$AB=0$ solving for matrices in MatLab
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I have the equation $AB = 0$, where $A$ and $B$ are matrices. And I know a matrix $B$ and want to find a matrix $A$.
How can I do it in MatLab?
matrices matlab matrix-equations
add a comment |
up vote
2
down vote
favorite
I have the equation $AB = 0$, where $A$ and $B$ are matrices. And I know a matrix $B$ and want to find a matrix $A$.
How can I do it in MatLab?
matrices matlab matrix-equations
Are there any hypothesis on $B$ ?
– nicomezi
Nov 13 at 11:16
I know only that matrix $A$ may be in view as $A=(I_k P)$
– Multifora
Nov 13 at 11:23
Typehelp null
into the command window.
– user1551
Nov 13 at 13:09
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have the equation $AB = 0$, where $A$ and $B$ are matrices. And I know a matrix $B$ and want to find a matrix $A$.
How can I do it in MatLab?
matrices matlab matrix-equations
I have the equation $AB = 0$, where $A$ and $B$ are matrices. And I know a matrix $B$ and want to find a matrix $A$.
How can I do it in MatLab?
matrices matlab matrix-equations
matrices matlab matrix-equations
edited Nov 13 at 11:13
amWhy
191k27223437
191k27223437
asked Nov 13 at 11:06
Multifora
162
162
Are there any hypothesis on $B$ ?
– nicomezi
Nov 13 at 11:16
I know only that matrix $A$ may be in view as $A=(I_k P)$
– Multifora
Nov 13 at 11:23
Typehelp null
into the command window.
– user1551
Nov 13 at 13:09
add a comment |
Are there any hypothesis on $B$ ?
– nicomezi
Nov 13 at 11:16
I know only that matrix $A$ may be in view as $A=(I_k P)$
– Multifora
Nov 13 at 11:23
Typehelp null
into the command window.
– user1551
Nov 13 at 13:09
Are there any hypothesis on $B$ ?
– nicomezi
Nov 13 at 11:16
Are there any hypothesis on $B$ ?
– nicomezi
Nov 13 at 11:16
I know only that matrix $A$ may be in view as $A=(I_k P)$
– Multifora
Nov 13 at 11:23
I know only that matrix $A$ may be in view as $A=(I_k P)$
– Multifora
Nov 13 at 11:23
Type
help null
into the command window.– user1551
Nov 13 at 13:09
Type
help null
into the command window.– user1551
Nov 13 at 13:09
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
Let us assume that all matrices involved are square ($n times n$).
You have first to convert your issue by transposition :
$B^T A^T = 0$
which amounts to say :
$$B^T [V_1 |V_2|...|V_n] = [0 | 0 | ... | 0] iff forall i, B^T V_i=0$$
which means that all $V_i$ should belong to the kernel of $B^T$, giving a manner to build such matrices.
Particular cases :
a) The kernel of $B^T$ is reduced to ${0}$ : only the null matrix is possible for $A$.
b) The kernel of $B^T$ has dimension 1, with basis $V$, then $A^T=[a_1V|a_2V|...|a_nV]$.
Let us take an example for case (b):
B=[1 2 3
2 2 4
3 5 8];% matrix with rank<3, here rank(B)=2, thus dim(ker(B'))=1
K=null(B'); % the columns of K are a basis of the Kernel
V=K(:,1); % in fact, in this case, V and K are identical
a=rand(1,3);
A=[a(1)*V,a(2)*V,a(3)*V]';% note the transposition operator
A*B,%test : must be zero
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Let us assume that all matrices involved are square ($n times n$).
You have first to convert your issue by transposition :
$B^T A^T = 0$
which amounts to say :
$$B^T [V_1 |V_2|...|V_n] = [0 | 0 | ... | 0] iff forall i, B^T V_i=0$$
which means that all $V_i$ should belong to the kernel of $B^T$, giving a manner to build such matrices.
Particular cases :
a) The kernel of $B^T$ is reduced to ${0}$ : only the null matrix is possible for $A$.
b) The kernel of $B^T$ has dimension 1, with basis $V$, then $A^T=[a_1V|a_2V|...|a_nV]$.
Let us take an example for case (b):
B=[1 2 3
2 2 4
3 5 8];% matrix with rank<3, here rank(B)=2, thus dim(ker(B'))=1
K=null(B'); % the columns of K are a basis of the Kernel
V=K(:,1); % in fact, in this case, V and K are identical
a=rand(1,3);
A=[a(1)*V,a(2)*V,a(3)*V]';% note the transposition operator
A*B,%test : must be zero
add a comment |
up vote
0
down vote
Let us assume that all matrices involved are square ($n times n$).
You have first to convert your issue by transposition :
$B^T A^T = 0$
which amounts to say :
$$B^T [V_1 |V_2|...|V_n] = [0 | 0 | ... | 0] iff forall i, B^T V_i=0$$
which means that all $V_i$ should belong to the kernel of $B^T$, giving a manner to build such matrices.
Particular cases :
a) The kernel of $B^T$ is reduced to ${0}$ : only the null matrix is possible for $A$.
b) The kernel of $B^T$ has dimension 1, with basis $V$, then $A^T=[a_1V|a_2V|...|a_nV]$.
Let us take an example for case (b):
B=[1 2 3
2 2 4
3 5 8];% matrix with rank<3, here rank(B)=2, thus dim(ker(B'))=1
K=null(B'); % the columns of K are a basis of the Kernel
V=K(:,1); % in fact, in this case, V and K are identical
a=rand(1,3);
A=[a(1)*V,a(2)*V,a(3)*V]';% note the transposition operator
A*B,%test : must be zero
add a comment |
up vote
0
down vote
up vote
0
down vote
Let us assume that all matrices involved are square ($n times n$).
You have first to convert your issue by transposition :
$B^T A^T = 0$
which amounts to say :
$$B^T [V_1 |V_2|...|V_n] = [0 | 0 | ... | 0] iff forall i, B^T V_i=0$$
which means that all $V_i$ should belong to the kernel of $B^T$, giving a manner to build such matrices.
Particular cases :
a) The kernel of $B^T$ is reduced to ${0}$ : only the null matrix is possible for $A$.
b) The kernel of $B^T$ has dimension 1, with basis $V$, then $A^T=[a_1V|a_2V|...|a_nV]$.
Let us take an example for case (b):
B=[1 2 3
2 2 4
3 5 8];% matrix with rank<3, here rank(B)=2, thus dim(ker(B'))=1
K=null(B'); % the columns of K are a basis of the Kernel
V=K(:,1); % in fact, in this case, V and K are identical
a=rand(1,3);
A=[a(1)*V,a(2)*V,a(3)*V]';% note the transposition operator
A*B,%test : must be zero
Let us assume that all matrices involved are square ($n times n$).
You have first to convert your issue by transposition :
$B^T A^T = 0$
which amounts to say :
$$B^T [V_1 |V_2|...|V_n] = [0 | 0 | ... | 0] iff forall i, B^T V_i=0$$
which means that all $V_i$ should belong to the kernel of $B^T$, giving a manner to build such matrices.
Particular cases :
a) The kernel of $B^T$ is reduced to ${0}$ : only the null matrix is possible for $A$.
b) The kernel of $B^T$ has dimension 1, with basis $V$, then $A^T=[a_1V|a_2V|...|a_nV]$.
Let us take an example for case (b):
B=[1 2 3
2 2 4
3 5 8];% matrix with rank<3, here rank(B)=2, thus dim(ker(B'))=1
K=null(B'); % the columns of K are a basis of the Kernel
V=K(:,1); % in fact, in this case, V and K are identical
a=rand(1,3);
A=[a(1)*V,a(2)*V,a(3)*V]';% note the transposition operator
A*B,%test : must be zero
edited Nov 15 at 13:19
answered Nov 15 at 13:08
Jean Marie
28k41848
28k41848
add a comment |
add a comment |
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Are there any hypothesis on $B$ ?
– nicomezi
Nov 13 at 11:16
I know only that matrix $A$ may be in view as $A=(I_k P)$
– Multifora
Nov 13 at 11:23
Type
help null
into the command window.– user1551
Nov 13 at 13:09