Find a Basis for $V$ if we know a basis for $V^perp$ The Next CEO of Stack Overflowfinding a basis for $W^perp$ and understanding it.Find the dimension of a subspace by find a basis for the null space.Finding a Basis for S$^perp$Find the Change of Basis MatrixHow to find a basis of $W$ given the span of $W^perp$?Find a basis for $S^⊥$Homework help - Find a basis where 4 vectors = 0Find a basis and dimension of the subspace R^n spanned by the following setsDetermine a basis for $U^perp.$Why is $W^perp$ the null space of the basis of $W$?
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Find a Basis for $V$ if we know a basis for $V^perp$
The Next CEO of Stack Overflowfinding a basis for $W^perp$ and understanding it.Find the dimension of a subspace by find a basis for the null space.Finding a Basis for S$^perp$Find the Change of Basis MatrixHow to find a basis of $W$ given the span of $W^perp$?Find a basis for $S^⊥$Homework help - Find a basis where 4 vectors = 0Find a basis and dimension of the subspace R^n spanned by the following setsDetermine a basis for $U^perp.$Why is $W^perp$ the null space of the basis of $W$?
$begingroup$
I tried to find a basis for a subspace $V$, but i know that: $(1,2,-1,0),(0,2,1,3)$ is basis of $V^perp $.
For this, i have the following theorem: Let $A$ a matrix, then we have:
$1.$ $Row(A)^perp$ $=$ $Null(A)$
So, my arguement is this: $Null(A) = Row(A)^perp$, and $Row(A) = V$, we have that $Null(A)^perp = (Row(A)^perp)^perp)$ but $(Row(A)^perp)^perp) = Row(A) = V$, so if we apply this theorem to $Null(A)^perp$ we must be find a basis for $V$, but it doesn't works.
I hope you can help me. Thanks.
linear-algebra
asked Mar 28 at 1:07
Mateo Soto ArangoMateo Soto Arango
114
$endgroup$
add a comment |
$begingroup$
I tried to find a basis for a subspace $V$, but i know that: $(1,2,-1,0),(0,2,1,3)$ is basis of $V^perp $.
For this, i have the following theorem: Let $A$ a matrix, then we have:
$1.$ $Row(A)^perp$ $=$ $Null(A)$
So, my arguement is this: $Null(A) = Row(A)^perp$, and $Row(A) = V$, we have that $Null(A)^perp = (Row(A)^perp)^perp)$ but $(Row(A)^perp)^perp) = Row(A) = V$, so if we apply this theorem to $Null(A)^perp$ we must be find a basis for $V$, but it doesn't works.
I hope you can help me. Thanks.
linear-algebra
asked Mar 28 at 1:07
Mateo Soto ArangoMateo Soto Arango
114
$endgroup$
$begingroup$
What if $Row(A)=V^perp$?
$endgroup$
– amd
Mar 28 at 1:12
$begingroup$
$Row(A)$ is is the subspace generated by the file vectors of A, in spanish is $Fil(A)$ the notation is wrong?
$endgroup$
– Mateo Soto Arango
Mar 28 at 1:20
$begingroup$
You asked for help. That was a hint. Try your reasoning again starting from that instead.
$endgroup$
– amd
Mar 28 at 4:47
add a comment |
$begingroup$
I tried to find a basis for a subspace $V$, but i know that: $(1,2,-1,0),(0,2,1,3)$ is basis of $V^perp $.
For this, i have the following theorem: Let $A$ a matrix, then we have:
$1.$ $Row(A)^perp$ $=$ $Null(A)$
So, my arguement is this: $Null(A) = Row(A)^perp$, and $Row(A) = V$, we have that $Null(A)^perp = (Row(A)^perp)^perp)$ but $(Row(A)^perp)^perp) = Row(A) = V$, so if we apply this theorem to $Null(A)^perp$ we must be find a basis for $V$, but it doesn't works.
I hope you can help me. Thanks.
linear-algebra
asked Mar 28 at 1:07
Mateo Soto ArangoMateo Soto Arango
114
$endgroup$
I tried to find a basis for a subspace $V$, but i know that: $(1,2,-1,0),(0,2,1,3)$ is basis of $V^perp $.
For this, i have the following theorem: Let $A$ a matrix, then we have:
$1.$ $Row(A)^perp$ $=$ $Null(A)$
So, my arguement is this: $Null(A) = Row(A)^perp$, and $Row(A) = V$, we have that $Null(A)^perp = (Row(A)^perp)^perp)$ but $(Row(A)^perp)^perp) = Row(A) = V$, so if we apply this theorem to $Null(A)^perp$ we must be find a basis for $V$, but it doesn't works.
I hope you can help me. Thanks.
linear-algebra
linear-algebra
asked Mar 28 at 1:07
Mateo Soto ArangoMateo Soto Arango
114
asked Mar 28 at 1:07
Mateo Soto ArangoMateo Soto Arango
114
asked Mar 28 at 1:07
Mateo Soto ArangoMateo Soto Arango
114
asked Mar 28 at 1:07
Mateo Soto ArangoMateo Soto Arango
114
asked Mar 28 at 1:07
Mateo Soto ArangoMateo Soto Arango
114
114
$begingroup$
What if $Row(A)=V^perp$?
$endgroup$
– amd
Mar 28 at 1:12
$begingroup$
$Row(A)$ is is the subspace generated by the file vectors of A, in spanish is $Fil(A)$ the notation is wrong?
$endgroup$
– Mateo Soto Arango
Mar 28 at 1:20
$begingroup$
You asked for help. That was a hint. Try your reasoning again starting from that instead.
$endgroup$
– amd
Mar 28 at 4:47
add a comment |
$begingroup$
What if $Row(A)=V^perp$?
$endgroup$
– amd
Mar 28 at 1:12
$begingroup$
$Row(A)$ is is the subspace generated by the file vectors of A, in spanish is $Fil(A)$ the notation is wrong?
$endgroup$
– Mateo Soto Arango
Mar 28 at 1:20
$begingroup$
You asked for help. That was a hint. Try your reasoning again starting from that instead.
$endgroup$
– amd
Mar 28 at 4:47
$begingroup$
What if $Row(A)=V^perp$?
$endgroup$
– amd
Mar 28 at 1:12
$begingroup$
What if $Row(A)=V^perp$?
$endgroup$
– amd
Mar 28 at 1:12
$begingroup$
$Row(A)$ is is the subspace generated by the file vectors of A, in spanish is $Fil(A)$ the notation is wrong?
$endgroup$
– Mateo Soto Arango
Mar 28 at 1:20
$begingroup$
$Row(A)$ is is the subspace generated by the file vectors of A, in spanish is $Fil(A)$ the notation is wrong?
$endgroup$
– Mateo Soto Arango
Mar 28 at 1:20
$begingroup$
You asked for help. That was a hint. Try your reasoning again starting from that instead.
$endgroup$
– amd
Mar 28 at 4:47
$begingroup$
You asked for help. That was a hint. Try your reasoning again starting from that instead.
$endgroup$
– amd
Mar 28 at 4:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I don't know what you mean by $Row$... However, one has $(Im,A)^perp = ker A^T$. So put your vectors as columns into a matrix $A$. Then you know that $Im,A = V^perp$. Equivalently, $(Im,A)^perp = V$. So, $V = ker A^T$. So, you have to find $ker A^T$.
answered Mar 28 at 1:14
amsmathamsmath
3,288420
$endgroup$
$begingroup$
$Row(A)$ is is the subspace generated by the file vectors of A, in spanish is $Fil(A)$ the notation is wrong?
$endgroup$
– Mateo Soto Arango
Mar 28 at 1:18
$begingroup$
It's the row vectors in English. But yeah, I've never seen it before.
$endgroup$
– amsmath
Mar 28 at 1:20
add a comment |
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1 Answer
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1 Answer
1
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votes
$begingroup$
I don't know what you mean by $Row$... However, one has $(Im,A)^perp = ker A^T$. So put your vectors as columns into a matrix $A$. Then you know that $Im,A = V^perp$. Equivalently, $(Im,A)^perp = V$. So, $V = ker A^T$. So, you have to find $ker A^T$.
answered Mar 28 at 1:14
amsmathamsmath
3,288420
$endgroup$
$begingroup$
$Row(A)$ is is the subspace generated by the file vectors of A, in spanish is $Fil(A)$ the notation is wrong?
$endgroup$
– Mateo Soto Arango
Mar 28 at 1:18
$begingroup$
It's the row vectors in English. But yeah, I've never seen it before.
$endgroup$
– amsmath
Mar 28 at 1:20
add a comment |
$begingroup$
I don't know what you mean by $Row$... However, one has $(Im,A)^perp = ker A^T$. So put your vectors as columns into a matrix $A$. Then you know that $Im,A = V^perp$. Equivalently, $(Im,A)^perp = V$. So, $V = ker A^T$. So, you have to find $ker A^T$.
answered Mar 28 at 1:14
amsmathamsmath
3,288420
$endgroup$
$begingroup$
$Row(A)$ is is the subspace generated by the file vectors of A, in spanish is $Fil(A)$ the notation is wrong?
$endgroup$
– Mateo Soto Arango
Mar 28 at 1:18
$begingroup$
It's the row vectors in English. But yeah, I've never seen it before.
$endgroup$
– amsmath
Mar 28 at 1:20
add a comment |
$begingroup$
I don't know what you mean by $Row$... However, one has $(Im,A)^perp = ker A^T$. So put your vectors as columns into a matrix $A$. Then you know that $Im,A = V^perp$. Equivalently, $(Im,A)^perp = V$. So, $V = ker A^T$. So, you have to find $ker A^T$.
answered Mar 28 at 1:14
amsmathamsmath
3,288420
$endgroup$
I don't know what you mean by $Row$... However, one has $(Im,A)^perp = ker A^T$. So put your vectors as columns into a matrix $A$. Then you know that $Im,A = V^perp$. Equivalently, $(Im,A)^perp = V$. So, $V = ker A^T$. So, you have to find $ker A^T$.
answered Mar 28 at 1:14
amsmathamsmath
3,288420
answered Mar 28 at 1:14
amsmathamsmath
3,288420
answered Mar 28 at 1:14
amsmathamsmath
3,288420
answered Mar 28 at 1:14
amsmathamsmath
3,288420
3,288420
$begingroup$
$Row(A)$ is is the subspace generated by the file vectors of A, in spanish is $Fil(A)$ the notation is wrong?
$endgroup$
– Mateo Soto Arango
Mar 28 at 1:18
$begingroup$
It's the row vectors in English. But yeah, I've never seen it before.
$endgroup$
– amsmath
Mar 28 at 1:20
add a comment |
$begingroup$
$Row(A)$ is is the subspace generated by the file vectors of A, in spanish is $Fil(A)$ the notation is wrong?
$endgroup$
– Mateo Soto Arango
Mar 28 at 1:18
$begingroup$
It's the row vectors in English. But yeah, I've never seen it before.
$endgroup$
– amsmath
Mar 28 at 1:20
$begingroup$
$Row(A)$ is is the subspace generated by the file vectors of A, in spanish is $Fil(A)$ the notation is wrong?
$endgroup$
– Mateo Soto Arango
Mar 28 at 1:18
$begingroup$
$Row(A)$ is is the subspace generated by the file vectors of A, in spanish is $Fil(A)$ the notation is wrong?
$endgroup$
– Mateo Soto Arango
Mar 28 at 1:18
$begingroup$
It's the row vectors in English. But yeah, I've never seen it before.
$endgroup$
– amsmath
Mar 28 at 1:20
$begingroup$
It's the row vectors in English. But yeah, I've never seen it before.
$endgroup$
– amsmath
Mar 28 at 1:20
add a comment |
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$begingroup$
What if $Row(A)=V^perp$?
$endgroup$
– amd
Mar 28 at 1:12
$begingroup$
$Row(A)$ is is the subspace generated by the file vectors of A, in spanish is $Fil(A)$ the notation is wrong?
$endgroup$
– Mateo Soto Arango
Mar 28 at 1:20
$begingroup$
You asked for help. That was a hint. Try your reasoning again starting from that instead.
$endgroup$
– amd
Mar 28 at 4:47