Finding a base of $S^bot$ is finding a solution for $Ax=0$ but for which matrix?
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Let $S$ is subspace of $mathbb R^4$ and $L((1,2,2,3),(1,3,3,2))=S$
a) Find one base of subspace $S^bot$
b) Finding that base is the same finding solution for $Ax=0$ but for which matrix?
a) It is easy I find a base $L((-5,1,0,1),(0,-1,1,0))$
b) For matrix which $ker(A)=L((-5,1,0,1),(0,-1,1,0))$ and they are orthogonal to every member of row space of matrix $A$. Is this the full answer?
linear-algebra matrices orthogonality
edited Sep 6 at 8:35
mechanodroid
24.6k62245
asked Sep 6 at 5:32
Marko Škorić
4008
add a comment |Â
up vote
1
down vote
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Let $S$ is subspace of $mathbb R^4$ and $L((1,2,2,3),(1,3,3,2))=S$
a) Find one base of subspace $S^bot$
b) Finding that base is the same finding solution for $Ax=0$ but for which matrix?
a) It is easy I find a base $L((-5,1,0,1),(0,-1,1,0))$
b) For matrix which $ker(A)=L((-5,1,0,1),(0,-1,1,0))$ and they are orthogonal to every member of row space of matrix $A$. Is this the full answer?
linear-algebra matrices orthogonality
edited Sep 6 at 8:35
mechanodroid
24.6k62245
asked Sep 6 at 5:32
Marko Škorić
4008
I would say "for a matrix $A$ whose rows are a basis of $S$".
– A.Γ.
Sep 6 at 5:42
How did you find $(-5,0,0,1)in S^bot$?
– A.Γ.
Sep 6 at 5:50
Sorry my mistake, i think that I have mistake in calculation, but I just need a full answer on last question
– Marko Å korić
Sep 6 at 5:54
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $S$ is subspace of $mathbb R^4$ and $L((1,2,2,3),(1,3,3,2))=S$
a) Find one base of subspace $S^bot$
b) Finding that base is the same finding solution for $Ax=0$ but for which matrix?
a) It is easy I find a base $L((-5,1,0,1),(0,-1,1,0))$
b) For matrix which $ker(A)=L((-5,1,0,1),(0,-1,1,0))$ and they are orthogonal to every member of row space of matrix $A$. Is this the full answer?
linear-algebra matrices orthogonality
edited Sep 6 at 8:35
mechanodroid
24.6k62245
asked Sep 6 at 5:32
Marko Škorić
4008
Let $S$ is subspace of $mathbb R^4$ and $L((1,2,2,3),(1,3,3,2))=S$
a) Find one base of subspace $S^bot$
b) Finding that base is the same finding solution for $Ax=0$ but for which matrix?
a) It is easy I find a base $L((-5,1,0,1),(0,-1,1,0))$
b) For matrix which $ker(A)=L((-5,1,0,1),(0,-1,1,0))$ and they are orthogonal to every member of row space of matrix $A$. Is this the full answer?
linear-algebra matrices orthogonality
linear-algebra matrices orthogonality
edited Sep 6 at 8:35
mechanodroid
24.6k62245
asked Sep 6 at 5:32
Marko Škorić
4008
edited Sep 6 at 8:35
mechanodroid
24.6k62245
asked Sep 6 at 5:32
Marko Škorić
4008
edited Sep 6 at 8:35
mechanodroid
24.6k62245
edited Sep 6 at 8:35
mechanodroid
24.6k62245
edited Sep 6 at 8:35
mechanodroid
24.6k62245
24.6k62245
asked Sep 6 at 5:32
Marko Škorić
4008
asked Sep 6 at 5:32
Marko Škorić
4008
asked Sep 6 at 5:32
Marko Škorić
4008
4008
I would say "for a matrix $A$ whose rows are a basis of $S$".
– A.Γ.
Sep 6 at 5:42
How did you find $(-5,0,0,1)in S^bot$?
– A.Γ.
Sep 6 at 5:50
Sorry my mistake, i think that I have mistake in calculation, but I just need a full answer on last question
– Marko Å korić
Sep 6 at 5:54
add a comment |Â
I would say "for a matrix $A$ whose rows are a basis of $S$".
– A.Γ.
Sep 6 at 5:42
How did you find $(-5,0,0,1)in S^bot$?
– A.Γ.
Sep 6 at 5:50
Sorry my mistake, i think that I have mistake in calculation, but I just need a full answer on last question
– Marko Å korić
Sep 6 at 5:54
I would say "for a matrix $A$ whose rows are a basis of $S$".
– A.Γ.
Sep 6 at 5:42
I would say "for a matrix $A$ whose rows are a basis of $S$".
– A.Γ.
Sep 6 at 5:42
How did you find $(-5,0,0,1)in S^bot$?
– A.Γ.
Sep 6 at 5:50
How did you find $(-5,0,0,1)in S^bot$?
– A.Γ.
Sep 6 at 5:50
Sorry my mistake, i think that I have mistake in calculation, but I just need a full answer on last question
– Marko Å korić
Sep 6 at 5:54
Sorry my mistake, i think that I have mistake in calculation, but I just need a full answer on last question
– Marko Å korić
Sep 6 at 5:54
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
We have
$$beginbmatrix 1 & 2 & 2 & 3 \ 1 & 3 & 3 & 2endbmatrix sim beginbmatrix 1 & 2 & 2 & 3 \ 0 & 1 & 1 & -1endbmatrix sim beginbmatrix 1 & 0 & 0 & 5 \ 0 & 1 & 1 & -1endbmatrix$$
so the basis for $S^perp$ is $(0,-1,1,0), (-5,1,0,1)$, as you said.
To find $S^perp$ we can solve $Ax = 0$ if and only if the matrix $A$ satisfies $ker A = S^perp$, or equivalently $R(A^T) = S$.
Therefore the rows of $A$ must span $S$.
answered Sep 6 at 8:34
mechanodroid
24.6k62245
soryy if I am wrong but I think that $R(A^T)=S$
– Marko Å korić
Sep 6 at 8:59
@MarkoŠkorić Of course, I meant the transpose.
– mechanodroid
Sep 6 at 9:00
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
We have
$$beginbmatrix 1 & 2 & 2 & 3 \ 1 & 3 & 3 & 2endbmatrix sim beginbmatrix 1 & 2 & 2 & 3 \ 0 & 1 & 1 & -1endbmatrix sim beginbmatrix 1 & 0 & 0 & 5 \ 0 & 1 & 1 & -1endbmatrix$$
so the basis for $S^perp$ is $(0,-1,1,0), (-5,1,0,1)$, as you said.
To find $S^perp$ we can solve $Ax = 0$ if and only if the matrix $A$ satisfies $ker A = S^perp$, or equivalently $R(A^T) = S$.
Therefore the rows of $A$ must span $S$.
answered Sep 6 at 8:34
mechanodroid
24.6k62245
soryy if I am wrong but I think that $R(A^T)=S$
– Marko Å korić
Sep 6 at 8:59
@MarkoŠkorić Of course, I meant the transpose.
– mechanodroid
Sep 6 at 9:00
add a comment |Â
up vote
0
down vote
accepted
We have
$$beginbmatrix 1 & 2 & 2 & 3 \ 1 & 3 & 3 & 2endbmatrix sim beginbmatrix 1 & 2 & 2 & 3 \ 0 & 1 & 1 & -1endbmatrix sim beginbmatrix 1 & 0 & 0 & 5 \ 0 & 1 & 1 & -1endbmatrix$$
so the basis for $S^perp$ is $(0,-1,1,0), (-5,1,0,1)$, as you said.
To find $S^perp$ we can solve $Ax = 0$ if and only if the matrix $A$ satisfies $ker A = S^perp$, or equivalently $R(A^T) = S$.
Therefore the rows of $A$ must span $S$.
answered Sep 6 at 8:34
mechanodroid
24.6k62245
soryy if I am wrong but I think that $R(A^T)=S$
– Marko Å korić
Sep 6 at 8:59
@MarkoŠkorić Of course, I meant the transpose.
– mechanodroid
Sep 6 at 9:00
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
We have
$$beginbmatrix 1 & 2 & 2 & 3 \ 1 & 3 & 3 & 2endbmatrix sim beginbmatrix 1 & 2 & 2 & 3 \ 0 & 1 & 1 & -1endbmatrix sim beginbmatrix 1 & 0 & 0 & 5 \ 0 & 1 & 1 & -1endbmatrix$$
so the basis for $S^perp$ is $(0,-1,1,0), (-5,1,0,1)$, as you said.
To find $S^perp$ we can solve $Ax = 0$ if and only if the matrix $A$ satisfies $ker A = S^perp$, or equivalently $R(A^T) = S$.
Therefore the rows of $A$ must span $S$.
answered Sep 6 at 8:34
mechanodroid
24.6k62245
We have
$$beginbmatrix 1 & 2 & 2 & 3 \ 1 & 3 & 3 & 2endbmatrix sim beginbmatrix 1 & 2 & 2 & 3 \ 0 & 1 & 1 & -1endbmatrix sim beginbmatrix 1 & 0 & 0 & 5 \ 0 & 1 & 1 & -1endbmatrix$$
so the basis for $S^perp$ is $(0,-1,1,0), (-5,1,0,1)$, as you said.
To find $S^perp$ we can solve $Ax = 0$ if and only if the matrix $A$ satisfies $ker A = S^perp$, or equivalently $R(A^T) = S$.
Therefore the rows of $A$ must span $S$.
answered Sep 6 at 8:34
mechanodroid
24.6k62245
edited Sep 6 at 9:00
answered Sep 6 at 8:34
mechanodroid
24.6k62245
answered Sep 6 at 8:34
mechanodroid
24.6k62245
answered Sep 6 at 8:34
mechanodroid
24.6k62245
24.6k62245
soryy if I am wrong but I think that $R(A^T)=S$
– Marko Å korić
Sep 6 at 8:59
@MarkoŠkorić Of course, I meant the transpose.
– mechanodroid
Sep 6 at 9:00
add a comment |Â
soryy if I am wrong but I think that $R(A^T)=S$
– Marko Å korić
Sep 6 at 8:59
@MarkoŠkorić Of course, I meant the transpose.
– mechanodroid
Sep 6 at 9:00
soryy if I am wrong but I think that $R(A^T)=S$
– Marko Å korić
Sep 6 at 8:59
soryy if I am wrong but I think that $R(A^T)=S$
– Marko Å korić
Sep 6 at 8:59
@MarkoŠkorić Of course, I meant the transpose.
– mechanodroid
Sep 6 at 9:00
@MarkoŠkorić Of course, I meant the transpose.
– mechanodroid
Sep 6 at 9:00
add a comment |Â
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I would say "for a matrix $A$ whose rows are a basis of $S$".
– A.Γ.
Sep 6 at 5:42
How did you find $(-5,0,0,1)in S^bot$?
– A.Γ.
Sep 6 at 5:50
Sorry my mistake, i think that I have mistake in calculation, but I just need a full answer on last question
– Marko Å korić
Sep 6 at 5:54