# Decompose a complex symmetric matrix to retain positive definitness

I have a complex symmetric matrix $A$, (i.e. non-Hermitian and obeying $A=A^T$), which is positive definite, in the sense that: $$\Re({z^HAz}) > 0$$ for any $z$. I am able to verify this numerically by noting that the Hermitian part of this matrix $(A + A^H)/2$ has all-positive eigenvalues (or that any negative parts are small enough to be attributable to rounding error and hopefully negligible)

I know that for my problem only a few eigenvalues of this matrix contribute to the solution, so I have performed a spectral decomposition of my matrix $$A = U\cdot diag(\lambda) \cdot V$$

with $V$ and $U = V^{-1}$ representing the (non-orthogonal) eigenvectors and $\lambda$ the eigenvalues. This then gives me a representation of $A$ in terms of "partial matrices", $$A = \sum_i \lambda_i U_i\otimes V_i$$

However, these partial matrices are no-longer positive definite (i.e. some of them have quite significant negative eigenvalues of their Hermitian parts), hence they cannot yield meaningful decompositions of the solution to my problem.

• Is there some explanation as to why these partial matrices are not positive definite?
• Is there an alternative decomposition which would preserve this property?

edit

As this is a complex symmetric system, there is the additional relationship $V=U^{T}$

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It's not misleading as long as you change your notion of equivalence. When a matrix represents a linear transformation $V \to V$, the correct notion of equivalence is similarity: $M \simeq B^{-1} MB$ where $B$ is invertible. When a matrix represents a bilinear form $V \times V \to \mathbb{R}$, the correct notion of equivalence is congruence: $M \simeq B^TMB$ where $B$ is invertible. As long as you keep this distinction in mind, you're fine.
Since you are dealing with a $z^HAz$, $A$ should be viewed as a bilinear form rather than a linear transformation. Therefore, if you want to find an equivalent matrix to $A$ via matrix decomposition, what you should use is not a similarity transform, but a matrix congruence. Actually, even in case of $\mathbb{R}$, a similarity transform in general wouldn't help. You may think of it this way: positive definiteness is a basis-independent property. If $x^TAx>0$ for all $x$ in the standard basis, by a change of basis $x\mapsto Sy$ ($S$ is an invertible matrix), $x^TAx=y^T(S^TAS)y$ is also positive for all $y$. Therefore, what preserves positive definiteness is the congruence $A\mapsto S^TAS$ but not a similarity transform $A\mapsto S^{-1}AS$.
Surely, when $A$ is real symmetric or Hermitian, $A$ is orthogonally/unitarily similar to a diagonal matrix. So, in this case, we have both congruence and similarity in one shot. For a general $A$, however, we are not that lucky. I am not sure what is the best way to go. This varies from scenario to scenario. Yet, if the preservation of positive definiteness is of utmost importance, I think the best one can do is perhaps just a Schur decomposition $A=UTU^H$, where $U$ is a unitary matrix and $T$ is upper triangular. You may still read off the eigenvalues from the diagonal of $T$, but $T$ is not a diagonal matrix so that in general you can only write $A$ as a linear combination of $n(n+1)/2$ tensor products.
Thanks for that. I haven't quite absorbed your answer fully but it's definitely pointed me in the right direction. In the case of complex matrices, should I consider congruence to be $A -> S^T A S$ (which I have) or $A -> S^H A S$ (valid for Hermitian matrices)? – DaveP Dec 11 '12 at 22:14
$A\mapsto S^HAS$ is called (by some authors) a $\ast$-congruence ("star-congruence"). Again, which of congruence or $\ast$-congruence is more appropriate depends on your needs. Star-congruence preserves positive definiteness, but congruence can bring you some nicely structured matrix. See Tagaki factorization for example. – user1551 Dec 13 '12 at 11:14