Mixing
I have the following matrix B and I need to find matrix A so that kerA=ImB and A has as few rows as possible
Every thing is considered in Z_3 (modulo 3).
B =begin{bmatrix}1&1&1\0&1&2\2&1&0\0&2&2end{bmatrix}
I think that the correct path is to show that ImB ⊆ KerA which is when AB=0 and KerA⊆ImB at the same time implies KerA=ImB. But I don't know when that is true. I also need to prove that A cannot have fewer rows.
linear-algebra
asked Nov 20 '18 at 21:45
Michal Rickwodo
62
add a comment |
Every thing is considered in Z_3 (modulo 3).
B =begin{bmatrix}1&1&1\0&1&2\2&1&0\0&2&2end{bmatrix}
I think that the correct path is to show that ImB ⊆ KerA which is when AB=0 and KerA⊆ImB at the same time implies KerA=ImB. But I don't know when that is true. I also need to prove that A cannot have fewer rows.
linear-algebra
asked Nov 20 '18 at 21:45
Michal Rickwodo
62
add a comment |
Every thing is considered in Z_3 (modulo 3).
B =begin{bmatrix}1&1&1\0&1&2\2&1&0\0&2&2end{bmatrix}
I think that the correct path is to show that ImB ⊆ KerA which is when AB=0 and KerA⊆ImB at the same time implies KerA=ImB. But I don't know when that is true. I also need to prove that A cannot have fewer rows.
linear-algebra
asked Nov 20 '18 at 21:45
Michal Rickwodo
62
Every thing is considered in Z_3 (modulo 3).
B =begin{bmatrix}1&1&1\0&1&2\2&1&0\0&2&2end{bmatrix}
I think that the correct path is to show that ImB ⊆ KerA which is when AB=0 and KerA⊆ImB at the same time implies KerA=ImB. But I don't know when that is true. I also need to prove that A cannot have fewer rows.
linear-algebra
linear-algebra
asked Nov 20 '18 at 21:45
Michal Rickwodo
62
asked Nov 20 '18 at 21:45
Michal Rickwodo
62
asked Nov 20 '18 at 21:45
Michal Rickwodo
62
asked Nov 20 '18 at 21:45
Michal Rickwodo
62
asked Nov 20 '18 at 21:45
Michal Rickwodo
62
62
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
The columns of $B $ are linearly independent so the dimension of the image of $B $ is $3$. $A $ has to have $4$ columns for multiplication to work. Hence by a theorem we have that the image of $A $ has dimension $1$. So $A $ can be taken as a $1times 4$ matrix. This row vector can be found by finding the orthogonal complement of image of $B $ in $mathbb {R}^4$.
One way to do this is to solve for $A=(x_1,x_2,x_3,x_4) $ so that $AB=0$. Equivalently $B^TA^T=0$.
answered Nov 20 '18 at 22:03
AnyAD
2,098812
Makes sence. What is the theorem by which I know that A has dimension 1?
– Michal Rickwodo
Nov 20 '18 at 22:10
Rank-nullity theorem
– AnyAD
Nov 20 '18 at 22:12
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
votes
active
oldest
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active
oldest
votes
The columns of $B $ are linearly independent so the dimension of the image of $B $ is $3$. $A $ has to have $4$ columns for multiplication to work. Hence by a theorem we have that the image of $A $ has dimension $1$. So $A $ can be taken as a $1times 4$ matrix. This row vector can be found by finding the orthogonal complement of image of $B $ in $mathbb {R}^4$.
One way to do this is to solve for $A=(x_1,x_2,x_3,x_4) $ so that $AB=0$. Equivalently $B^TA^T=0$.
answered Nov 20 '18 at 22:03
AnyAD
2,098812
Makes sence. What is the theorem by which I know that A has dimension 1?
– Michal Rickwodo
Nov 20 '18 at 22:10
Rank-nullity theorem
– AnyAD
Nov 20 '18 at 22:12
add a comment |
The columns of $B $ are linearly independent so the dimension of the image of $B $ is $3$. $A $ has to have $4$ columns for multiplication to work. Hence by a theorem we have that the image of $A $ has dimension $1$. So $A $ can be taken as a $1times 4$ matrix. This row vector can be found by finding the orthogonal complement of image of $B $ in $mathbb {R}^4$.
One way to do this is to solve for $A=(x_1,x_2,x_3,x_4) $ so that $AB=0$. Equivalently $B^TA^T=0$.
answered Nov 20 '18 at 22:03
AnyAD
2,098812
Makes sence. What is the theorem by which I know that A has dimension 1?
– Michal Rickwodo
Nov 20 '18 at 22:10
Rank-nullity theorem
– AnyAD
Nov 20 '18 at 22:12
add a comment |
The columns of $B $ are linearly independent so the dimension of the image of $B $ is $3$. $A $ has to have $4$ columns for multiplication to work. Hence by a theorem we have that the image of $A $ has dimension $1$. So $A $ can be taken as a $1times 4$ matrix. This row vector can be found by finding the orthogonal complement of image of $B $ in $mathbb {R}^4$.
One way to do this is to solve for $A=(x_1,x_2,x_3,x_4) $ so that $AB=0$. Equivalently $B^TA^T=0$.
answered Nov 20 '18 at 22:03
AnyAD
2,098812
The columns of $B $ are linearly independent so the dimension of the image of $B $ is $3$. $A $ has to have $4$ columns for multiplication to work. Hence by a theorem we have that the image of $A $ has dimension $1$. So $A $ can be taken as a $1times 4$ matrix. This row vector can be found by finding the orthogonal complement of image of $B $ in $mathbb {R}^4$.
One way to do this is to solve for $A=(x_1,x_2,x_3,x_4) $ so that $AB=0$. Equivalently $B^TA^T=0$.
answered Nov 20 '18 at 22:03
AnyAD
2,098812
edited Nov 20 '18 at 22:08
answered Nov 20 '18 at 22:03
AnyAD
2,098812
answered Nov 20 '18 at 22:03
AnyAD
2,098812
answered Nov 20 '18 at 22:03
AnyAD
2,098812
2,098812
Makes sence. What is the theorem by which I know that A has dimension 1?
– Michal Rickwodo
Nov 20 '18 at 22:10
Rank-nullity theorem
– AnyAD
Nov 20 '18 at 22:12
add a comment |
Makes sence. What is the theorem by which I know that A has dimension 1?
– Michal Rickwodo
Nov 20 '18 at 22:10
Rank-nullity theorem
– AnyAD
Nov 20 '18 at 22:12
Makes sence. What is the theorem by which I know that A has dimension 1?
– Michal Rickwodo
Nov 20 '18 at 22:10
Makes sence. What is the theorem by which I know that A has dimension 1?
– Michal Rickwodo
Nov 20 '18 at 22:10
Rank-nullity theorem
– AnyAD
Nov 20 '18 at 22:12
Rank-nullity theorem
– AnyAD
Nov 20 '18 at 22:12
add a comment |
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