Linear independence over $mathbb{Z}_p^r$
Linearly independent vectors $X_1, X_2, ..., X_n$ over $mathbb{Q}^r$ have integer coordinates. Prove they are linearly independent over $mathbb{Z}_p^r$ for almost every prime $p$. I've been thinking about this problem for a few days and couldn't find any solution.
linear-algebra vector-spaces
asked Nov 23 '18 at 7:43
Quo Si ThanQuo Si Than
1477
add a comment |
Linearly independent vectors $X_1, X_2, ..., X_n$ over $mathbb{Q}^r$ have integer coordinates. Prove they are linearly independent over $mathbb{Z}_p^r$ for almost every prime $p$. I've been thinking about this problem for a few days and couldn't find any solution.
linear-algebra vector-spaces
asked Nov 23 '18 at 7:43
Quo Si ThanQuo Si Than
1477
Are you suggesting that $(frac{1}{2},frac{1}{2})$ and $(frac{1}{2},frac{1}{3})$ are linearly dependent over $Bbb{Q}$?
– Servaes
Nov 23 '18 at 8:21
Vectors consist only of integers
– Quo Si Than
Nov 23 '18 at 9:41
I don't understand the close votes, the question is quite clear.
– Slade
Nov 23 '18 at 9:46
add a comment |
Linearly independent vectors $X_1, X_2, ..., X_n$ over $mathbb{Q}^r$ have integer coordinates. Prove they are linearly independent over $mathbb{Z}_p^r$ for almost every prime $p$. I've been thinking about this problem for a few days and couldn't find any solution.
linear-algebra vector-spaces
asked Nov 23 '18 at 7:43
Quo Si ThanQuo Si Than
1477
Linearly independent vectors $X_1, X_2, ..., X_n$ over $mathbb{Q}^r$ have integer coordinates. Prove they are linearly independent over $mathbb{Z}_p^r$ for almost every prime $p$. I've been thinking about this problem for a few days and couldn't find any solution.
linear-algebra vector-spaces
linear-algebra vector-spaces
asked Nov 23 '18 at 7:43
Quo Si ThanQuo Si Than
1477
asked Nov 23 '18 at 7:43
Quo Si ThanQuo Si Than
1477
asked Nov 23 '18 at 7:43
Quo Si ThanQuo Si Than
1477
asked Nov 23 '18 at 7:43
Quo Si ThanQuo Si Than
1477
asked Nov 23 '18 at 7:43
Quo Si ThanQuo Si Than
1477
1477
Are you suggesting that $(frac{1}{2},frac{1}{2})$ and $(frac{1}{2},frac{1}{3})$ are linearly dependent over $Bbb{Q}$?
– Servaes
Nov 23 '18 at 8:21
Vectors consist only of integers
– Quo Si Than
Nov 23 '18 at 9:41
I don't understand the close votes, the question is quite clear.
– Slade
Nov 23 '18 at 9:46
add a comment |
Are you suggesting that $(frac{1}{2},frac{1}{2})$ and $(frac{1}{2},frac{1}{3})$ are linearly dependent over $Bbb{Q}$?
– Servaes
Nov 23 '18 at 8:21
Vectors consist only of integers
– Quo Si Than
Nov 23 '18 at 9:41
I don't understand the close votes, the question is quite clear.
– Slade
Nov 23 '18 at 9:46
Are you suggesting that $(frac{1}{2},frac{1}{2})$ and $(frac{1}{2},frac{1}{3})$ are linearly dependent over $Bbb{Q}$?
– Servaes
Nov 23 '18 at 8:21
Are you suggesting that $(frac{1}{2},frac{1}{2})$ and $(frac{1}{2},frac{1}{3})$ are linearly dependent over $Bbb{Q}$?
– Servaes
Nov 23 '18 at 8:21
Vectors consist only of integers
– Quo Si Than
Nov 23 '18 at 9:41
Vectors consist only of integers
– Quo Si Than
Nov 23 '18 at 9:41
I don't understand the close votes, the question is quite clear.
– Slade
Nov 23 '18 at 9:46
I don't understand the close votes, the question is quite clear.
– Slade
Nov 23 '18 at 9:46
add a comment |
1 Answer
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If $M$ denotes the matrix sending the standard basis vector $E_i$ to $X_i$, then linear independence is equivalent to $det M neq 0$.
What happens if $det M equiv 0 pmod{p}$ for infinitely many primes $p$?
answered Nov 23 '18 at 9:44
SladeSlade
24.9k12564
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
If $M$ denotes the matrix sending the standard basis vector $E_i$ to $X_i$, then linear independence is equivalent to $det M neq 0$.
What happens if $det M equiv 0 pmod{p}$ for infinitely many primes $p$?
answered Nov 23 '18 at 9:44
SladeSlade
24.9k12564
add a comment |
If $M$ denotes the matrix sending the standard basis vector $E_i$ to $X_i$, then linear independence is equivalent to $det M neq 0$.
What happens if $det M equiv 0 pmod{p}$ for infinitely many primes $p$?
answered Nov 23 '18 at 9:44
SladeSlade
24.9k12564
add a comment |
If $M$ denotes the matrix sending the standard basis vector $E_i$ to $X_i$, then linear independence is equivalent to $det M neq 0$.
What happens if $det M equiv 0 pmod{p}$ for infinitely many primes $p$?
answered Nov 23 '18 at 9:44
SladeSlade
24.9k12564
If $M$ denotes the matrix sending the standard basis vector $E_i$ to $X_i$, then linear independence is equivalent to $det M neq 0$.
What happens if $det M equiv 0 pmod{p}$ for infinitely many primes $p$?
answered Nov 23 '18 at 9:44
SladeSlade
24.9k12564
answered Nov 23 '18 at 9:44
SladeSlade
24.9k12564
answered Nov 23 '18 at 9:44
SladeSlade
24.9k12564
answered Nov 23 '18 at 9:44
SladeSlade
24.9k12564
24.9k12564
add a comment |
add a comment |
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Are you suggesting that $(frac{1}{2},frac{1}{2})$ and $(frac{1}{2},frac{1}{3})$ are linearly dependent over $Bbb{Q}$?
– Servaes
Nov 23 '18 at 8:21
Vectors consist only of integers
– Quo Si Than
Nov 23 '18 at 9:41
I don't understand the close votes, the question is quite clear.
– Slade
Nov 23 '18 at 9:46