The set of positive eigenvalues $\Lambda$ may be separated from the negative using only Cholesky factoring. The following shows how. Any constructive comments/answers are appreciated.

Starting with a symmetric and real matrix $A$ and applying $LDU$ factoring gives the Cholesky factoring when choosing appropriate diagonal matrtix $D$. If the matrix $A$ is positive definite then $D$ has real values. If $A$ is indefinite (positive and negative eigenvalues) then $D$ may still be chosen to have pure real or pure imaginary values to have: $$A=\pmatrix{X & iY}\pmatrix{X & iY}^T$$ where i denotes $\sqrt{-1}$. A more thourough explanation to this point may be seen in my previous question Characterizing a real symmetric matrix $A$ as $A = XX^T - YY^T$.

What remains in order for this to separate the negative eigenvalues from the positive is to have $X$ and $Y$ orthogonal, and what follows shows the procedure. (Note that I do not commute scalars as I treat them as $1 \times 1$ matrices, e.g. c a scalar gives $\mathbf{x}c \ne c\mathbf{x}$ as the dimensions are not conformable).

Let $\mathbf{x}$ be a column of $X$ and $\mathbf{y}$ be a column of $Y$. The desire is to find a Givens rotation matrix (in order to preserve symmetry) $U$ such that the following orthogonalizes the $\mathbf{x}$ and $\mathbf{y}$ to have $\mathbf{x}^T\mathbf{y}=0$: $$\pmatrix{\mathbf{x} & \mathbf{y}i}U =\pmatrix{\mathbf{x} & \mathbf{y}i} \pmatrix{c & -is \\ is & c}$$ $$= \pmatrix{\mathbf{x}c - \mathbf{y}s & -\mathbf{x}si + \mathbf{y}ci}$$ Notice the preservation of the complex vs. real, $\mathbf{x}$ remains purely real and $\mathbf{y}i$ remains purely imaginary. The new inner product$\langle \mathbf{x},\mathbf{y}i\rangle$ then becomes: $$ (-is\mathbf{x}^T + ic\mathbf{y}^T)(\mathbf{x}c - \mathbf{y}s)$$ $$ = -csi(\mathbf{x}^T\mathbf{x}) +c^2i\mathbf{y}^T\mathbf{x} + s^2i\mathbf{x}^T\mathbf{y} - csi\mathbf{y}^T\mathbf{y}$$ $$ = -csi(\mathbf{x}^T\mathbf{x} + \mathbf{y}^T\mathbf{y}) + (c^2+s^2)i\mathbf{x}^T\mathbf{y}$$

Where the last line uses $\mathbf{x}^T\mathbf{y} = \mathbf{y}^T\mathbf{x}$ since the vectors are real. Remember that here $c^2-s^2=1$ not $c^2+s^2=1$.

If the dot product zero is desired, then this somewhat simply boils down to a final equation of the form $$\frac{cs}{c^2+s^2}=\frac{\mathbf{x}^T\mathbf{y}}{\mathbf{x}^T\mathbf{x} +\mathbf{y}^T\mathbf{y}}$$ Consider $$\lim_{c\rightarrow +\infty}\frac{cs}{c^2+s^2}=\frac{1}{2}$$ and $$\lim_{c\rightarrow -\infty}\frac{cs}{c^2+s^2}=\frac{-1}{2}$$ when the constraint of $c^2-s^2=1$ is held, it looks to be possible here to orthogonalize using the complex form of Givens rotation. The dot product divided by the sum of magnitudes will always be less than $\frac{1}{2}$. This then shows the possibility of orthogonalizing the representation of the symmetric matrix $A$.

If have $A=XX^T - YY^T$ with $X^TY=\mathbf{0}$ then the spectrum is separated into $X$ and $Y$ as such: $$X^TA = X^T(XX^T - YY^T) = X^TXX^T - \mathbf{0} = \mathbf\Lambda_x X^T$$ and $$Y^TA = Y^T(XX^T - YY^T) = \mathbf{0} - Y^TYY^T = -\mathbf\Lambda_y Y^T$$


1 Answer 1


The idea of being able to split the positive eigenvalues from the negative with only the computational cost of matrix factoring and orthogonalizing is an exciting one. However, the procedure shown here can not orthogonalize an entire set of vectors from another set and the reason is that a Givens rotation changes BOTH vectors. It is interesting that it can even be done with two vectors using the fact that one is purely real and the other is purely imaginary. But because both vectors are changed and inter-mixed, it is not possible to keep even just one orthogonal from a set in this manner.

Dealing with the full matrix version of your equations: $$\pmatrix{X & iY}\pmatrix{C & iS \\ -iS & C}=\pmatrix{XC+YS & iXS + iYC}$$ with the unitary matrix : $$\pmatrix{C & iS \\ -iS & C}\pmatrix{C^T & -iS^T \\ iS^T & C^T}=\pmatrix{I & \mathbf{0} \\ \mathbf{0} & I}$$

It then becomes the full matrix equivalent of the equations in the question, and we saw how difficult that was to solve even when commutations could be used :)

Specifically, with $C$ and $S$ from unitary matrix above, to orthogonalize $X$ and $Y$ (or $iY$ same difference since it is just a scaling of $Y$) then this equation must be solved: $$C^TX^TXS+C^TX^TYC+S^TY^TXS+S^TY^TYC = \mathbf{0}$$


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