Mixing
Rank of an integer matrix modulo a prime
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Suppose I have a matrix A with integer elements and a matrix A mod p, where p is a prime number. In what circumstances rank(A)=rank(A mod p)?
Based on my understanding,
1) when det(A) is nonzero but not divisible by p, rank(A)=rank(A mod p)
2) when det(A) is zero, rank(A)=rank(A mod p)
If this is correct, how to prove it? Where can I find the relevant references?
Recently I am developing an algorithm where an integer matrix is computed, using modular arithmetic (mod p) can control the growth of integers in the matrix and thus improve the efficiency of my algorithm. The success of the algorithm requires rank of the matrix with mod does not change from the rank of the original matrix. This is where I will apply the problem. If someone can help me, I will really appreciate it.
modular-arithmetic matrix-rank
asked Jan 9 at 17:26
Xiaodong ShiXiaodong Shi
61
$endgroup$
add a comment |
$begingroup$
Suppose I have a matrix A with integer elements and a matrix A mod p, where p is a prime number. In what circumstances rank(A)=rank(A mod p)?
Based on my understanding,
1) when det(A) is nonzero but not divisible by p, rank(A)=rank(A mod p)
2) when det(A) is zero, rank(A)=rank(A mod p)
If this is correct, how to prove it? Where can I find the relevant references?
Recently I am developing an algorithm where an integer matrix is computed, using modular arithmetic (mod p) can control the growth of integers in the matrix and thus improve the efficiency of my algorithm. The success of the algorithm requires rank of the matrix with mod does not change from the rank of the original matrix. This is where I will apply the problem. If someone can help me, I will really appreciate it.
modular-arithmetic matrix-rank
asked Jan 9 at 17:26
Xiaodong ShiXiaodong Shi
61
$endgroup$
$begingroup$
The matrix with rows $(2,0)$ and $(0,0)$ has rank $1$ over the integers, rank $0$ mod $2$, so the second assertion is wrong.
$endgroup$
– Ethan Bolker
Jan 9 at 17:33
$begingroup$
I am comparing the determinants. The matrix with rows (2,0) and (0,0) has rank 1 so it is not full rank; Its determinant is zero which also shows it is not full rank. Now if calculate its determinant mod 2, it is zero, which means det(A)=det(A mod 2)=0. The results coincides.
$endgroup$
– Xiaodong Shi
Jan 10 at 17:27
$begingroup$
Your question says "when det $A$ is $0$, rank $A$ = rank $A pmod{p}$". My example shows that's not true. You never mention "full rank". Your last paragraph doesn't either, so it suggests that your algorithm will fail. The rank is not preserved.
$endgroup$
– Ethan Bolker
Jan 10 at 20:55
add a comment |
$begingroup$
Suppose I have a matrix A with integer elements and a matrix A mod p, where p is a prime number. In what circumstances rank(A)=rank(A mod p)?
Based on my understanding,
1) when det(A) is nonzero but not divisible by p, rank(A)=rank(A mod p)
2) when det(A) is zero, rank(A)=rank(A mod p)
If this is correct, how to prove it? Where can I find the relevant references?
Recently I am developing an algorithm where an integer matrix is computed, using modular arithmetic (mod p) can control the growth of integers in the matrix and thus improve the efficiency of my algorithm. The success of the algorithm requires rank of the matrix with mod does not change from the rank of the original matrix. This is where I will apply the problem. If someone can help me, I will really appreciate it.
modular-arithmetic matrix-rank
asked Jan 9 at 17:26
Xiaodong ShiXiaodong Shi
61
$endgroup$
Suppose I have a matrix A with integer elements and a matrix A mod p, where p is a prime number. In what circumstances rank(A)=rank(A mod p)?
Based on my understanding,
1) when det(A) is nonzero but not divisible by p, rank(A)=rank(A mod p)
2) when det(A) is zero, rank(A)=rank(A mod p)
If this is correct, how to prove it? Where can I find the relevant references?
Recently I am developing an algorithm where an integer matrix is computed, using modular arithmetic (mod p) can control the growth of integers in the matrix and thus improve the efficiency of my algorithm. The success of the algorithm requires rank of the matrix with mod does not change from the rank of the original matrix. This is where I will apply the problem. If someone can help me, I will really appreciate it.
modular-arithmetic matrix-rank
modular-arithmetic matrix-rank
asked Jan 9 at 17:26
Xiaodong ShiXiaodong Shi
61
asked Jan 9 at 17:26
Xiaodong ShiXiaodong Shi
61
asked Jan 9 at 17:26
Xiaodong ShiXiaodong Shi
61
asked Jan 9 at 17:26
Xiaodong ShiXiaodong Shi
61
asked Jan 9 at 17:26
Xiaodong ShiXiaodong Shi
61
61
$begingroup$
The matrix with rows $(2,0)$ and $(0,0)$ has rank $1$ over the integers, rank $0$ mod $2$, so the second assertion is wrong.
$endgroup$
– Ethan Bolker
Jan 9 at 17:33
$begingroup$
I am comparing the determinants. The matrix with rows (2,0) and (0,0) has rank 1 so it is not full rank; Its determinant is zero which also shows it is not full rank. Now if calculate its determinant mod 2, it is zero, which means det(A)=det(A mod 2)=0. The results coincides.
$endgroup$
– Xiaodong Shi
Jan 10 at 17:27
$begingroup$
Your question says "when det $A$ is $0$, rank $A$ = rank $A pmod{p}$". My example shows that's not true. You never mention "full rank". Your last paragraph doesn't either, so it suggests that your algorithm will fail. The rank is not preserved.
$endgroup$
– Ethan Bolker
Jan 10 at 20:55
add a comment |
$begingroup$
The matrix with rows $(2,0)$ and $(0,0)$ has rank $1$ over the integers, rank $0$ mod $2$, so the second assertion is wrong.
$endgroup$
– Ethan Bolker
Jan 9 at 17:33
$begingroup$
I am comparing the determinants. The matrix with rows (2,0) and (0,0) has rank 1 so it is not full rank; Its determinant is zero which also shows it is not full rank. Now if calculate its determinant mod 2, it is zero, which means det(A)=det(A mod 2)=0. The results coincides.
$endgroup$
– Xiaodong Shi
Jan 10 at 17:27
$begingroup$
Your question says "when det $A$ is $0$, rank $A$ = rank $A pmod{p}$". My example shows that's not true. You never mention "full rank". Your last paragraph doesn't either, so it suggests that your algorithm will fail. The rank is not preserved.
$endgroup$
– Ethan Bolker
Jan 10 at 20:55
$begingroup$
The matrix with rows $(2,0)$ and $(0,0)$ has rank $1$ over the integers, rank $0$ mod $2$, so the second assertion is wrong.
$endgroup$
– Ethan Bolker
Jan 9 at 17:33
$begingroup$
The matrix with rows $(2,0)$ and $(0,0)$ has rank $1$ over the integers, rank $0$ mod $2$, so the second assertion is wrong.
$endgroup$
– Ethan Bolker
Jan 9 at 17:33
$begingroup$
I am comparing the determinants. The matrix with rows (2,0) and (0,0) has rank 1 so it is not full rank; Its determinant is zero which also shows it is not full rank. Now if calculate its determinant mod 2, it is zero, which means det(A)=det(A mod 2)=0. The results coincides.
$endgroup$
– Xiaodong Shi
Jan 10 at 17:27
$begingroup$
I am comparing the determinants. The matrix with rows (2,0) and (0,0) has rank 1 so it is not full rank; Its determinant is zero which also shows it is not full rank. Now if calculate its determinant mod 2, it is zero, which means det(A)=det(A mod 2)=0. The results coincides.
$endgroup$
– Xiaodong Shi
Jan 10 at 17:27
$begingroup$
Your question says "when det $A$ is $0$, rank $A$ = rank $A pmod{p}$". My example shows that's not true. You never mention "full rank". Your last paragraph doesn't either, so it suggests that your algorithm will fail. The rank is not preserved.
$endgroup$
– Ethan Bolker
Jan 10 at 20:55
$begingroup$
Your question says "when det $A$ is $0$, rank $A$ = rank $A pmod{p}$". My example shows that's not true. You never mention "full rank". Your last paragraph doesn't either, so it suggests that your algorithm will fail. The rank is not preserved.
$endgroup$
– Ethan Bolker
Jan 10 at 20:55
add a comment |
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$begingroup$
The matrix with rows $(2,0)$ and $(0,0)$ has rank $1$ over the integers, rank $0$ mod $2$, so the second assertion is wrong.
$endgroup$
– Ethan Bolker
Jan 9 at 17:33
$begingroup$
I am comparing the determinants. The matrix with rows (2,0) and (0,0) has rank 1 so it is not full rank; Its determinant is zero which also shows it is not full rank. Now if calculate its determinant mod 2, it is zero, which means det(A)=det(A mod 2)=0. The results coincides.
$endgroup$
– Xiaodong Shi
Jan 10 at 17:27
$begingroup$
Your question says "when det $A$ is $0$, rank $A$ = rank $A pmod{p}$". My example shows that's not true. You never mention "full rank". Your last paragraph doesn't either, so it suggests that your algorithm will fail. The rank is not preserved.
$endgroup$
– Ethan Bolker
Jan 10 at 20:55