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
asked Nov 22 at 13:32
JorgeeFG
2661315
add a comment |
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0
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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
asked Nov 22 at 13:32
JorgeeFG
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
add a comment |
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0
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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
asked Nov 22 at 13:32
JorgeeFG
2661315
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
asked Nov 22 at 13:32
JorgeeFG
2661315
asked Nov 22 at 13:32
JorgeeFG
2661315
edited Nov 22 at 13:48
asked Nov 22 at 13:32
JorgeeFG
2661315
asked Nov 22 at 13:32
JorgeeFG
2661315
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
add a comment |
$S^perp={(2,1,5)}$ 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
add a comment |
1 Answer
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1
down 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 $S$...
answered Nov 22 at 13:39
GreginGre
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
add a comment |
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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 $S$...
answered Nov 22 at 13:39
GreginGre
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
add a comment |
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 $S$...
answered Nov 22 at 13:39
GreginGre
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
add a comment |
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 $S$...
answered Nov 22 at 13:39
GreginGre
1,33828
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 $S$...
answered Nov 22 at 13:39
GreginGre
1,33828
answered Nov 22 at 13:39
GreginGre
1,33828
answered Nov 22 at 13:39
GreginGre
1,33828
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
add a comment |
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
add a comment |
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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