# Determination of transition from non-singular matrix to singular matrix

I have the following matrix as a biproduct of inverting a matrix sum by the Woodbury matrix identity:

$$\mathbf{A} = -(g\mathbf{G})^{-1} + \mathbf{W}^T \mathbf{K}^{-1} \mathbf{W}$$

where $$g$$ is an arbitrary but positive scalar, $$\mathbf{G}$$ is a $$m\times m$$ diagonal matrix with positive entities, $$\mathbf{W}$$ is an $$n \times m$$ array with each column having a single $$1$$ or $$-1$$ entity and otherwise zeros and $$\mathbf{K}$$ is a $$n \times n$$ symmetric, real and positive definite matrix, representing the stiffness matrix of an elastic solid. Further more $$m << n$$, typically $$10^4 and $$m < 20$$.

When increasing $$g$$ towards $$g_0$$ ($$0) the matrix $$\mathbf{A}$$ will at some point become singular and the system loses its stability. I am to determine $$g_0$$ which indentifies instability.

I am trying to set up a robust way to determine $$\mathbf{A}$$'s transition from non-singular to singular. Right now I am calculating the determinant of $$\mathbf{A}$$ with Matlab, and it seems that as $$\mathbf{A}$$ becomes 'less and less' positive definite, the determinant decreases and becomes $$0$$ when singular. For $$g_0 it seems that the determinant becomes negative, which I guess is a result of $$\mathbf{A}$$ then being negative definite?

Is it correct that as $$g \rightarrow \infty$$ $$\mathbf{A}$$ goes from being positive definite to singular to negative definite along with the determinant going from positive through $$0$$ to negative with $$det(\mathbf{A})=0$$ identifying the point when $$\mathbf{A}$$ becomes singular? Is there a precise mathematical argument for the determinant to behave in that way?

Thanks for any comments.

• Look at the corner case: $W=G=I_2$ and $K = a\,I_2$. The determinant is always positive or null. – Damien Jan 3 at 10:24
• I believe it should be $gG^{-1}$, and not $(gG)^{-1}=(1/g)G^{-1}$. Otherwise, for small values of $g$, $A$ is obviously negative definite and becomes positive definite as $g\to\infty$. – obareey Jan 3 at 14:14
• @obareey The equation is correct as it is, the problem then is my interpretation of the determinant of $\mathbf{A}$. In the case of $m = 2$ then $det(\mathbf{A})\geq 0$, which as far as I know truely is the case for negative definite matrices with even $m$. So my statement that $\mathbf{A}$ goes from positive definite to negative definite is probably not true? – DavidH Jan 3 at 14:54
• If $m$ is even, for both negative and positive (semi-)definite matrices $\operatorname{det}A \geq 0$ as determinant is the multiplication of the eigenvalues. So, you cannot infer positive or negative definiteness from the determinant. But I think your statement is true but in the reverse, i.e. it goes from negative definite to positive definite. – obareey Jan 3 at 14:58
• Also the transition from negative-to-positive definiteness (or vice versa) may not be smooth, or strict. There might be some indefinite matrices in between. For example $A=tI + \begin{bmatrix}0 & 0\\0&-1\end{bmatrix}$ for $t\in\mathbb{R}$. – obareey Jan 3 at 15:10