Find a Matrix A on the ring of integers modulo 3 so that KerA=ImB.
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B={{1,1,1},{0,1,2},{2,1,0},{0,2,2}}
I understand that each vector from then span of column vectors of B is a solution for Ax=o and that matrix A should have four columns. However I don't know how many rows it should have and how to find it.
linear-algebra matrices modular-arithmetic
asked Nov 17 at 10:02
Oleksandr
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up vote
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B={{1,1,1},{0,1,2},{2,1,0},{0,2,2}}
I understand that each vector from then span of column vectors of B is a solution for Ax=o and that matrix A should have four columns. However I don't know how many rows it should have and how to find it.
linear-algebra matrices modular-arithmetic
asked Nov 17 at 10:02
Oleksandr
362
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
B={{1,1,1},{0,1,2},{2,1,0},{0,2,2}}
I understand that each vector from then span of column vectors of B is a solution for Ax=o and that matrix A should have four columns. However I don't know how many rows it should have and how to find it.
linear-algebra matrices modular-arithmetic
asked Nov 17 at 10:02
Oleksandr
362
B={{1,1,1},{0,1,2},{2,1,0},{0,2,2}}
I understand that each vector from then span of column vectors of B is a solution for Ax=o and that matrix A should have four columns. However I don't know how many rows it should have and how to find it.
linear-algebra matrices modular-arithmetic
linear-algebra matrices modular-arithmetic
asked Nov 17 at 10:02
Oleksandr
362
asked Nov 17 at 10:02
Oleksandr
362
asked Nov 17 at 10:02
Oleksandr
362
asked Nov 17 at 10:02
Oleksandr
362
asked Nov 17 at 10:02
Oleksandr
362
362
add a comment |
add a comment |
1 Answer
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Suppose $A$ has $r$ rows. As $mathbb Z_3$ is a field, by rank-nullity theorem,
$$
4 =text{dimension of domain of $A$}=operatorname{rank}(A)+operatorname{nullity}(A)=operatorname{rank}(A)+operatorname{rank}(B).
$$
Hence $r:=operatorname{rank}(A)=4-operatorname{rank}(B)$ and $A$ must have at least $r$ rows.
answered Nov 17 at 10:53
user1551
70.2k566125
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Suppose $A$ has $r$ rows. As $mathbb Z_3$ is a field, by rank-nullity theorem,
$$
4 =text{dimension of domain of $A$}=operatorname{rank}(A)+operatorname{nullity}(A)=operatorname{rank}(A)+operatorname{rank}(B).
$$
Hence $r:=operatorname{rank}(A)=4-operatorname{rank}(B)$ and $A$ must have at least $r$ rows.
answered Nov 17 at 10:53
user1551
70.2k566125
add a comment |
up vote
1
down vote
Suppose $A$ has $r$ rows. As $mathbb Z_3$ is a field, by rank-nullity theorem,
$$
4 =text{dimension of domain of $A$}=operatorname{rank}(A)+operatorname{nullity}(A)=operatorname{rank}(A)+operatorname{rank}(B).
$$
Hence $r:=operatorname{rank}(A)=4-operatorname{rank}(B)$ and $A$ must have at least $r$ rows.
answered Nov 17 at 10:53
user1551
70.2k566125
add a comment |
up vote
1
down vote
up vote
1
down vote
Suppose $A$ has $r$ rows. As $mathbb Z_3$ is a field, by rank-nullity theorem,
$$
4 =text{dimension of domain of $A$}=operatorname{rank}(A)+operatorname{nullity}(A)=operatorname{rank}(A)+operatorname{rank}(B).
$$
Hence $r:=operatorname{rank}(A)=4-operatorname{rank}(B)$ and $A$ must have at least $r$ rows.
answered Nov 17 at 10:53
user1551
70.2k566125
Suppose $A$ has $r$ rows. As $mathbb Z_3$ is a field, by rank-nullity theorem,
$$
4 =text{dimension of domain of $A$}=operatorname{rank}(A)+operatorname{nullity}(A)=operatorname{rank}(A)+operatorname{rank}(B).
$$
Hence $r:=operatorname{rank}(A)=4-operatorname{rank}(B)$ and $A$ must have at least $r$ rows.
answered Nov 17 at 10:53
user1551
70.2k566125
answered Nov 17 at 10:53
user1551
70.2k566125
answered Nov 17 at 10:53
user1551
70.2k566125
answered Nov 17 at 10:53
user1551
70.2k566125
70.2k566125
add a comment |
add a comment |
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