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 $10^4<n$ and $m < 20$.

When increasing $g$ towards $g_0$ ($0<g\leq g_0$) 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 $g_0<g$ 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.

  • $\begingroup$ Look at the corner case: $W=G=I_2$ and $K = a\,I_2$. The determinant is always positive or null. $\endgroup$ – Damien Jan 3 at 10:24
  • $\begingroup$ 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 $g\to\infty$. $\endgroup$ – obareey Jan 3 at 14:14
  • $\begingroup$ @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? $\endgroup$ – DavidH Jan 3 at 14:54
  • $\begingroup$ 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. $\endgroup$ – obareey Jan 3 at 14:58
  • $\begingroup$ 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 $t\in\mathbb{R}$. $\endgroup$ – obareey Jan 3 at 15:10

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