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







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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














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







share|cite|improve this question























  • 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












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







share|cite|improve this question















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






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share|cite|improve this question













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edited Sep 6 at 8:35









mechanodroid

24.6k62245




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asked Sep 6 at 5:32









Marko Škorić

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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
















  • 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










1 Answer
1






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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 $S$.






share|cite|improve this answer






















  • soryy if I am wrong but I think that $R(A^T)=S$
    – Marko Å korić
    Sep 6 at 8:59










  • @MarkoÅ korić Of course, I meant the transpose.
    – mechanodroid
    Sep 6 at 9:00










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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 $S$.






share|cite|improve this answer






















  • soryy if I am wrong but I think that $R(A^T)=S$
    – Marko Å korić
    Sep 6 at 8:59










  • @MarkoÅ korić Of course, I meant the transpose.
    – mechanodroid
    Sep 6 at 9:00














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 $S$.






share|cite|improve this answer






















  • soryy if I am wrong but I think that $R(A^T)=S$
    – Marko Å korić
    Sep 6 at 8:59










  • @MarkoÅ korić Of course, I meant the transpose.
    – mechanodroid
    Sep 6 at 9:00












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 $S$.






share|cite|improve this answer














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 $S$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Sep 6 at 9:00

























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










  • @MarkoÅ korić 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










  • @MarkoÅ korić 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

















 

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