Gram Schmidt procedure
$begingroup$
For the subspace on the left in the image I am to find an orthogonal basis for the subspace. The answer is to the right in the image, and when I follow the procedure I get those vectors, however I also get one vector to be (0, 0, -120/15, 0). Why is that not included in the basis?
linear-algebra gram-schmidt
asked Dec 12 '18 at 15:58
P.ythonP.ython
205
$endgroup$
|
show 1 more comment
$begingroup$
For the subspace on the left in the image I am to find an orthogonal basis for the subspace. The answer is to the right in the image, and when I follow the procedure I get those vectors, however I also get one vector to be (0, 0, -120/15, 0). Why is that not included in the basis?
linear-algebra gram-schmidt
asked Dec 12 '18 at 15:58
P.ythonP.ython
205
$endgroup$
$begingroup$
Are you trying to find an orthogonal basis for the subspace spanned by the vectors on the left (which is $3$-dimensional)? Or the orthogonal complement (which is $1$-dimensional)?
$endgroup$
– David Hill
Dec 12 '18 at 16:04
$begingroup$
Sorry, I should say orthogonal basis for the subspace!
$endgroup$
– P.ython
Dec 12 '18 at 16:05
$begingroup$
OK, then you made a mistake. You should not be able to get $4$ vectors.
$endgroup$
– David Hill
Dec 12 '18 at 16:25
$begingroup$
If I update the question with a description of the procedure our particular textbook uses, could it have something to do with the last part, where they write "w3 was already in S2" does that reasoning somehow apply to this problem?
$endgroup$
– P.ython
Dec 12 '18 at 16:36
$begingroup$
That is the precise reasoning.
$endgroup$
– David Hill
Dec 12 '18 at 17:09
|
show 1 more comment
$begingroup$
For the subspace on the left in the image I am to find an orthogonal basis for the subspace. The answer is to the right in the image, and when I follow the procedure I get those vectors, however I also get one vector to be (0, 0, -120/15, 0). Why is that not included in the basis?
linear-algebra gram-schmidt
asked Dec 12 '18 at 15:58
P.ythonP.ython
205
$endgroup$
For the subspace on the left in the image I am to find an orthogonal basis for the subspace. The answer is to the right in the image, and when I follow the procedure I get those vectors, however I also get one vector to be (0, 0, -120/15, 0). Why is that not included in the basis?
linear-algebra gram-schmidt
linear-algebra gram-schmidt
asked Dec 12 '18 at 15:58
P.ythonP.ython
205
asked Dec 12 '18 at 15:58
P.ythonP.ython
205
edited Dec 12 '18 at 16:36
P.ython
asked Dec 12 '18 at 15:58
P.ythonP.ython
205
asked Dec 12 '18 at 15:58
P.ythonP.ython
205
asked Dec 12 '18 at 15:58
P.ythonP.ython
205
205
$begingroup$
Are you trying to find an orthogonal basis for the subspace spanned by the vectors on the left (which is $3$-dimensional)? Or the orthogonal complement (which is $1$-dimensional)?
$endgroup$
– David Hill
Dec 12 '18 at 16:04
$begingroup$
Sorry, I should say orthogonal basis for the subspace!
$endgroup$
– P.ython
Dec 12 '18 at 16:05
$begingroup$
OK, then you made a mistake. You should not be able to get $4$ vectors.
$endgroup$
– David Hill
Dec 12 '18 at 16:25
$begingroup$
If I update the question with a description of the procedure our particular textbook uses, could it have something to do with the last part, where they write "w3 was already in S2" does that reasoning somehow apply to this problem?
$endgroup$
– P.ython
Dec 12 '18 at 16:36
$begingroup$
That is the precise reasoning.
$endgroup$
– David Hill
Dec 12 '18 at 17:09
|
show 1 more comment
$begingroup$
Are you trying to find an orthogonal basis for the subspace spanned by the vectors on the left (which is $3$-dimensional)? Or the orthogonal complement (which is $1$-dimensional)?
$endgroup$
– David Hill
Dec 12 '18 at 16:04
$begingroup$
Sorry, I should say orthogonal basis for the subspace!
$endgroup$
– P.ython
Dec 12 '18 at 16:05
$begingroup$
OK, then you made a mistake. You should not be able to get $4$ vectors.
$endgroup$
– David Hill
Dec 12 '18 at 16:25
$begingroup$
If I update the question with a description of the procedure our particular textbook uses, could it have something to do with the last part, where they write "w3 was already in S2" does that reasoning somehow apply to this problem?
$endgroup$
– P.ython
Dec 12 '18 at 16:36
$begingroup$
That is the precise reasoning.
$endgroup$
– David Hill
Dec 12 '18 at 17:09
$begingroup$
Are you trying to find an orthogonal basis for the subspace spanned by the vectors on the left (which is $3$-dimensional)? Or the orthogonal complement (which is $1$-dimensional)?
$endgroup$
– David Hill
Dec 12 '18 at 16:04
$begingroup$
Are you trying to find an orthogonal basis for the subspace spanned by the vectors on the left (which is $3$-dimensional)? Or the orthogonal complement (which is $1$-dimensional)?
$endgroup$
– David Hill
Dec 12 '18 at 16:04
$begingroup$
Sorry, I should say orthogonal basis for the subspace!
$endgroup$
– P.ython
Dec 12 '18 at 16:05
$begingroup$
Sorry, I should say orthogonal basis for the subspace!
$endgroup$
– P.ython
Dec 12 '18 at 16:05
$begingroup$
OK, then you made a mistake. You should not be able to get $4$ vectors.
$endgroup$
– David Hill
Dec 12 '18 at 16:25
$begingroup$
OK, then you made a mistake. You should not be able to get $4$ vectors.
$endgroup$
– David Hill
Dec 12 '18 at 16:25
$begingroup$
If I update the question with a description of the procedure our particular textbook uses, could it have something to do with the last part, where they write "w3 was already in S2" does that reasoning somehow apply to this problem?
$endgroup$
– P.ython
Dec 12 '18 at 16:36
$begingroup$
If I update the question with a description of the procedure our particular textbook uses, could it have something to do with the last part, where they write "w3 was already in S2" does that reasoning somehow apply to this problem?
$endgroup$
– P.ython
Dec 12 '18 at 16:36
$begingroup$
That is the precise reasoning.
$endgroup$
– David Hill
Dec 12 '18 at 17:09
$begingroup$
That is the precise reasoning.
$endgroup$
– David Hill
Dec 12 '18 at 17:09
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
In the third step, $mathbb{S}_2=mathrm{span}{v_1,v_2}$ also equals $mathbb{S}_2=mathrm{span}{w_1,w_2}$. The Gram-Schmidt process produces a vector $v_3=mathrm{perp}_{mathbb{S}_2}w_3in mathbb{S}_2^perp$ with $mathrm{span}{v_1,v_2,v_3}=mathrm{span}{w_1,w_2,w_3}$. Since $v_3=0$,
$$
mathrm{span}{w_1,w_2,w_3}=mathrm{span}{v_1,v_2,v_3}=mathrm{span}{v_1,v_2,0}=mathrm{span}{v_1,v_2}=mathrm{span}{w_1,w_2}.
$$
Therefore, $w_3inmathrm{span}{w_1,w_2}=mathbb{S}_2$.
answered Dec 12 '18 at 17:22
David HillDavid Hill
9,5361619
$endgroup$
$begingroup$
Thank you David! Is this just because w3 gives the zero vector and therefore can be left out? Im guessing not because then why would (0,0,-120,0) from the example not be in the basis? If you are tired of explaining thank you for your time!
$endgroup$
– P.ython
Dec 12 '18 at 17:35
$begingroup$
I think you mean $v_3$. Also, you must have made a mistake in your calculation.
$endgroup$
– David Hill
Dec 12 '18 at 18:00
add a comment |
$begingroup$
You made a mistake somewhere. The four original vectors span a $3$-dimensional space and therefore there is no way of obtaining a basis with $4$ elements. Exactly which error you made is something I cannot tell you without seeing your computations.
answered Dec 12 '18 at 16:15
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
$begingroup$
Hello, Jose. I updated my question after the first post as I wrote the question incorrectly. I am trying to compute the orthogonal basis.
$endgroup$
– P.ython
Dec 12 '18 at 16:24
$begingroup$
I see no reason to change my answer.
$endgroup$
– José Carlos Santos
Dec 12 '18 at 16:25
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In the third step, $mathbb{S}_2=mathrm{span}{v_1,v_2}$ also equals $mathbb{S}_2=mathrm{span}{w_1,w_2}$. The Gram-Schmidt process produces a vector $v_3=mathrm{perp}_{mathbb{S}_2}w_3in mathbb{S}_2^perp$ with $mathrm{span}{v_1,v_2,v_3}=mathrm{span}{w_1,w_2,w_3}$. Since $v_3=0$,
$$
mathrm{span}{w_1,w_2,w_3}=mathrm{span}{v_1,v_2,v_3}=mathrm{span}{v_1,v_2,0}=mathrm{span}{v_1,v_2}=mathrm{span}{w_1,w_2}.
$$
Therefore, $w_3inmathrm{span}{w_1,w_2}=mathbb{S}_2$.
answered Dec 12 '18 at 17:22
David HillDavid Hill
9,5361619
$endgroup$
$begingroup$
Thank you David! Is this just because w3 gives the zero vector and therefore can be left out? Im guessing not because then why would (0,0,-120,0) from the example not be in the basis? If you are tired of explaining thank you for your time!
$endgroup$
– P.ython
Dec 12 '18 at 17:35
$begingroup$
I think you mean $v_3$. Also, you must have made a mistake in your calculation.
$endgroup$
– David Hill
Dec 12 '18 at 18:00
add a comment |
$begingroup$
In the third step, $mathbb{S}_2=mathrm{span}{v_1,v_2}$ also equals $mathbb{S}_2=mathrm{span}{w_1,w_2}$. The Gram-Schmidt process produces a vector $v_3=mathrm{perp}_{mathbb{S}_2}w_3in mathbb{S}_2^perp$ with $mathrm{span}{v_1,v_2,v_3}=mathrm{span}{w_1,w_2,w_3}$. Since $v_3=0$,
$$
mathrm{span}{w_1,w_2,w_3}=mathrm{span}{v_1,v_2,v_3}=mathrm{span}{v_1,v_2,0}=mathrm{span}{v_1,v_2}=mathrm{span}{w_1,w_2}.
$$
Therefore, $w_3inmathrm{span}{w_1,w_2}=mathbb{S}_2$.
answered Dec 12 '18 at 17:22
David HillDavid Hill
9,5361619
$endgroup$
$begingroup$
Thank you David! Is this just because w3 gives the zero vector and therefore can be left out? Im guessing not because then why would (0,0,-120,0) from the example not be in the basis? If you are tired of explaining thank you for your time!
$endgroup$
– P.ython
Dec 12 '18 at 17:35
$begingroup$
I think you mean $v_3$. Also, you must have made a mistake in your calculation.
$endgroup$
– David Hill
Dec 12 '18 at 18:00
add a comment |
$begingroup$
In the third step, $mathbb{S}_2=mathrm{span}{v_1,v_2}$ also equals $mathbb{S}_2=mathrm{span}{w_1,w_2}$. The Gram-Schmidt process produces a vector $v_3=mathrm{perp}_{mathbb{S}_2}w_3in mathbb{S}_2^perp$ with $mathrm{span}{v_1,v_2,v_3}=mathrm{span}{w_1,w_2,w_3}$. Since $v_3=0$,
$$
mathrm{span}{w_1,w_2,w_3}=mathrm{span}{v_1,v_2,v_3}=mathrm{span}{v_1,v_2,0}=mathrm{span}{v_1,v_2}=mathrm{span}{w_1,w_2}.
$$
Therefore, $w_3inmathrm{span}{w_1,w_2}=mathbb{S}_2$.
answered Dec 12 '18 at 17:22
David HillDavid Hill
9,5361619
$endgroup$
In the third step, $mathbb{S}_2=mathrm{span}{v_1,v_2}$ also equals $mathbb{S}_2=mathrm{span}{w_1,w_2}$. The Gram-Schmidt process produces a vector $v_3=mathrm{perp}_{mathbb{S}_2}w_3in mathbb{S}_2^perp$ with $mathrm{span}{v_1,v_2,v_3}=mathrm{span}{w_1,w_2,w_3}$. Since $v_3=0$,
$$
mathrm{span}{w_1,w_2,w_3}=mathrm{span}{v_1,v_2,v_3}=mathrm{span}{v_1,v_2,0}=mathrm{span}{v_1,v_2}=mathrm{span}{w_1,w_2}.
$$
Therefore, $w_3inmathrm{span}{w_1,w_2}=mathbb{S}_2$.
answered Dec 12 '18 at 17:22
David HillDavid Hill
9,5361619
answered Dec 12 '18 at 17:22
David HillDavid Hill
9,5361619
answered Dec 12 '18 at 17:22
David HillDavid Hill
9,5361619
answered Dec 12 '18 at 17:22
David HillDavid Hill
9,5361619
9,5361619
$begingroup$
Thank you David! Is this just because w3 gives the zero vector and therefore can be left out? Im guessing not because then why would (0,0,-120,0) from the example not be in the basis? If you are tired of explaining thank you for your time!
$endgroup$
– P.ython
Dec 12 '18 at 17:35
$begingroup$
I think you mean $v_3$. Also, you must have made a mistake in your calculation.
$endgroup$
– David Hill
Dec 12 '18 at 18:00
add a comment |
$begingroup$
Thank you David! Is this just because w3 gives the zero vector and therefore can be left out? Im guessing not because then why would (0,0,-120,0) from the example not be in the basis? If you are tired of explaining thank you for your time!
$endgroup$
– P.ython
Dec 12 '18 at 17:35
$begingroup$
I think you mean $v_3$. Also, you must have made a mistake in your calculation.
$endgroup$
– David Hill
Dec 12 '18 at 18:00
$begingroup$
Thank you David! Is this just because w3 gives the zero vector and therefore can be left out? Im guessing not because then why would (0,0,-120,0) from the example not be in the basis? If you are tired of explaining thank you for your time!
$endgroup$
– P.ython
Dec 12 '18 at 17:35
$begingroup$
Thank you David! Is this just because w3 gives the zero vector and therefore can be left out? Im guessing not because then why would (0,0,-120,0) from the example not be in the basis? If you are tired of explaining thank you for your time!
$endgroup$
– P.ython
Dec 12 '18 at 17:35
$begingroup$
I think you mean $v_3$. Also, you must have made a mistake in your calculation.
$endgroup$
– David Hill
Dec 12 '18 at 18:00
$begingroup$
I think you mean $v_3$. Also, you must have made a mistake in your calculation.
$endgroup$
– David Hill
Dec 12 '18 at 18:00
add a comment |
$begingroup$
You made a mistake somewhere. The four original vectors span a $3$-dimensional space and therefore there is no way of obtaining a basis with $4$ elements. Exactly which error you made is something I cannot tell you without seeing your computations.
answered Dec 12 '18 at 16:15
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
$begingroup$
Hello, Jose. I updated my question after the first post as I wrote the question incorrectly. I am trying to compute the orthogonal basis.
$endgroup$
– P.ython
Dec 12 '18 at 16:24
$begingroup$
I see no reason to change my answer.
$endgroup$
– José Carlos Santos
Dec 12 '18 at 16:25
add a comment |
$begingroup$
You made a mistake somewhere. The four original vectors span a $3$-dimensional space and therefore there is no way of obtaining a basis with $4$ elements. Exactly which error you made is something I cannot tell you without seeing your computations.
answered Dec 12 '18 at 16:15
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
$begingroup$
Hello, Jose. I updated my question after the first post as I wrote the question incorrectly. I am trying to compute the orthogonal basis.
$endgroup$
– P.ython
Dec 12 '18 at 16:24
$begingroup$
I see no reason to change my answer.
$endgroup$
– José Carlos Santos
Dec 12 '18 at 16:25
add a comment |
$begingroup$
You made a mistake somewhere. The four original vectors span a $3$-dimensional space and therefore there is no way of obtaining a basis with $4$ elements. Exactly which error you made is something I cannot tell you without seeing your computations.
answered Dec 12 '18 at 16:15
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
You made a mistake somewhere. The four original vectors span a $3$-dimensional space and therefore there is no way of obtaining a basis with $4$ elements. Exactly which error you made is something I cannot tell you without seeing your computations.
answered Dec 12 '18 at 16:15
José Carlos SantosJosé Carlos Santos
172k22132239
answered Dec 12 '18 at 16:15
José Carlos SantosJosé Carlos Santos
172k22132239
answered Dec 12 '18 at 16:15
José Carlos SantosJosé Carlos Santos
172k22132239
answered Dec 12 '18 at 16:15
José Carlos SantosJosé Carlos Santos
172k22132239
172k22132239
$begingroup$
Hello, Jose. I updated my question after the first post as I wrote the question incorrectly. I am trying to compute the orthogonal basis.
$endgroup$
– P.ython
Dec 12 '18 at 16:24
$begingroup$
I see no reason to change my answer.
$endgroup$
– José Carlos Santos
Dec 12 '18 at 16:25
add a comment |
$begingroup$
Hello, Jose. I updated my question after the first post as I wrote the question incorrectly. I am trying to compute the orthogonal basis.
$endgroup$
– P.ython
Dec 12 '18 at 16:24
$begingroup$
I see no reason to change my answer.
$endgroup$
– José Carlos Santos
Dec 12 '18 at 16:25
$begingroup$
Hello, Jose. I updated my question after the first post as I wrote the question incorrectly. I am trying to compute the orthogonal basis.
$endgroup$
– P.ython
Dec 12 '18 at 16:24
$begingroup$
Hello, Jose. I updated my question after the first post as I wrote the question incorrectly. I am trying to compute the orthogonal basis.
$endgroup$
– P.ython
Dec 12 '18 at 16:24
$begingroup$
I see no reason to change my answer.
$endgroup$
– José Carlos Santos
Dec 12 '18 at 16:25
$begingroup$
I see no reason to change my answer.
$endgroup$
– José Carlos Santos
Dec 12 '18 at 16:25
add a comment |
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$begingroup$
Are you trying to find an orthogonal basis for the subspace spanned by the vectors on the left (which is $3$-dimensional)? Or the orthogonal complement (which is $1$-dimensional)?
$endgroup$
– David Hill
Dec 12 '18 at 16:04
$begingroup$
Sorry, I should say orthogonal basis for the subspace!
$endgroup$
– P.ython
Dec 12 '18 at 16:05
$begingroup$
OK, then you made a mistake. You should not be able to get $4$ vectors.
$endgroup$
– David Hill
Dec 12 '18 at 16:25
$begingroup$
If I update the question with a description of the procedure our particular textbook uses, could it have something to do with the last part, where they write "w3 was already in S2" does that reasoning somehow apply to this problem?
$endgroup$
– P.ython
Dec 12 '18 at 16:36
$begingroup$
That is the precise reasoning.
$endgroup$
– David Hill
Dec 12 '18 at 17:09