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--Updated description--

I'm trying to investigate the stability of tensegrity structures, and this question is related to the second order test.

Suppose there is a vector space $W=\operatorname{span}\{\vec{w_1},\vec{w_2},...,\vec{w_s}\}$, and $\vec{w_1},\vec{w_2},...\vec{w_s}$ is a basis. Similarly, we have a vector space $V=\operatorname{span}\{\vec{v_1},\vec{v_2},...\vec{v_m}\}$. In literature, $W$ is the nullspace of the equilibrium matrix, whereas $V$ is the left-nullspace of the equilibrium matrix. $\{\mathrm{w}\}\in W$ is called a self-stress, and $\{\mathrm{v}\}\in V$ is the initial nodal velocities of the structure, also called a mechanism.

A linear operator $L$ produces a generalized Laplacian matrix $[\mathbf{L}]$ from a vector in $W$, with a fixed graph. A generalized Laplacian matrix is a symmetric one whose row sums and column sums are all $0$. The diagonal terms are the additive inverse of the sum of other entries on the corresponding row, respectively. (Other entries depend on the specific topology of the graph.)

$$\begin{array}{rccl} 4 & --- & --- & 3\\ | & \diagdown & \diagup & | \\ | & \diagup & \diagdown & | \\ 1 & --- & --- &2 \\ \end{array}$$

For example, the structure above has 4 nodes and 6 members. Sorry for the horrible diagram. I can't upload images. Member 1-2, 2-3, 3-4 and 1-4 are cables. In the middle of the rectangular, there are 2 strut members (that's not 4 struts): Member 1-3 and Member 2-4. Each member is assigned a number, with cables only getting positive numbers and struts negative ones. These numbers, put into a vector is the self-stress. The equilibrium matrix of this example is

$$[\mathbf{A}]=\begin{bmatrix} x_{1}-x_2 & 0 & 0 & x_1-x_4 & x_1-x_3 & 0\\ y_{1}-y_2 & 0 & 0 & y_1-y_4 & y_1-y_3 & 0\\ x_2-x_1 & x_2-x_3 & 0 & 0 & 0 & x_2-x_4\\ y_2-y_1 & y_2-y_3 & 0 & 0 & 0 & y_2-y_4\\ 0 & x_3-x_2 & x_3-x_4 & 0 & x_3-x_1 & 0\\ 0 & y_3-y_2 & y_3-y_4 & 0 & y_3-y_1 & 0\\ 0 & 0 & x_4-x_3 & x_4-x_1 & 0 & x_4-x_2\\ 0 & 0 & y_4-y_3 & y_4-y_1 & 0 & y_4-y_2\\ \end{bmatrix}$$

where $x$ and $y$ are coordinates of nodes.

The nullspace of $[\mathbf{A}]$ is $W$, whereas the left-nullspace is $V$.

Suppose we have ourselves a self-stress vector $\{\mathrm{w}\}$.

$$\{\mathrm{w}\}=\begin{Bmatrix} w_{12}\\ w_{23}\\ w_{34}\\ w_{14}\\ w_{13}\\ w_{24}\\ \end{Bmatrix}$$

Then the Laplacian matrix is constructed in the following way:

  • If there is a member connecting Node $i$ and Node $j$, then $L_{ij}$ equals the negative value of the member's self-stress.
  • If there is no member between two nodes, that entry is $0$.
  • The diagonal terms are the sums of the self-stresses of the members connected to the nodes.

For this example, its Laplacian matrix is

$$[\mathbf{L}]=\begin{bmatrix} w_{12}+w_{14}+w_{13} & -w_{12} & -w_{13} & -w_{14}\\ -w_{12} & w_{12}+w_{23}+w_{24} & -w_{23} & -w_{24}\\ -w_{13} & -w_{23} & w_{13}+w_{23}+w_{34} & -w_{34}\\ -w_{14} & -w_{24} & -w_{34} & w_{14}+w_{24}+w_{34}\\ \end{bmatrix}$$

And we can denote $L$ as a function maps $\{\mathrm{w}\}$ to a $[\mathbf{L}_w]$. As long as the structure is the same, the function $L$ is the same.

Now a sufficient condition for a tensegrity to be second-order stable is:

for every $\{\mathrm{v}\}$ in $V$, there is a $\{\mathrm{w}\}$ in $W$ so that $\{\mathrm{v}\}^\mathbf{T}([\mathbf{L}]\otimes[\mathbf{I}_d])\{\mathrm{v}\}\geq0$ (only when $\{\mathrm{v}\}$ represents a rigid body motion can $\{\mathrm{v}\}^\mathbf{T}([\mathbf{L}]\otimes[\mathbf{I}_d])\{\mathrm{v}\}=0$)

where $[\mathbf{I}_d]$ is a identity matrix of order $d$ which is the dimension of the structure ($d=3$ for 3D problems).

So my question is, given $W$, $V$ and $L$, and the dimension of $W$ is more than 1, how do we test the above condition? Are there any related studies?

--Original description--

Suppose we have a space of symmetric, real matrices, M, with some particular structure. And there is a vector space V (or it could be a subspace of V). Matrices in M may be positive-definite, may be not. But for every vector {v} in V, there is always a matrix [M] in M, so that the quadratic form is positive. Could somebody tell me what this subject is called, please? And where can I find more information about it? I've searched the web but they are all about positive-definite matrices, instead of a positive-definite space whose matrices don't have to be positive-definite.

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    $\begingroup$ As I understand your question, you have a set $M$ of square matrices (with no specific structure) acting on a Euclidean space $V$, and the question is whether for all $v\in V$ there is $m\in M$ so that $v^Tmv\geq0$. This is awfully broad and I fear not much can be said unless you give more constraints. Can you narrow down your question and tell what applications you have in mind? $\endgroup$ Dec 18, 2014 at 16:47
  • $\begingroup$ @JoonasIlmavirta Thank you. I've just edit my description. The matrices are generalized Laplacian matrices as used in knot theory. And the vector doesn't really have a special structure, but the basis of the vector space is known. $\endgroup$
    – wenru
    Dec 19, 2014 at 15:34
  • $\begingroup$ If these matrices are symmetric, then the real spectral theorem gives us an orthogonal basis of eigenvectors. If you look at just the subspace generated by the eigenvectors with positive eigenvalues, then all of the vectors in this subspace have $v^\top L v \geq 0$, with equality only for $v=0$. I am not sure there is too much more to say. Are you interested in the existence of positive eigenvalues? You would need to tell us more about the structure of these matrices to determine anything about that. $\endgroup$ Dec 19, 2014 at 15:40
  • $\begingroup$ I think I misunderstood the question actually. Can you tell us, exactly, how a vector $w$ generates $L_w$ (I do think a subscript like this is needed to keep things clear)? $\endgroup$ Dec 19, 2014 at 15:43
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    $\begingroup$ Excellent! Now the question makes more sense (although is too far from my expertise for me to answer). A little suggestion: include the dependence on $w$ in the notation for $\mathbf{L}$. The function you study seems to be bilinear in $V$ and linear in $W$ but the notation does not show all (or I misunderstood something). $\endgroup$ Dec 19, 2014 at 15:43

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