Mixing
Vector space over finite field: finding the plane perpendicular the vector $(1,1,1)$
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For some 3 dimensional vector space over a finite field $F_{p^k}$ how can I determine (and visualise) the elements which are in a plane which is perpendicular to the vector $(1, 1, 1)$? For an $n$ dimensional vector space over $F_{p^k}$ how can I determine/visualise the points which are in the $n-1$ dimensional space perpendicular to the vector $(1,1,dots,1)$?
In the case of a 2 dimensional vector space the line perpendicular to the unit vector $(1,1)$ which passes through the point $(1,1)$ can be visualised as a diagonal line from $(0, p^k - 1)$ to $(p^k - 1, 0)$ but I'm struggling to visualise even the 3 dimensional case.
vector-spaces finite-fields
asked Jan 12 at 15:45
Chris RussellChris Russell
205
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show 7 more comments
$begingroup$
For some 3 dimensional vector space over a finite field $F_{p^k}$ how can I determine (and visualise) the elements which are in a plane which is perpendicular to the vector $(1, 1, 1)$? For an $n$ dimensional vector space over $F_{p^k}$ how can I determine/visualise the points which are in the $n-1$ dimensional space perpendicular to the vector $(1,1,dots,1)$?
In the case of a 2 dimensional vector space the line perpendicular to the unit vector $(1,1)$ which passes through the point $(1,1)$ can be visualised as a diagonal line from $(0, p^k - 1)$ to $(p^k - 1, 0)$ but I'm struggling to visualise even the 3 dimensional case.
vector-spaces finite-fields
asked Jan 12 at 15:45
Chris RussellChris Russell
205
$endgroup$
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What is THE unit vector? And for which norm?
$endgroup$
– Bernard
Jan 12 at 15:54
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Oops. I mean the vector $(1, 1,dots, 1)$, which is not a unit vector.
$endgroup$
– Chris Russell
Jan 12 at 15:55
$begingroup$
Well, such a hyperplane has equation $x_1+x_2+dots+x_n=0$. What else can you say?
$endgroup$
– Bernard
Jan 12 at 16:02
$begingroup$
Is this plane really perpendicular to the vector? What if $n = p^k$ then $x_1 + x_2 + cdots + x_n = 0$ mod $p^k$ contains both the points $(0,0,dots,0)$ and $(1,1,dots, 1)$. It is meant to be perpendicular to the vector $(1,1,dots,1)$ yet it contains two points where line between them is parallel to the vector $(1,1,dots,1)$.
$endgroup$
– Chris Russell
Jan 12 at 16:09
$begingroup$
Actually, you have to define an inner product on your vector space.
$endgroup$
– Bernard
Jan 12 at 16:15
|
show 7 more comments
$begingroup$
For some 3 dimensional vector space over a finite field $F_{p^k}$ how can I determine (and visualise) the elements which are in a plane which is perpendicular to the vector $(1, 1, 1)$? For an $n$ dimensional vector space over $F_{p^k}$ how can I determine/visualise the points which are in the $n-1$ dimensional space perpendicular to the vector $(1,1,dots,1)$?
In the case of a 2 dimensional vector space the line perpendicular to the unit vector $(1,1)$ which passes through the point $(1,1)$ can be visualised as a diagonal line from $(0, p^k - 1)$ to $(p^k - 1, 0)$ but I'm struggling to visualise even the 3 dimensional case.
vector-spaces finite-fields
asked Jan 12 at 15:45
Chris RussellChris Russell
205
$endgroup$
For some 3 dimensional vector space over a finite field $F_{p^k}$ how can I determine (and visualise) the elements which are in a plane which is perpendicular to the vector $(1, 1, 1)$? For an $n$ dimensional vector space over $F_{p^k}$ how can I determine/visualise the points which are in the $n-1$ dimensional space perpendicular to the vector $(1,1,dots,1)$?
In the case of a 2 dimensional vector space the line perpendicular to the unit vector $(1,1)$ which passes through the point $(1,1)$ can be visualised as a diagonal line from $(0, p^k - 1)$ to $(p^k - 1, 0)$ but I'm struggling to visualise even the 3 dimensional case.
vector-spaces finite-fields
vector-spaces finite-fields
asked Jan 12 at 15:45
Chris RussellChris Russell
205
asked Jan 12 at 15:45
Chris RussellChris Russell
205
edited Jan 12 at 15:56
Chris Russell
asked Jan 12 at 15:45
Chris RussellChris Russell
205
asked Jan 12 at 15:45
Chris RussellChris Russell
205
asked Jan 12 at 15:45
Chris RussellChris Russell
205
205
$begingroup$
What is THE unit vector? And for which norm?
$endgroup$
– Bernard
Jan 12 at 15:54
$begingroup$
Oops. I mean the vector $(1, 1,dots, 1)$, which is not a unit vector.
$endgroup$
– Chris Russell
Jan 12 at 15:55
$begingroup$
Well, such a hyperplane has equation $x_1+x_2+dots+x_n=0$. What else can you say?
$endgroup$
– Bernard
Jan 12 at 16:02
$begingroup$
Is this plane really perpendicular to the vector? What if $n = p^k$ then $x_1 + x_2 + cdots + x_n = 0$ mod $p^k$ contains both the points $(0,0,dots,0)$ and $(1,1,dots, 1)$. It is meant to be perpendicular to the vector $(1,1,dots,1)$ yet it contains two points where line between them is parallel to the vector $(1,1,dots,1)$.
$endgroup$
– Chris Russell
Jan 12 at 16:09
$begingroup$
Actually, you have to define an inner product on your vector space.
$endgroup$
– Bernard
Jan 12 at 16:15
|
show 7 more comments
$begingroup$
What is THE unit vector? And for which norm?
$endgroup$
– Bernard
Jan 12 at 15:54
$begingroup$
Oops. I mean the vector $(1, 1,dots, 1)$, which is not a unit vector.
$endgroup$
– Chris Russell
Jan 12 at 15:55
$begingroup$
Well, such a hyperplane has equation $x_1+x_2+dots+x_n=0$. What else can you say?
$endgroup$
– Bernard
Jan 12 at 16:02
$begingroup$
Is this plane really perpendicular to the vector? What if $n = p^k$ then $x_1 + x_2 + cdots + x_n = 0$ mod $p^k$ contains both the points $(0,0,dots,0)$ and $(1,1,dots, 1)$. It is meant to be perpendicular to the vector $(1,1,dots,1)$ yet it contains two points where line between them is parallel to the vector $(1,1,dots,1)$.
$endgroup$
– Chris Russell
Jan 12 at 16:09
$begingroup$
Actually, you have to define an inner product on your vector space.
$endgroup$
– Bernard
Jan 12 at 16:15
$begingroup$
What is THE unit vector? And for which norm?
$endgroup$
– Bernard
Jan 12 at 15:54
$begingroup$
What is THE unit vector? And for which norm?
$endgroup$
– Bernard
Jan 12 at 15:54
$begingroup$
Oops. I mean the vector $(1, 1,dots, 1)$, which is not a unit vector.
$endgroup$
– Chris Russell
Jan 12 at 15:55
$begingroup$
Oops. I mean the vector $(1, 1,dots, 1)$, which is not a unit vector.
$endgroup$
– Chris Russell
Jan 12 at 15:55
$begingroup$
Well, such a hyperplane has equation $x_1+x_2+dots+x_n=0$. What else can you say?
$endgroup$
– Bernard
Jan 12 at 16:02
$begingroup$
Well, such a hyperplane has equation $x_1+x_2+dots+x_n=0$. What else can you say?
$endgroup$
– Bernard
Jan 12 at 16:02
$begingroup$
Is this plane really perpendicular to the vector? What if $n = p^k$ then $x_1 + x_2 + cdots + x_n = 0$ mod $p^k$ contains both the points $(0,0,dots,0)$ and $(1,1,dots, 1)$. It is meant to be perpendicular to the vector $(1,1,dots,1)$ yet it contains two points where line between them is parallel to the vector $(1,1,dots,1)$.
$endgroup$
– Chris Russell
Jan 12 at 16:09
$begingroup$
Is this plane really perpendicular to the vector? What if $n = p^k$ then $x_1 + x_2 + cdots + x_n = 0$ mod $p^k$ contains both the points $(0,0,dots,0)$ and $(1,1,dots, 1)$. It is meant to be perpendicular to the vector $(1,1,dots,1)$ yet it contains two points where line between them is parallel to the vector $(1,1,dots,1)$.
$endgroup$
– Chris Russell
Jan 12 at 16:09
$begingroup$
Actually, you have to define an inner product on your vector space.
$endgroup$
– Bernard
Jan 12 at 16:15
$begingroup$
Actually, you have to define an inner product on your vector space.
$endgroup$
– Bernard
Jan 12 at 16:15
|
show 7 more comments
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$begingroup$
What is THE unit vector? And for which norm?
$endgroup$
– Bernard
Jan 12 at 15:54
$begingroup$
Oops. I mean the vector $(1, 1,dots, 1)$, which is not a unit vector.
$endgroup$
– Chris Russell
Jan 12 at 15:55
$begingroup$
Well, such a hyperplane has equation $x_1+x_2+dots+x_n=0$. What else can you say?
$endgroup$
– Bernard
Jan 12 at 16:02
$begingroup$
Is this plane really perpendicular to the vector? What if $n = p^k$ then $x_1 + x_2 + cdots + x_n = 0$ mod $p^k$ contains both the points $(0,0,dots,0)$ and $(1,1,dots, 1)$. It is meant to be perpendicular to the vector $(1,1,dots,1)$ yet it contains two points where line between them is parallel to the vector $(1,1,dots,1)$.
$endgroup$
– Chris Russell
Jan 12 at 16:09
$begingroup$
Actually, you have to define an inner product on your vector space.
$endgroup$
– Bernard
Jan 12 at 16:15