Show that these vectors are linearly independent almost surely So I'm doing research in control theory and I have been stuck with this problem for a while. Let me explain my issue, then my proposal, and finally my concrete question.
Problem: I have $m<n$ real $n\times n$ positive definite matrices $P_1,\dotsc,P_m$. These define ellipsoids $E_i=\{x\in\mathbb{R}^n\mathrel:x^TP_ix=1\}$. I'm interested in the points that lie in the intersection of all these ellipsoids (let's call it $E\mathrel{:=}\bigcap_{i=1}^mE_i$ for short). However, there are points which are troublesome (let's called them non regular). A point $x$ is non regular if

*

*$x\in E$.

*The vectors $\{P_1x,\dotsc,P_mx\}$ are linearly dependent.

So, matrices $P_1,\dotsc,P_m$ that induce non regular points are problematic in my case. And ideally I would want to show that these matrices can be perturbed a little such that these non regular points disappear.
Proposal: I want to show that given $P_1,\dotsc,P_m$, if you substitute $P_i\leftarrow P_i+\varepsilon_i$ with $\varepsilon_i$ a random matrix with elements uniformly distributed in $[-\epsilon,\epsilon]$ (or some other distribution if desired),  the probability of a point $x\in E$ to be nonregular is 0. Or equivalently that $x$ is "regular" almost surely for any $\epsilon>0$.
Now, for $\{P_1x,\dotsc,P_mx\}$ to be linearly independent (with $P_i$ taking into account the random matrices $\varepsilon_i$) we require the existence of coefficients $\alpha_1,\dotsc,\alpha_m$ such that the matrix $H=\sum_{i=1}^m \alpha_i P_i$ is singular. Thus, non regular points $x$ must lie in the intersection of $E$ and the kernel of $H$. Moreover, since $x^THx=0$ and $x^TP_ix=1$, thus $\sum_{i=1}^m\alpha_i=0$. Existence of coefficients $\alpha_i$ with these characteristics (with $\sum_{i=1}^m\alpha_i=0$ and $H$ singular) are guaranteed as described in the answer of this other question of mine here, so my hope to show that $H$ is non-singular almost surely vanished. But I can still try to show that the probability $P(\text{$\{x\in E\}$ and $\{x\in\operatorname{ker} H\}$})=0$.
More context: This question arises since I'm looking for a "special" point $x^*\in E$ (say optimal in the sense of an objective $x^TA_0x$ with $A_0$ positive definite) and I want to make sure (or at least almost sure) that $x^*$ is regular.
Questions:

*

*Do you think this approach is correct in order to make (almost) sure that the "optimal point" $x^*$ is regular?


*Do you think such statement (that we can show that disturbing a little the matrices will make non regular points disappear) is true?


*Do you have any idea how I can approach this problem to show such a statement?


*Do you suggest any other approach?
Hopefully this is the right forum to ask this question. I'm not trying to find a full solution here of course. At this point, any suggestion is valuable for me.
P.S. I'm quite new to this site, so I'll try to choose the best tags. However, I would I appreciate if someone can add/remove tags if needed.
EDIT: EXAMPLE
This example appeals mostly to intuition, sorry if I'm not too rigorous in this part:
Consider $P_1=\text{diag}(1,1,2)$ and $P_2=\text{diag}(1,1,4)$. Clearly, the intersection of $E_1$ and $E_2$ is the unit circle on the "floor". This is, $E=\{x\in\mathbb{R}^3: x = (x_1,x_2,0)^T, x_1^2+x_2^2=1\}$. Moreover, any point in $E$ have $P_1x=P_2x$. Hence, all points in $E$ are non regular. However, it should be easy to show (in this example) that disturbing a little $P_1,P_2$ will prevent the intersection to be full of non regular points, since $P_1,P_2$ were very carefully chosen (are a very degenerate case). They (non regular points) may not disappear completely, but my intuition is that they will lie in a set of measure zero, and therefore any $x\in E$ will be regular almost surely. However, I don't have any intuition on what might happen in higher dimensions.
 A: We denote $L_{\epsilon}(x):=\{(P_{1}+\epsilon_{1})x,\cdots,(P_{m}+\epsilon_{m})x\text{ linearly independant}\}$
First we have that for any fixed $x\in\mathbb{R}^{n}$, $$\mathbb{P}(L_{\epsilon}(x))=0.$$
Indeed if you only consider the $m$ first entries of these $m$ vectors you have get an $m\times m$ random matrix $M_{\epsilon}(x)$ with independant random entries. And then $\det(M_{\epsilon}(x))$ is a smooth random variable on \mathbb{R} so \mathbb{P}$(\det(M_{\epsilon}(x))=0)=0$. In fact we used that $$\{(\epsilon_{i})\in(\mathbb{R}^{n\times n})^{m}:\det(M_{\epsilon}(x))=0\}$$ has Lebesgue measure $0$.
The question now is what about $\mathbb{P}(L_\epsilon(x))$ but given the condition that $x\in E(\epsilon)$.
I will consider slitly different random matrices $$\tilde{\epsilon}_{i}=\epsilon_{i}+s_{i}I_{n}$$ where $s_{i}$ independent random variable in $[-\epsilon,\epsilon]$ with continuous density and $I_{n}$ the identity matrix. The proof below works as well with just $\epsilon_{i}$ but it is a bit more complicate and I guess that this random law work as well for what you want. Then we can write $$x\in E_{i}(\epsilon)=\{x\in\mathbb{R}^{n}:s_{i}=\frac{1}{\|x\|^{2}}(1-x^{T}(P+\epsilon_{i})x)\}$$
In a way we have decoupled the two events :${x\in E(\epsilon)}$ is a random event that depends on the variabl}e $s_{i}$, whereas $L_{\epsilon}(x)$ is a random event that depends on $\epsilon_i$.
We denote $\rho(s)$ the density for the $s$ variables, $\mu(\epsilon)$ the density for the $\epsilon$ variables and $\sigma_{E(\tilde{\epsilon})}$ the measure on the surface $E(\tilde{\epsilon})$. Then we have
\begin{align*}
& \mathbb{P}(\{\tilde{\epsilon}:\sigma_{E(\tilde{\epsilon})}(L_{\epsilon}(x))=0\})=0 \\
& \Leftrightarrow \int_{[-\epsilon,\epsilon]^{*}}\mu(\epsilon)d\epsilon\int_{[-\epsilon,\epsilon]^{m}}\rho(s)d^{m}s\int_{E(\epsilon)}1_{L_{\epsilon}(x)}d\sigma(x)=0
\\ & \Leftrightarrow \int_{[-\epsilon,\epsilon]^{*}}\mu(\epsilon)d\epsilon\int_{\mathbb{R}^{n}}1_{L_{\epsilon}(x)}\rho(\frac{1}{\|x\|^{2}}(1-x^{T}(P+\epsilon_{i})x))d^{n}x=0
\\ & \Leftrightarrow \int_{\mathbb{R}^{n}}\int_{[-\epsilon,\epsilon]^{*}}\mu(\epsilon)1_{L_{\epsilon}(x)}\rho(\frac{1}{\|x\|^{2}}(1-x^{T}(P+\epsilon_{i})x))d\epsilon d^{n}x=0
\end{align*}
And the last equality is satisfied because as we proved before for any $x$, $\{\epsilon:L_\epsilon(x)\}$ is Lebesgue measure $0$.
Conclusion, with probability 1 on $\tilde{\epsilon}$, The set of non regular point on $E(\tilde{\epsilon})$ is of measure 0.
