Orthogonal complement not resulting ok











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I have the subspaces:



$$S = langle(2,1,-1), (-1,2,0)rangle, qquad T = { X + Y + 2Z =0; X - Y - Z = 0}.$$



I got that $T = langle(1, 3,-2)rangle$.



All vectors are linearly independent, so $S + T = mathbb{R}^3.$
Then I tried to calculate the orthogonal complement for both.



It is my understanding that the orthogonal complement for $S$ should give a vector that is a multiple of $(1,3,-2)$ (generates the same subspace as $T$) and the orthogonal complement for $T$ should give vectors that are linearly dependent from the ones in $S$ (generates the same subspace).



However, complement of $S = t(2,1,3)$ from my calculations, and complement of $T = t(-3,1,0) + s(2,0,1)$.



What I did was resolving the system



$$2x + y - z = 0 \ - x + 2y = 0$$



For $S$, and for $T$:



$$x + 3y - 2z = 0 .$$



Am I doing something wrong?



Edit: Maybe I'm understanding wrong a part of the theory, but I read that $S$ and its orthogonal complement will generate $mathbb{R}^n$. The orthogonal complement for T is indeed resulting in $0$ for the scalar product. But if $dim(S) + dim(S^bot) = 3$, and $dim(S) + dim(T) = 3$, shouldn't $T$ and $S^bot$ be the same subspace?






linear-algebra orthogonality







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  • $S^perp={(2,1,5)}$ and $T^perp={(1,1,2)}$. Orthogonal complements need not be unique.
    – Yadati Kiran
    Nov 22 at 13:45

















up vote
0
down vote

favorite












I have the subspaces:



$$S = langle(2,1,-1), (-1,2,0)rangle, qquad T = { X + Y + 2Z =0; X - Y - Z = 0}.$$



I got that $T = langle(1, 3,-2)rangle$.



All vectors are linearly independent, so $S + T = mathbb{R}^3.$
Then I tried to calculate the orthogonal complement for both.



It is my understanding that the orthogonal complement for $S$ should give a vector that is a multiple of $(1,3,-2)$ (generates the same subspace as $T$) and the orthogonal complement for $T$ should give vectors that are linearly dependent from the ones in $S$ (generates the same subspace).



However, complement of $S = t(2,1,3)$ from my calculations, and complement of $T = t(-3,1,0) + s(2,0,1)$.



What I did was resolving the system



$$2x + y - z = 0 \ - x + 2y = 0$$



For $S$, and for $T$:



$$x + 3y - 2z = 0 .$$



Am I doing something wrong?



Edit: Maybe I'm understanding wrong a part of the theory, but I read that $S$ and its orthogonal complement will generate $mathbb{R}^n$. The orthogonal complement for T is indeed resulting in $0$ for the scalar product. But if $dim(S) + dim(S^bot) = 3$, and $dim(S) + dim(T) = 3$, shouldn't $T$ and $S^bot$ be the same subspace?






linear-algebra orthogonality







share|cite|improve this question
























  • $S^perp={(2,1,5)}$ and $T^perp={(1,1,2)}$. Orthogonal complements need not be unique.
    – Yadati Kiran
    Nov 22 at 13:45















up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have the subspaces:



$$S = langle(2,1,-1), (-1,2,0)rangle, qquad T = { X + Y + 2Z =0; X - Y - Z = 0}.$$



I got that $T = langle(1, 3,-2)rangle$.



All vectors are linearly independent, so $S + T = mathbb{R}^3.$
Then I tried to calculate the orthogonal complement for both.



It is my understanding that the orthogonal complement for $S$ should give a vector that is a multiple of $(1,3,-2)$ (generates the same subspace as $T$) and the orthogonal complement for $T$ should give vectors that are linearly dependent from the ones in $S$ (generates the same subspace).



However, complement of $S = t(2,1,3)$ from my calculations, and complement of $T = t(-3,1,0) + s(2,0,1)$.



What I did was resolving the system



$$2x + y - z = 0 \ - x + 2y = 0$$



For $S$, and for $T$:



$$x + 3y - 2z = 0 .$$



Am I doing something wrong?



Edit: Maybe I'm understanding wrong a part of the theory, but I read that $S$ and its orthogonal complement will generate $mathbb{R}^n$. The orthogonal complement for T is indeed resulting in $0$ for the scalar product. But if $dim(S) + dim(S^bot) = 3$, and $dim(S) + dim(T) = 3$, shouldn't $T$ and $S^bot$ be the same subspace?






linear-algebra orthogonality







share|cite|improve this question















I have the subspaces:



$$S = langle(2,1,-1), (-1,2,0)rangle, qquad T = { X + Y + 2Z =0; X - Y - Z = 0}.$$



I got that $T = langle(1, 3,-2)rangle$.



All vectors are linearly independent, so $S + T = mathbb{R}^3.$
Then I tried to calculate the orthogonal complement for both.



It is my understanding that the orthogonal complement for $S$ should give a vector that is a multiple of $(1,3,-2)$ (generates the same subspace as $T$) and the orthogonal complement for $T$ should give vectors that are linearly dependent from the ones in $S$ (generates the same subspace).



However, complement of $S = t(2,1,3)$ from my calculations, and complement of $T = t(-3,1,0) + s(2,0,1)$.



What I did was resolving the system



$$2x + y - z = 0 \ - x + 2y = 0$$



For $S$, and for $T$:



$$x + 3y - 2z = 0 .$$



Am I doing something wrong?



Edit: Maybe I'm understanding wrong a part of the theory, but I read that $S$ and its orthogonal complement will generate $mathbb{R}^n$. The orthogonal complement for T is indeed resulting in $0$ for the scalar product. But if $dim(S) + dim(S^bot) = 3$, and $dim(S) + dim(T) = 3$, shouldn't $T$ and $S^bot$ be the same subspace?






linear-algebra orthogonality




linear-algebra orthogonality






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edited Nov 22 at 13:48

























asked Nov 22 at 13:32









JorgeeFG

2661315




2661315












  • $S^perp={(2,1,5)}$ and $T^perp={(1,1,2)}$. Orthogonal complements need not be unique.
    – Yadati Kiran
    Nov 22 at 13:45




















  • $S^perp={(2,1,5)}$ and $T^perp={(1,1,2)}$. Orthogonal complements need not be unique.
    – Yadati Kiran
    Nov 22 at 13:45


















$S^perp={(2,1,5)}$ and $T^perp={(1,1,2)}$. Orthogonal complements need not be unique.
– Yadati Kiran
Nov 22 at 13:45






$S^perp={(2,1,5)}$ and $T^perp={(1,1,2)}$. Orthogonal complements need not be unique.
– Yadati Kiran
Nov 22 at 13:45












1 Answer
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There is absolutely no reason why a vector orthogonal to $S$ should be an element of $T$. This would be true if $T$ were actually the orthogonal complement of $S$, but it is not. Clearly $(1,3,-2)$ is not orthogonal to $S$.



You seem to believe that a complement of $S$ is unique. It is not the case (take the line generated by any vector outside $S$).



PS. Note that your calculations for the complement of $S$ are certainly wrong, since the vector you obtain is not orthogonal to the generators of $S$...






share|cite|improve this answer





















  • Please see my edit
    – JorgeeFG
    Nov 22 at 13:43










  • Your example shows that the answer to your edit is NO. Once again, $T$ is NOT the orthogonal complement of $S$ because it is NOT orthogonal to $S$. Ypu have infinitely many possible complements for $S$.
    – GreginGre
    Nov 22 at 13:46











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up vote
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There is absolutely no reason why a vector orthogonal to $S$ should be an element of $T$. This would be true if $T$ were actually the orthogonal complement of $S$, but it is not. Clearly $(1,3,-2)$ is not orthogonal to $S$.



You seem to believe that a complement of $S$ is unique. It is not the case (take the line generated by any vector outside $S$).



PS. Note that your calculations for the complement of $S$ are certainly wrong, since the vector you obtain is not orthogonal to the generators of $S$...






share|cite|improve this answer





















  • Please see my edit
    – JorgeeFG
    Nov 22 at 13:43










  • Your example shows that the answer to your edit is NO. Once again, $T$ is NOT the orthogonal complement of $S$ because it is NOT orthogonal to $S$. Ypu have infinitely many possible complements for $S$.
    – GreginGre
    Nov 22 at 13:46















up vote
1
down vote



accepted










There is absolutely no reason why a vector orthogonal to $S$ should be an element of $T$. This would be true if $T$ were actually the orthogonal complement of $S$, but it is not. Clearly $(1,3,-2)$ is not orthogonal to $S$.



You seem to believe that a complement of $S$ is unique. It is not the case (take the line generated by any vector outside $S$).



PS. Note that your calculations for the complement of $S$ are certainly wrong, since the vector you obtain is not orthogonal to the generators of $S$...






share|cite|improve this answer





















  • Please see my edit
    – JorgeeFG
    Nov 22 at 13:43










  • Your example shows that the answer to your edit is NO. Once again, $T$ is NOT the orthogonal complement of $S$ because it is NOT orthogonal to $S$. Ypu have infinitely many possible complements for $S$.
    – GreginGre
    Nov 22 at 13:46













up vote
1
down vote



accepted







up vote
1
down vote



accepted






There is absolutely no reason why a vector orthogonal to $S$ should be an element of $T$. This would be true if $T$ were actually the orthogonal complement of $S$, but it is not. Clearly $(1,3,-2)$ is not orthogonal to $S$.



You seem to believe that a complement of $S$ is unique. It is not the case (take the line generated by any vector outside $S$).



PS. Note that your calculations for the complement of $S$ are certainly wrong, since the vector you obtain is not orthogonal to the generators of $S$...






share|cite|improve this answer












There is absolutely no reason why a vector orthogonal to $S$ should be an element of $T$. This would be true if $T$ were actually the orthogonal complement of $S$, but it is not. Clearly $(1,3,-2)$ is not orthogonal to $S$.



You seem to believe that a complement of $S$ is unique. It is not the case (take the line generated by any vector outside $S$).



PS. Note that your calculations for the complement of $S$ are certainly wrong, since the vector you obtain is not orthogonal to the generators of $S$...







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 22 at 13:39









GreginGre

1,33828




1,33828












  • Please see my edit
    – JorgeeFG
    Nov 22 at 13:43










  • Your example shows that the answer to your edit is NO. Once again, $T$ is NOT the orthogonal complement of $S$ because it is NOT orthogonal to $S$. Ypu have infinitely many possible complements for $S$.
    – GreginGre
    Nov 22 at 13:46


















  • Please see my edit
    – JorgeeFG
    Nov 22 at 13:43










  • Your example shows that the answer to your edit is NO. Once again, $T$ is NOT the orthogonal complement of $S$ because it is NOT orthogonal to $S$. Ypu have infinitely many possible complements for $S$.
    – GreginGre
    Nov 22 at 13:46
















Please see my edit
– JorgeeFG
Nov 22 at 13:43




Please see my edit
– JorgeeFG
Nov 22 at 13:43












Your example shows that the answer to your edit is NO. Once again, $T$ is NOT the orthogonal complement of $S$ because it is NOT orthogonal to $S$. Ypu have infinitely many possible complements for $S$.
– GreginGre
Nov 22 at 13:46




Your example shows that the answer to your edit is NO. Once again, $T$ is NOT the orthogonal complement of $S$ because it is NOT orthogonal to $S$. Ypu have infinitely many possible complements for $S$.
– GreginGre
Nov 22 at 13:46


















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