Mixing
Dimension of a vector space $W={f in V; f circ pi = f}$
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Let $N={1,2,3,4,5,6,7}$ and $pi$ a permutation $pi: N rightarrow N$.$$pi=(14)(235)(67)$$ Let $V$ be the vector space of all functions $N rightarrow mathbb{R}$. Let $W$ be the following set, $$W={f in V; f circ pi = f}$$ What is the dimension of $W$?
linear-algebra vector-spaces
asked Jan 21 at 21:22
nene123nene123
284
$endgroup$
add a comment |
$begingroup$
Let $N={1,2,3,4,5,6,7}$ and $pi$ a permutation $pi: N rightarrow N$.$$pi=(14)(235)(67)$$ Let $V$ be the vector space of all functions $N rightarrow mathbb{R}$. Let $W$ be the following set, $$W={f in V; f circ pi = f}$$ What is the dimension of $W$?
linear-algebra vector-spaces
asked Jan 21 at 21:22
nene123nene123
284
$endgroup$
$begingroup$
What have you tried and where did you get stuck? This is just a problem statement, which is generally not received well.
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– Servaes
Jan 21 at 21:23
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Also, how can $pi$ permute $7$ when $7notin N$?
$endgroup$
– Servaes
Jan 21 at 21:23
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Please add at least a modicum of context to tell Readers where your interest or difficulty in this problem lies. Are you able to make any progress (such as an upper or lower bound on the dimension)?
$endgroup$
– hardmath
Jan 22 at 3:53
add a comment |
$begingroup$
Let $N={1,2,3,4,5,6,7}$ and $pi$ a permutation $pi: N rightarrow N$.$$pi=(14)(235)(67)$$ Let $V$ be the vector space of all functions $N rightarrow mathbb{R}$. Let $W$ be the following set, $$W={f in V; f circ pi = f}$$ What is the dimension of $W$?
linear-algebra vector-spaces
asked Jan 21 at 21:22
nene123nene123
284
$endgroup$
Let $N={1,2,3,4,5,6,7}$ and $pi$ a permutation $pi: N rightarrow N$.$$pi=(14)(235)(67)$$ Let $V$ be the vector space of all functions $N rightarrow mathbb{R}$. Let $W$ be the following set, $$W={f in V; f circ pi = f}$$ What is the dimension of $W$?
linear-algebra vector-spaces
linear-algebra vector-spaces
asked Jan 21 at 21:22
nene123nene123
284
asked Jan 21 at 21:22
nene123nene123
284
edited Jan 21 at 21:24
nene123
asked Jan 21 at 21:22
nene123nene123
284
asked Jan 21 at 21:22
nene123nene123
284
asked Jan 21 at 21:22
nene123nene123
284
284
$begingroup$
What have you tried and where did you get stuck? This is just a problem statement, which is generally not received well.
$endgroup$
– Servaes
Jan 21 at 21:23
$begingroup$
Also, how can $pi$ permute $7$ when $7notin N$?
$endgroup$
– Servaes
Jan 21 at 21:23
$begingroup$
Please add at least a modicum of context to tell Readers where your interest or difficulty in this problem lies. Are you able to make any progress (such as an upper or lower bound on the dimension)?
$endgroup$
– hardmath
Jan 22 at 3:53
add a comment |
$begingroup$
What have you tried and where did you get stuck? This is just a problem statement, which is generally not received well.
$endgroup$
– Servaes
Jan 21 at 21:23
$begingroup$
Also, how can $pi$ permute $7$ when $7notin N$?
$endgroup$
– Servaes
Jan 21 at 21:23
$begingroup$
Please add at least a modicum of context to tell Readers where your interest or difficulty in this problem lies. Are you able to make any progress (such as an upper or lower bound on the dimension)?
$endgroup$
– hardmath
Jan 22 at 3:53
$begingroup$
What have you tried and where did you get stuck? This is just a problem statement, which is generally not received well.
$endgroup$
– Servaes
Jan 21 at 21:23
$begingroup$
What have you tried and where did you get stuck? This is just a problem statement, which is generally not received well.
$endgroup$
– Servaes
Jan 21 at 21:23
$begingroup$
Also, how can $pi$ permute $7$ when $7notin N$?
$endgroup$
– Servaes
Jan 21 at 21:23
$begingroup$
Also, how can $pi$ permute $7$ when $7notin N$?
$endgroup$
– Servaes
Jan 21 at 21:23
$begingroup$
Please add at least a modicum of context to tell Readers where your interest or difficulty in this problem lies. Are you able to make any progress (such as an upper or lower bound on the dimension)?
$endgroup$
– hardmath
Jan 22 at 3:53
$begingroup$
Please add at least a modicum of context to tell Readers where your interest or difficulty in this problem lies. Are you able to make any progress (such as an upper or lower bound on the dimension)?
$endgroup$
– hardmath
Jan 22 at 3:53
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint
Let $f in W$, then $f(pi(1))=f(1) implies f(4)=f(1)$. Likewise $f(2)=f(3)=f(5)$ and $f(6)=f(7)$. So there is only so much freedom to choose the values for $f$.......???
answered Jan 21 at 21:27
Anurag AAnurag A
26.3k12251
$endgroup$
add a comment |
$begingroup$
We have
$$f(1)=f(4)$$
$$f(2)=f(3)=f(5)$$
$$f(6)=f(7)$$
The subspace satisfying these equations has dimension $3$.
answered Jan 21 at 21:29
Matt SamuelMatt Samuel
38.8k63769
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint
Let $f in W$, then $f(pi(1))=f(1) implies f(4)=f(1)$. Likewise $f(2)=f(3)=f(5)$ and $f(6)=f(7)$. So there is only so much freedom to choose the values for $f$.......???
answered Jan 21 at 21:27
Anurag AAnurag A
26.3k12251
$endgroup$
add a comment |
$begingroup$
Hint
Let $f in W$, then $f(pi(1))=f(1) implies f(4)=f(1)$. Likewise $f(2)=f(3)=f(5)$ and $f(6)=f(7)$. So there is only so much freedom to choose the values for $f$.......???
answered Jan 21 at 21:27
Anurag AAnurag A
26.3k12251
$endgroup$
add a comment |
$begingroup$
Hint
Let $f in W$, then $f(pi(1))=f(1) implies f(4)=f(1)$. Likewise $f(2)=f(3)=f(5)$ and $f(6)=f(7)$. So there is only so much freedom to choose the values for $f$.......???
answered Jan 21 at 21:27
Anurag AAnurag A
26.3k12251
$endgroup$
Hint
Let $f in W$, then $f(pi(1))=f(1) implies f(4)=f(1)$. Likewise $f(2)=f(3)=f(5)$ and $f(6)=f(7)$. So there is only so much freedom to choose the values for $f$.......???
answered Jan 21 at 21:27
Anurag AAnurag A
26.3k12251
answered Jan 21 at 21:27
Anurag AAnurag A
26.3k12251
answered Jan 21 at 21:27
Anurag AAnurag A
26.3k12251
answered Jan 21 at 21:27
Anurag AAnurag A
26.3k12251
26.3k12251
add a comment |
add a comment |
$begingroup$
We have
$$f(1)=f(4)$$
$$f(2)=f(3)=f(5)$$
$$f(6)=f(7)$$
The subspace satisfying these equations has dimension $3$.
answered Jan 21 at 21:29
Matt SamuelMatt Samuel
38.8k63769
$endgroup$
add a comment |
$begingroup$
We have
$$f(1)=f(4)$$
$$f(2)=f(3)=f(5)$$
$$f(6)=f(7)$$
The subspace satisfying these equations has dimension $3$.
answered Jan 21 at 21:29
Matt SamuelMatt Samuel
38.8k63769
$endgroup$
add a comment |
$begingroup$
We have
$$f(1)=f(4)$$
$$f(2)=f(3)=f(5)$$
$$f(6)=f(7)$$
The subspace satisfying these equations has dimension $3$.
answered Jan 21 at 21:29
Matt SamuelMatt Samuel
38.8k63769
$endgroup$
We have
$$f(1)=f(4)$$
$$f(2)=f(3)=f(5)$$
$$f(6)=f(7)$$
The subspace satisfying these equations has dimension $3$.
answered Jan 21 at 21:29
Matt SamuelMatt Samuel
38.8k63769
answered Jan 21 at 21:29
Matt SamuelMatt Samuel
38.8k63769
answered Jan 21 at 21:29
Matt SamuelMatt Samuel
38.8k63769
answered Jan 21 at 21:29
Matt SamuelMatt Samuel
38.8k63769
38.8k63769
add a comment |
add a comment |
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$begingroup$
What have you tried and where did you get stuck? This is just a problem statement, which is generally not received well.
$endgroup$
– Servaes
Jan 21 at 21:23
$begingroup$
Also, how can $pi$ permute $7$ when $7notin N$?
$endgroup$
– Servaes
Jan 21 at 21:23
$begingroup$
Please add at least a modicum of context to tell Readers where your interest or difficulty in this problem lies. Are you able to make any progress (such as an upper or lower bound on the dimension)?
$endgroup$
– hardmath
Jan 22 at 3:53