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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}](img2493.png) |
(195) |
there exists a complex symmetric
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}](img2494.png) |
(196) |
The only eigenvalue of this matrix is
, with
algebraic multiplicity
but geometric multiplicity
.
In fact, the Jordan normal form of
is as follows:
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}](img2497.png) |
(197) |
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
, 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}](img2499.png) |
(198) |
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}](img2500.png) |
(199) |
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
of
would be
of the form
, where
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
spans the one-dimensional eigenspace associated with
and it satisfies (7.93).
Next: Properties of the Algorithm
Up: Lanczos Method for Complex
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Susan Blackford
2000-11-20