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I posted this question some time ago here. I started a bounty for it and received an answer which helped me a lot. However, I still have some issues I wan't to discuss regarding it. Unfortunately I can't contact the contributor that answered before, so I decided to ask a different question. Through this question I will be referring to the answer I received here just as the "answer".

The main issue summary is the following: 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). A point $x$ is non regular if

• $x\in E$.
• The vectors $\{P_1x,\dotsc,P_mx\}$ are linearly dependent.

The problem is to show that if we make the change $P_i\leftarrow P_i+\epsilon_i$ with $\epsilon_i$ 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 $\epsilon>0$.

In the previous question, a contributor considered slightly 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. Then we can write $$x\in E_{i}(\epsilon)=\{x\in\mathbb{R}^{n}:s_{i}=\frac{1}{\lVert x\rVert^{2}}(1-x^{T}(P_i+\epsilon_{i})x)\}.$$

The contributor claimed that we decoupled the two events: ${x\in E(\epsilon)}$ is a random event that depends on the variable $s_{i}$, whereas $L_{\epsilon}(x):=\{(P_{1}+\epsilon_{1})x,\dotsc,(P_{m}+\epsilon_{m})x\text{ linearly independent}\}$ is a random event that depends on $\epsilon_i$. Then proceeded to compute the measure of the set of points in $L_{\epsilon}(x)$ and ${x\in E(\epsilon)}$.

However, I have some issues with this answer:

• Shouldn't $L_{\epsilon}(x):=\{(P_{1}+\epsilon_{1}+s_{i}I_{n})x,\dotsc,(P_{m}+\epsilon_{m}+s_{i}I_{n})x\text{ linearly *dependent*}\}$ with ‘dependent’ instead of ‘independent’?
• Since we care about vectors $\{(P_{1}+\epsilon_{1}+s_{i}I_{n})x,\dotsc,(P_{m}+\epsilon_{m}+s_{i}I_{n})x\}$ and not $\{(P_{1}+\epsilon_{1})x,\dotsc,(P_{m}+\epsilon_{m})x\}$ thus, we haven't truly decoupled both events ($L_{\epsilon}(x)$ and ${x\in E(\epsilon)}$) right? Is the proof still valid?
• Why is it that \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 \end{align*} at the end of the answer? Can you help me understand the last part of the proof? Do you think this proof is correct?

I posted this question some time ago here. I started a bounty for it and received an answer which helped me a lot. However, I still have some issues I want to discuss regarding it. Unfortunately I can't contact the contributor that answered before, so I decided to ask a different question. Through this question I will be referring to the answer I received here just as the "answer".

The main issue summary is the following: 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). A point $x$ is non regular if

• $x\in E$.
• The vectors $\{P_1x,\dotsc,P_mx\}$ are linearly dependent.

The problem is to show that if we make the change $P_i\leftarrow P_i+\epsilon_i$ with $\epsilon_i$ 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 $\epsilon>0$.

In the previous question, a contributor considered slightly 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. Then we can write $$x\in E_{i}(\epsilon)=\{x\in\mathbb{R}^{n}:s_{i}=\frac{1}{\lVert x\rVert^{2}}(1-x^{T}(P_i+\epsilon_{i})x)\}.$$

The contributor claimed that we decoupled the two events: ${x\in E(\epsilon)}$ is a random event that depends on the variable $s_{i}$, whereas $L_{\epsilon}(x):=\{(P_{1}+\epsilon_{1})x,\dotsc,(P_{m}+\epsilon_{m})x\text{ linearly independent}\}$ is a random event that depends on $\epsilon_i$. Then proceeded to compute the measure of the set of points in $L_{\epsilon}(x)$ and ${x\in E(\epsilon)}$.

However, I have some issues with this answer:

• Shouldn't $L_{\epsilon}(x):=\{(P_{1}+\epsilon_{1}+s_{i}I_{n})x,\dotsc,(P_{m}+\epsilon_{m}+s_{i}I_{n})x\text{ linearly *dependent*}\}$ with ‘dependent’ instead of ‘independent’?
• Since we care about vectors $\{(P_{1}+\epsilon_{1}+s_{i}I_{n})x,\dotsc,(P_{m}+\epsilon_{m}+s_{i}I_{n})x\}$ and not $\{(P_{1}+\epsilon_{1})x,\dotsc,(P_{m}+\epsilon_{m})x\}$ thus, we haven't truly decoupled both events ($L_{\epsilon}(x)$ and ${x\in E(\epsilon)}$) right? Is the proof still valid?
• Why is it that \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 \end{align*} at the end of the answer? Can you help me understand the last part of the proof? Do you think this proof is correct?

I posted this question some time ago here. I started a bounty for it and received an answer which helped me a lot. However, I still have some issues I wan't to discuss regarding it. Unfortunately I can't contact the contributor that answered before, so I decided to ask a different question. Through this question I will be referring to the answer I received here just as the "answer".

The main issue summary is the following: 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). A point $x$ is non regular if

• $x\in E$.
• The vectors $\{P_1x,\dotsc,P_mx\}$ are linearly dependent.

The problem is to show that if we make the change $P_i\leftarrow P_i+\epsilon_i$ with $\epsilon_i$ 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 $\epsilon>0$.

In the previous question, a contributor considered slightly 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. Then we can write $$x\in E_{i}(\epsilon)=\{x\in\mathbb{R}^{n}:s_{i}=\frac{1}{\|x\|^{2}}(1-x^{T}(P_i+\epsilon_{i})x)\}$$

The contributor claimed that we decoupled the two events  :${x\in E(\epsilon)}$ is a random event that depends on the variable $s_{i}$, whereas $L_{\epsilon}(x):=\{(P_{1}+\epsilon_{1})x,\cdots,(P_{m}+\epsilon_{m})x\text{ linearly independant}\}$ is a random event that depends on $\epsilon_i$. Then proceeded to compute the measure of the set of points in $L_{\epsilon}(x)$ and ${x\in E(\epsilon)}$.

However, I have some issues with this answer:

• Shouldn't $L_{\epsilon}(x):=\{(P_{1}+\epsilon_{1}+s_{i}I_{n})x,\cdots,(P_{m}+\epsilon_{m}+s_{i}I_{n})x\text{ linearly *dependent*}\}$? with dependent instead of independent.
• Since we care about vectors $\{(P_{1}+\epsilon_{1}+s_{i}I_{n})x,\cdots,(P_{m}+\epsilon_{m}+s_{i}I_{n})x\}$ and not $\{(P_{1}+\epsilon_{1})x,\cdots,(P_{m}+\epsilon_{m})x\}$ thus, we haven't truly decoupled both events ($L_{\epsilon}(x)$ and ${x\in E(\epsilon)}$) right? is the proof still valid?
• Why is it that \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 \end{align*} at the end of the answer? Can you help me understand the last part of the proof? Do you think this proof its correct?

I posted this question some time ago here. I started a bounty for it and received an answer which helped me a lot. However, I still have some issues I wan't to discuss regarding it. Unfortunately I can't contact the contributor that answered before, so I decided to ask a different question. Through this question I will be referring to the answer I received here just as the "answer".

The main issue summary is the following: 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). A point $x$ is non regular if

• $x\in E$.
• The vectors $\{P_1x,\dotsc,P_mx\}$ are linearly dependent.

The problem is to show that if we make the change $P_i\leftarrow P_i+\epsilon_i$ with $\epsilon_i$ 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 $\epsilon>0$.

In the previous question, a contributor considered slightly 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. Then we can write $$x\in E_{i}(\epsilon)=\{x\in\mathbb{R}^{n}:s_{i}=\frac{1}{\lVert x\rVert^{2}}(1-x^{T}(P_i+\epsilon_{i})x)\}.$$

The contributor claimed that we decoupled the two events:  ${x\in E(\epsilon)}$ is a random event that depends on the variable $s_{i}$, whereas $L_{\epsilon}(x):=\{(P_{1}+\epsilon_{1})x,\dotsc,(P_{m}+\epsilon_{m})x\text{ linearly independent}\}$ is a random event that depends on $\epsilon_i$. Then proceeded to compute the measure of the set of points in $L_{\epsilon}(x)$ and ${x\in E(\epsilon)}$.

However, I have some issues with this answer:

• Shouldn't $L_{\epsilon}(x):=\{(P_{1}+\epsilon_{1}+s_{i}I_{n})x,\dotsc,(P_{m}+\epsilon_{m}+s_{i}I_{n})x\text{ linearly *dependent*}\}$ with ‘dependent’ instead of ‘independent’?
• Since we care about vectors $\{(P_{1}+\epsilon_{1}+s_{i}I_{n})x,\dotsc,(P_{m}+\epsilon_{m}+s_{i}I_{n})x\}$ and not $\{(P_{1}+\epsilon_{1})x,\dotsc,(P_{m}+\epsilon_{m})x\}$ thus, we haven't truly decoupled both events ($L_{\epsilon}(x)$ and ${x\in E(\epsilon)}$) right? Is the proof still valid?
• Why is it that \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 \end{align*} at the end of the answer? Can you help me understand the last part of the proof? Do you think this proof is correct?

I posted this question some time ago here. I started a bounty for it and received an answer which helped me a lot. However, I still have some issues I wan't to discuss regarding it. Unfortunately I can't contact the contributor that answered before, so I decided to ask a different question. Through this question I will be referring to the answer I received here just as the "answer".

The main issue summary is the following: 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). A point $x$ is non regular if

• $x\in E$.
• The vectors $\{P_1x,\dotsc,P_mx\}$ are linearly dependent.

The problem is to show that if we make the change $P_i\leftarrow P_i+\varepsilon_i$ 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 $\epsilon>0$.

In the previous question, a contributor considered slightly 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. Then we can write $$x\in E_{i}(\epsilon)=\{x\in\mathbb{R}^{n}:s_{i}=\frac{1}{\|x\|^{2}}(1-x^{T}(P+\epsilon_{i})x)\}$$

The contributor claimed that we decoupled the two events :${x\in E(\epsilon)}$ is a random event that depends on the variable $s_{i}$, whereas $L_{\epsilon}(x):=\{(P_{1}+\epsilon_{1})x,\cdots,(P_{m}+\epsilon_{m})x\text{ linearly independant}\}$ is a random event that depends on $\epsilon_i$. Then proceeded to compute the measure of the set of points in $L_{\epsilon}(x)$ and ${x\in E(\epsilon)}$.

However, I have some issues with this answer:

• Shouldn't $L_{\epsilon}(x):=\{(P_{1}+\epsilon_{1}+s_{i}I_{n})x,\cdots,(P_{m}+\epsilon_{m}+s_{i}I_{n})x\text{ linearly *dependent*}\}$? with dependent instead of independent.
• Since we care about vectors $\{(P_{1}+\epsilon_{1}+s_{i}I_{n})x,\cdots,(P_{m}+\epsilon_{m}+s_{i}I_{n})x\}$ and not $\{(P_{1}+\epsilon_{1})x,\cdots,(P_{m}+\epsilon_{m})x\}$ thus, we haven't truly decoupled both events ($L_{\epsilon}(x)$ and ${x\in E(\epsilon)}$) right? is the proof still valid?
• Why is it that \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 \end{align*} at the end of the question? Can you help me understand the last part of the proof? Do you think this proof its correct?

I posted this question some time ago here. I started a bounty for it and received an answer which helped me a lot. However, I still have some issues I wan't to discuss regarding it. Unfortunately I can't contact the contributor that answered before, so I decided to ask a different question. Through this question I will be referring to the answer I received here just as the "answer".

The main issue summary is the following: 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). A point $x$ is non regular if

• $x\in E$.
• The vectors $\{P_1x,\dotsc,P_mx\}$ are linearly dependent.

The problem is to show that if we make the change $P_i\leftarrow P_i+\epsilon_i$ with $\epsilon_i$ 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 $\epsilon>0$.

In the previous question, a contributor considered slightly 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. Then we can write $$x\in E_{i}(\epsilon)=\{x\in\mathbb{R}^{n}:s_{i}=\frac{1}{\|x\|^{2}}(1-x^{T}(P_i+\epsilon_{i})x)\}$$

The contributor claimed that we decoupled the two events :${x\in E(\epsilon)}$ is a random event that depends on the variable $s_{i}$, whereas $L_{\epsilon}(x):=\{(P_{1}+\epsilon_{1})x,\cdots,(P_{m}+\epsilon_{m})x\text{ linearly independant}\}$ is a random event that depends on $\epsilon_i$. Then proceeded to compute the measure of the set of points in $L_{\epsilon}(x)$ and ${x\in E(\epsilon)}$.

However, I have some issues with this answer:

• Shouldn't $L_{\epsilon}(x):=\{(P_{1}+\epsilon_{1}+s_{i}I_{n})x,\cdots,(P_{m}+\epsilon_{m}+s_{i}I_{n})x\text{ linearly *dependent*}\}$? with dependent instead of independent.
• Since we care about vectors $\{(P_{1}+\epsilon_{1}+s_{i}I_{n})x,\cdots,(P_{m}+\epsilon_{m}+s_{i}I_{n})x\}$ and not $\{(P_{1}+\epsilon_{1})x,\cdots,(P_{m}+\epsilon_{m})x\}$ thus, we haven't truly decoupled both events ($L_{\epsilon}(x)$ and ${x\in E(\epsilon)}$) right? is the proof still valid?
• Why is it that \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 \end{align*} at the end of the answer? Can you help me understand the last part of the proof? Do you think this proof its correct?