Mixing
Prove that exist a matrix $X in mathbb C^{m,n}$ for which $B=AX$
$begingroup$
There are a matrices $A in mathbb C^{m,m}$ and $B in mathbb C^{m,n}$ for which $rank[A | B]=rank A$. Prove that exist a matrix $X in mathbb C^{m,n}$ for which $B=AX$.
I know that probably this task is really easy because the thesis seems quite obvious. However I really try to do it for a long time and I still can not prove it.
Can you get some tips?
linear-algebra
asked Feb 2 at 12:56
MP3129MP3129
829211
$endgroup$
add a comment |
$begingroup$
There are a matrices $A in mathbb C^{m,m}$ and $B in mathbb C^{m,n}$ for which $rank[A | B]=rank A$. Prove that exist a matrix $X in mathbb C^{m,n}$ for which $B=AX$.
I know that probably this task is really easy because the thesis seems quite obvious. However I really try to do it for a long time and I still can not prove it.
Can you get some tips?
linear-algebra
asked Feb 2 at 12:56
MP3129MP3129
829211
$endgroup$
$begingroup$
What have you tried?
$endgroup$
– Christoph
Feb 2 at 13:07
$begingroup$
@Christoph If I have $rank[A | B]=rank A$ then I know that where I have the zero rows of matrix $A$ there are also zeroes of the matrix $B$ and I think that vectors in lines of matrix $A$ and $B$ are linearly dependent, so I have $B=AX$ but this proof it's not good and I do not know how to prove it in a better way.
$endgroup$
– MP3129
Feb 2 at 13:15
1
$begingroup$
You claim "so I have $B=AX$" without any justification, so this is not a proof at all. See my answer for an approach.
$endgroup$
– Christoph
Feb 2 at 13:24
add a comment |
$begingroup$
There are a matrices $A in mathbb C^{m,m}$ and $B in mathbb C^{m,n}$ for which $rank[A | B]=rank A$. Prove that exist a matrix $X in mathbb C^{m,n}$ for which $B=AX$.
I know that probably this task is really easy because the thesis seems quite obvious. However I really try to do it for a long time and I still can not prove it.
Can you get some tips?
linear-algebra
asked Feb 2 at 12:56
MP3129MP3129
829211
$endgroup$
There are a matrices $A in mathbb C^{m,m}$ and $B in mathbb C^{m,n}$ for which $rank[A | B]=rank A$. Prove that exist a matrix $X in mathbb C^{m,n}$ for which $B=AX$.
I know that probably this task is really easy because the thesis seems quite obvious. However I really try to do it for a long time and I still can not prove it.
Can you get some tips?
linear-algebra
linear-algebra
asked Feb 2 at 12:56
MP3129MP3129
829211
asked Feb 2 at 12:56
MP3129MP3129
829211
asked Feb 2 at 12:56
MP3129MP3129
829211
asked Feb 2 at 12:56
MP3129MP3129
829211
asked Feb 2 at 12:56
MP3129MP3129
829211
829211
$begingroup$
What have you tried?
$endgroup$
– Christoph
Feb 2 at 13:07
$begingroup$
@Christoph If I have $rank[A | B]=rank A$ then I know that where I have the zero rows of matrix $A$ there are also zeroes of the matrix $B$ and I think that vectors in lines of matrix $A$ and $B$ are linearly dependent, so I have $B=AX$ but this proof it's not good and I do not know how to prove it in a better way.
$endgroup$
– MP3129
Feb 2 at 13:15
1
$begingroup$
You claim "so I have $B=AX$" without any justification, so this is not a proof at all. See my answer for an approach.
$endgroup$
– Christoph
Feb 2 at 13:24
add a comment |
$begingroup$
What have you tried?
$endgroup$
– Christoph
Feb 2 at 13:07
$begingroup$
@Christoph If I have $rank[A | B]=rank A$ then I know that where I have the zero rows of matrix $A$ there are also zeroes of the matrix $B$ and I think that vectors in lines of matrix $A$ and $B$ are linearly dependent, so I have $B=AX$ but this proof it's not good and I do not know how to prove it in a better way.
$endgroup$
– MP3129
Feb 2 at 13:15
1
$begingroup$
You claim "so I have $B=AX$" without any justification, so this is not a proof at all. See my answer for an approach.
$endgroup$
– Christoph
Feb 2 at 13:24
$begingroup$
What have you tried?
$endgroup$
– Christoph
Feb 2 at 13:07
$begingroup$
What have you tried?
$endgroup$
– Christoph
Feb 2 at 13:07
$begingroup$
@Christoph If I have $rank[A | B]=rank A$ then I know that where I have the zero rows of matrix $A$ there are also zeroes of the matrix $B$ and I think that vectors in lines of matrix $A$ and $B$ are linearly dependent, so I have $B=AX$ but this proof it's not good and I do not know how to prove it in a better way.
$endgroup$
– MP3129
Feb 2 at 13:15
$begingroup$
@Christoph If I have $rank[A | B]=rank A$ then I know that where I have the zero rows of matrix $A$ there are also zeroes of the matrix $B$ and I think that vectors in lines of matrix $A$ and $B$ are linearly dependent, so I have $B=AX$ but this proof it's not good and I do not know how to prove it in a better way.
$endgroup$
– MP3129
Feb 2 at 13:15
1
1
$begingroup$
You claim "so I have $B=AX$" without any justification, so this is not a proof at all. See my answer for an approach.
$endgroup$
– Christoph
Feb 2 at 13:24
$begingroup$
You claim "so I have $B=AX$" without any justification, so this is not a proof at all. See my answer for an approach.
$endgroup$
– Christoph
Feb 2 at 13:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: The rank of a matrix $Min K^{mtimes n}$ with columns $M_1,dots,M_nin K^m$ is the dimension of $langle M_1,dots,M_n rangle$ (the subspace of $K^m$ spanned by $M_1,dots,M_n$). Since $operatorname{rank}(A) = operatorname{rank}(A|B)$ you have
$$
langle A_1,dots,A_m rangle = langle A_1,dots,A_m,B_1,dots,B_nrangle.
$$
Can you take it from here?
answered Feb 2 at 13:23
ChristophChristoph
12.5k1642
$endgroup$
$begingroup$
So columns $B$ are linearly dependent on $A$ but I don't know what I can do the next
$endgroup$
– MP3129
Feb 2 at 21:56
1
$begingroup$
Write each $B_i$ as a linear combination of the $A_i$. Try to combine all these equations into a matrix identity.
$endgroup$
– Christoph
Feb 2 at 23:05
$begingroup$
Ok - Do you mean that I should create linear transformation and find matrix of transformation? Ok, that works bot I am not sure if I can do just like that
$endgroup$
– MP3129
Feb 2 at 23:13
1
$begingroup$
No, I don't mean that. I mean exactly what I said. There are real number $x_{ij}$ such that $B_i = sum_{j=1}^m x_{ji} A_j$. Letting $X=(x_{ij})$ yields $B = AX$.
$endgroup$
– Christoph
Feb 2 at 23:18
$begingroup$
Okay, I mean about something similar - thanks for help!
$endgroup$
– MP3129
Feb 2 at 23:21
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
Hint: The rank of a matrix $Min K^{mtimes n}$ with columns $M_1,dots,M_nin K^m$ is the dimension of $langle M_1,dots,M_n rangle$ (the subspace of $K^m$ spanned by $M_1,dots,M_n$). Since $operatorname{rank}(A) = operatorname{rank}(A|B)$ you have
$$
langle A_1,dots,A_m rangle = langle A_1,dots,A_m,B_1,dots,B_nrangle.
$$
Can you take it from here?
answered Feb 2 at 13:23
ChristophChristoph
12.5k1642
$endgroup$
$begingroup$
So columns $B$ are linearly dependent on $A$ but I don't know what I can do the next
$endgroup$
– MP3129
Feb 2 at 21:56
1
$begingroup$
Write each $B_i$ as a linear combination of the $A_i$. Try to combine all these equations into a matrix identity.
$endgroup$
– Christoph
Feb 2 at 23:05
$begingroup$
Ok - Do you mean that I should create linear transformation and find matrix of transformation? Ok, that works bot I am not sure if I can do just like that
$endgroup$
– MP3129
Feb 2 at 23:13
1
$begingroup$
No, I don't mean that. I mean exactly what I said. There are real number $x_{ij}$ such that $B_i = sum_{j=1}^m x_{ji} A_j$. Letting $X=(x_{ij})$ yields $B = AX$.
$endgroup$
– Christoph
Feb 2 at 23:18
$begingroup$
Okay, I mean about something similar - thanks for help!
$endgroup$
– MP3129
Feb 2 at 23:21
add a comment |
$begingroup$
Hint: The rank of a matrix $Min K^{mtimes n}$ with columns $M_1,dots,M_nin K^m$ is the dimension of $langle M_1,dots,M_n rangle$ (the subspace of $K^m$ spanned by $M_1,dots,M_n$). Since $operatorname{rank}(A) = operatorname{rank}(A|B)$ you have
$$
langle A_1,dots,A_m rangle = langle A_1,dots,A_m,B_1,dots,B_nrangle.
$$
Can you take it from here?
answered Feb 2 at 13:23
ChristophChristoph
12.5k1642
$endgroup$
$begingroup$
So columns $B$ are linearly dependent on $A$ but I don't know what I can do the next
$endgroup$
– MP3129
Feb 2 at 21:56
1
$begingroup$
Write each $B_i$ as a linear combination of the $A_i$. Try to combine all these equations into a matrix identity.
$endgroup$
– Christoph
Feb 2 at 23:05
$begingroup$
Ok - Do you mean that I should create linear transformation and find matrix of transformation? Ok, that works bot I am not sure if I can do just like that
$endgroup$
– MP3129
Feb 2 at 23:13
1
$begingroup$
No, I don't mean that. I mean exactly what I said. There are real number $x_{ij}$ such that $B_i = sum_{j=1}^m x_{ji} A_j$. Letting $X=(x_{ij})$ yields $B = AX$.
$endgroup$
– Christoph
Feb 2 at 23:18
$begingroup$
Okay, I mean about something similar - thanks for help!
$endgroup$
– MP3129
Feb 2 at 23:21
add a comment |
$begingroup$
Hint: The rank of a matrix $Min K^{mtimes n}$ with columns $M_1,dots,M_nin K^m$ is the dimension of $langle M_1,dots,M_n rangle$ (the subspace of $K^m$ spanned by $M_1,dots,M_n$). Since $operatorname{rank}(A) = operatorname{rank}(A|B)$ you have
$$
langle A_1,dots,A_m rangle = langle A_1,dots,A_m,B_1,dots,B_nrangle.
$$
Can you take it from here?
answered Feb 2 at 13:23
ChristophChristoph
12.5k1642
$endgroup$
Hint: The rank of a matrix $Min K^{mtimes n}$ with columns $M_1,dots,M_nin K^m$ is the dimension of $langle M_1,dots,M_n rangle$ (the subspace of $K^m$ spanned by $M_1,dots,M_n$). Since $operatorname{rank}(A) = operatorname{rank}(A|B)$ you have
$$
langle A_1,dots,A_m rangle = langle A_1,dots,A_m,B_1,dots,B_nrangle.
$$
Can you take it from here?
answered Feb 2 at 13:23
ChristophChristoph
12.5k1642
answered Feb 2 at 13:23
ChristophChristoph
12.5k1642
answered Feb 2 at 13:23
ChristophChristoph
12.5k1642
answered Feb 2 at 13:23
ChristophChristoph
12.5k1642
12.5k1642
$begingroup$
So columns $B$ are linearly dependent on $A$ but I don't know what I can do the next
$endgroup$
– MP3129
Feb 2 at 21:56
1
$begingroup$
Write each $B_i$ as a linear combination of the $A_i$. Try to combine all these equations into a matrix identity.
$endgroup$
– Christoph
Feb 2 at 23:05
$begingroup$
Ok - Do you mean that I should create linear transformation and find matrix of transformation? Ok, that works bot I am not sure if I can do just like that
$endgroup$
– MP3129
Feb 2 at 23:13
1
$begingroup$
No, I don't mean that. I mean exactly what I said. There are real number $x_{ij}$ such that $B_i = sum_{j=1}^m x_{ji} A_j$. Letting $X=(x_{ij})$ yields $B = AX$.
$endgroup$
– Christoph
Feb 2 at 23:18
$begingroup$
Okay, I mean about something similar - thanks for help!
$endgroup$
– MP3129
Feb 2 at 23:21
add a comment |
$begingroup$
So columns $B$ are linearly dependent on $A$ but I don't know what I can do the next
$endgroup$
– MP3129
Feb 2 at 21:56
1
$begingroup$
Write each $B_i$ as a linear combination of the $A_i$. Try to combine all these equations into a matrix identity.
$endgroup$
– Christoph
Feb 2 at 23:05
$begingroup$
Ok - Do you mean that I should create linear transformation and find matrix of transformation? Ok, that works bot I am not sure if I can do just like that
$endgroup$
– MP3129
Feb 2 at 23:13
1
$begingroup$
No, I don't mean that. I mean exactly what I said. There are real number $x_{ij}$ such that $B_i = sum_{j=1}^m x_{ji} A_j$. Letting $X=(x_{ij})$ yields $B = AX$.
$endgroup$
– Christoph
Feb 2 at 23:18
$begingroup$
Okay, I mean about something similar - thanks for help!
$endgroup$
– MP3129
Feb 2 at 23:21
$begingroup$
So columns $B$ are linearly dependent on $A$ but I don't know what I can do the next
$endgroup$
– MP3129
Feb 2 at 21:56
$begingroup$
So columns $B$ are linearly dependent on $A$ but I don't know what I can do the next
$endgroup$
– MP3129
Feb 2 at 21:56
1
1
$begingroup$
Write each $B_i$ as a linear combination of the $A_i$. Try to combine all these equations into a matrix identity.
$endgroup$
– Christoph
Feb 2 at 23:05
$begingroup$
Write each $B_i$ as a linear combination of the $A_i$. Try to combine all these equations into a matrix identity.
$endgroup$
– Christoph
Feb 2 at 23:05
$begingroup$
Ok - Do you mean that I should create linear transformation and find matrix of transformation? Ok, that works bot I am not sure if I can do just like that
$endgroup$
– MP3129
Feb 2 at 23:13
$begingroup$
Ok - Do you mean that I should create linear transformation and find matrix of transformation? Ok, that works bot I am not sure if I can do just like that
$endgroup$
– MP3129
Feb 2 at 23:13
1
1
$begingroup$
No, I don't mean that. I mean exactly what I said. There are real number $x_{ij}$ such that $B_i = sum_{j=1}^m x_{ji} A_j$. Letting $X=(x_{ij})$ yields $B = AX$.
$endgroup$
– Christoph
Feb 2 at 23:18
$begingroup$
No, I don't mean that. I mean exactly what I said. There are real number $x_{ij}$ such that $B_i = sum_{j=1}^m x_{ji} A_j$. Letting $X=(x_{ij})$ yields $B = AX$.
$endgroup$
– Christoph
Feb 2 at 23:18
$begingroup$
Okay, I mean about something similar - thanks for help!
$endgroup$
– MP3129
Feb 2 at 23:21
$begingroup$
Okay, I mean about something similar - thanks for help!
$endgroup$
– MP3129
Feb 2 at 23:21
add a comment |
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$begingroup$
What have you tried?
$endgroup$
– Christoph
Feb 2 at 13:07
$begingroup$
@Christoph If I have $rank[A | B]=rank A$ then I know that where I have the zero rows of matrix $A$ there are also zeroes of the matrix $B$ and I think that vectors in lines of matrix $A$ and $B$ are linearly dependent, so I have $B=AX$ but this proof it's not good and I do not know how to prove it in a better way.
$endgroup$
– MP3129
Feb 2 at 13:15
1
$begingroup$
You claim "so I have $B=AX$" without any justification, so this is not a proof at all. See my answer for an approach.
$endgroup$
– Christoph
Feb 2 at 13:24