensional eigenspace and the vector $z$ spanning that space satisfies (7.93). Then one of the columns of any eigenvector matrix $Z$ of $A$ would be of the form $z_k = \gamma z$, where $\gamma\not=0$ is a scalar. Then, by (7.93), $z_k^T z_k = 0$, while the complex orthogonality condition, $Z^T Z = I_n$, in (7.91) would imply $z_k^T z_k =1$. 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).