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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,

(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

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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

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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 , the eigenvector matrix
can always be chosen to be unitary:

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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

(199) 
Indeed, suppose has an eigenvalue with a onedimensional 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 onedimensional eigenspace associated with
and it satisfies (7.93).
Next: Properties of the Algorithm
Up: Lanczos Method for Complex
Previous: Lanczos Method for Complex
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Susan Blackford
20001120