From here, if $rad(A)$ is maximal I think I'd have a contradiction, but I don't know if that's true. If not, does anyone have another strategy I could try?

bedblogi38am
2022-05-11
Answered

Let x not be quasinilpotent, so $li{m}_{n\to \mathrm{\infty}}||{x}^{n}|{|}^{1/n}=\lambda \ne 0$. Let $\pi (x)=\overline{x}$, where $\pi (x):A\to A/rad(A)$ is the canonical quotient map. Suppose that $||{\overline{x}}^{n}|{|}^{1/n}=0$. Then it's spectrum $\sigma (\overline{x})=0$, so $\overline{x}$ is not invertible in $A/rad(A)$, and thus generates a proper ideal in $A/rad(A)$. So then ${\pi}^{-1}(\overline{x})$ generates a proper ideal in $A$ containing $rad(A)$.

From here, if $rad(A)$ is maximal I think I'd have a contradiction, but I don't know if that's true. If not, does anyone have another strategy I could try?

From here, if $rad(A)$ is maximal I think I'd have a contradiction, but I don't know if that's true. If not, does anyone have another strategy I could try?

You can still ask an expert for help

Bentyrchjurvk

Answered 2022-05-12
Author has **15** answers

Assume $x\in A$ such that $x+rad(A)$ is nilpotent in $A/rad(A)$. This implies that for all $\epsilon >0$ there is an $n\in \mathbb{N}$ and $r\in rad(A)$ for which $|{x}^{n}-r|\le {\epsilon}^{n}$.

$r$ is quasi-nilpotent so there exists $M\in \mathbb{N}$ s.t. for all $m>M$, $|{r}^{m}|\le {\epsilon}^{nm}$

We then show by induction that there exists $A>0$ such that $|{x}^{nm}|\le A{2}^{m}{\epsilon}^{nm}$ for all $m$. We choose $A\ge 1$ such that this is true for all $m\in [0,M]$.

Then if $m>M$, we have

${x}^{nm}-{r}^{m}=({x}^{n}-r)({x}^{n(m-1)}+\dots +{r}^{m-1}),$

so

$|{x}^{nm}|\le {\epsilon}^{n}(A{2}^{m-1}{\epsilon}^{n(m-1)}+\dots +A{\epsilon}^{n(m-1)})+{\epsilon}^{nm}\le A{\epsilon}^{nm}(1+1+\dots {2}^{m-1})=A{2}^{m}{\epsilon}^{nm}.$

So we can conclude that $\mathrm{\forall}\epsilon >0\phantom{\rule{mediummathspace}{0ex}}lim\u2006inf|{x}^{n}{|}^{1/n}<2\epsilon $, which means that $x\in rad(A)$.

$r$ is quasi-nilpotent so there exists $M\in \mathbb{N}$ s.t. for all $m>M$, $|{r}^{m}|\le {\epsilon}^{nm}$

We then show by induction that there exists $A>0$ such that $|{x}^{nm}|\le A{2}^{m}{\epsilon}^{nm}$ for all $m$. We choose $A\ge 1$ such that this is true for all $m\in [0,M]$.

Then if $m>M$, we have

${x}^{nm}-{r}^{m}=({x}^{n}-r)({x}^{n(m-1)}+\dots +{r}^{m-1}),$

so

$|{x}^{nm}|\le {\epsilon}^{n}(A{2}^{m-1}{\epsilon}^{n(m-1)}+\dots +A{\epsilon}^{n(m-1)})+{\epsilon}^{nm}\le A{\epsilon}^{nm}(1+1+\dots {2}^{m-1})=A{2}^{m}{\epsilon}^{nm}.$

So we can conclude that $\mathrm{\forall}\epsilon >0\phantom{\rule{mediummathspace}{0ex}}lim\u2006inf|{x}^{n}{|}^{1/n}<2\epsilon $, which means that $x\in rad(A)$.

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Now in linear algebra this is obviously not true (equations of the form y=mx+q are not necessarily linear).

Am i mistaking something or is this just a point where statisticans and mathematicians use the same words for different things?

$\mathbb{R}\mapsto \mathbb{R}:y(x)=mx+q$

Now in linear algebra this is obviously not true (equations of the form y=mx+q are not necessarily linear).

Am i mistaking something or is this just a point where statisticans and mathematicians use the same words for different things?

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Find the vector form of the equation of the line in ${\mathbb{R}}^{3}$ that passes through the point $P=(-1,1,3)$ and is perpendicular to the plane with general equation $x-3y+2z=5$.

I know the normal vector $\overrightarrow{n}=[1,-3,2]$ from the general equation but i'm not sure how it helps me. Based on the question it would seem this would be parallel to line i'm trying to find?

My thought was to find 2 other points on the plane, say Q and R, by satisfying the general equation and use those points to find the direction vectors $\overrightarrow{u}=\overrightarrow{P}Q$ and $\overrightarrow{v}=\overrightarrow{P}R$. Then I would use those to form my vector equation

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Is this the proper way to solve this type of problem? Or would this only work if the line is parallel to the plane? The parallel/perpendicular to the plane concept has me really confused.

I just started this course and i'm struggling, so i'm doing my best to try and understand problems outside of the assigned homework (one which I cannot look up an answer for). If someone could give me the proper advice for how to approach these types of problems, I would be extremely grateful. Thanks.

Find the vector form of the equation of the line in ${\mathbb{R}}^{3}$ that passes through the point $P=(-1,1,3)$ and is perpendicular to the plane with general equation $x-3y+2z=5$.

I know the normal vector $\overrightarrow{n}=[1,-3,2]$ from the general equation but i'm not sure how it helps me. Based on the question it would seem this would be parallel to line i'm trying to find?

My thought was to find 2 other points on the plane, say Q and R, by satisfying the general equation and use those points to find the direction vectors $\overrightarrow{u}=\overrightarrow{P}Q$ and $\overrightarrow{v}=\overrightarrow{P}R$. Then I would use those to form my vector equation

$\overrightarrow{x}=\overrightarrow{p}+s\overrightarrow{u}+t\overrightarrow{v}$

Is this the proper way to solve this type of problem? Or would this only work if the line is parallel to the plane? The parallel/perpendicular to the plane concept has me really confused.

I just started this course and i'm struggling, so i'm doing my best to try and understand problems outside of the assigned homework (one which I cannot look up an answer for). If someone could give me the proper advice for how to approach these types of problems, I would be extremely grateful. Thanks.

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