Mixing
Finding orthogonal basis in $mathbb R^4$ from given vectors
$begingroup$
I have two subsets of $mathbb R^4$
$S=((-1,0,1,1),(0,1,1,1),(1,0,0,1))$
and
$T=(x,y,z,x-y+2z)$
I've proved that T is a subspace of $mathbb R^4$ and that S is a basis for T. So far, so good!
I have to show that 2 of the vectors are orthogonal, which I have done, namely $(-1,0,1,1)$ and $(1,0,0,1)$ as their product is equal to zero. I now need to find an orthogonal basis for "T".
I presume I need to use Gram-Schmidt process, but am struggling as to how to find two more orthogonal vectors to start with. Would any starting vector need to be in the form $(x,y,z,x-y+2z)$?
Any hints/help much appreciated!
orthogonality
edited Jan 14 at 18:28
Andrei
12k21126
asked Jan 14 at 18:23
Andrea BrabrookAndrea Brabrook
33
$endgroup$
add a comment |
$begingroup$
I have two subsets of $mathbb R^4$
$S=((-1,0,1,1),(0,1,1,1),(1,0,0,1))$
and
$T=(x,y,z,x-y+2z)$
I've proved that T is a subspace of $mathbb R^4$ and that S is a basis for T. So far, so good!
I have to show that 2 of the vectors are orthogonal, which I have done, namely $(-1,0,1,1)$ and $(1,0,0,1)$ as their product is equal to zero. I now need to find an orthogonal basis for "T".
I presume I need to use Gram-Schmidt process, but am struggling as to how to find two more orthogonal vectors to start with. Would any starting vector need to be in the form $(x,y,z,x-y+2z)$?
Any hints/help much appreciated!
orthogonality
edited Jan 14 at 18:28
Andrei
12k21126
asked Jan 14 at 18:23
Andrea BrabrookAndrea Brabrook
33
$endgroup$
add a comment |
$begingroup$
I have two subsets of $mathbb R^4$
$S=((-1,0,1,1),(0,1,1,1),(1,0,0,1))$
and
$T=(x,y,z,x-y+2z)$
I've proved that T is a subspace of $mathbb R^4$ and that S is a basis for T. So far, so good!
I have to show that 2 of the vectors are orthogonal, which I have done, namely $(-1,0,1,1)$ and $(1,0,0,1)$ as their product is equal to zero. I now need to find an orthogonal basis for "T".
I presume I need to use Gram-Schmidt process, but am struggling as to how to find two more orthogonal vectors to start with. Would any starting vector need to be in the form $(x,y,z,x-y+2z)$?
Any hints/help much appreciated!
orthogonality
edited Jan 14 at 18:28
Andrei
12k21126
asked Jan 14 at 18:23
Andrea BrabrookAndrea Brabrook
33
$endgroup$
I have two subsets of $mathbb R^4$
$S=((-1,0,1,1),(0,1,1,1),(1,0,0,1))$
and
$T=(x,y,z,x-y+2z)$
I've proved that T is a subspace of $mathbb R^4$ and that S is a basis for T. So far, so good!
I have to show that 2 of the vectors are orthogonal, which I have done, namely $(-1,0,1,1)$ and $(1,0,0,1)$ as their product is equal to zero. I now need to find an orthogonal basis for "T".
I presume I need to use Gram-Schmidt process, but am struggling as to how to find two more orthogonal vectors to start with. Would any starting vector need to be in the form $(x,y,z,x-y+2z)$?
Any hints/help much appreciated!
orthogonality
orthogonality
edited Jan 14 at 18:28
Andrei
12k21126
asked Jan 14 at 18:23
Andrea BrabrookAndrea Brabrook
33
edited Jan 14 at 18:28
Andrei
12k21126
asked Jan 14 at 18:23
Andrea BrabrookAndrea Brabrook
33
edited Jan 14 at 18:28
Andrei
12k21126
edited Jan 14 at 18:28
Andrei
12k21126
edited Jan 14 at 18:28
Andrei
12k21126
12k21126
asked Jan 14 at 18:23
Andrea BrabrookAndrea Brabrook
33
asked Jan 14 at 18:23
Andrea BrabrookAndrea Brabrook
33
asked Jan 14 at 18:23
Andrea BrabrookAndrea Brabrook
33
33
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Why two more vectors? Since $dim T=3$, you only need one more vector. So, apply Gram-Schmidt to $e_1=(-1,0,1,1)$, $e_2=(1,0,0,1)$, and $e_3=(0,1,1,1)$. The fact that $e_1$ and $e_2$ are orthogonal will make your computations easier.
answered Jan 14 at 18:27
José Carlos SantosJosé Carlos Santos
162k22128232
$endgroup$
$begingroup$
Thank you so much, I had tried this, done my calculations slightly wrong and ended up with something not orthogonal as the answer, so had presumed it was the wrong way to do it! Done again, and have an orthogonal basis! Worth asking for clarification..thanks again!
$endgroup$
– Andrea Brabrook
Jan 14 at 18:37
$begingroup$
Then you should have obtained$$frac1{sqrt3}left(-1,0,0,1right), frac1{sqrt2}(1,0,0,1)text{, and }frac1{sqrt{42}}(1,6,2,-1).$$
$endgroup$
– José Carlos Santos
Jan 14 at 18:43
$begingroup$
actually i had a different answer but presuming that yours is the orthonormal basis?
$endgroup$
– Andrea Brabrook
Jan 14 at 18:51
$begingroup$
This basis is the one that is obtained by applying Gram-Schmidt to the basis from my answer.
$endgroup$
– José Carlos Santos
Jan 14 at 18:54
add a comment |
$begingroup$
If "S is basis for T" and S has only 3 vektor in it, then you can assume, that dim(T) = 3 so you don't need two more vektors.
Just use the Gram-Schmidt process on those 3 basevektors in S.
But as we know that two of them are orthogonal, you just need to sove the last one.
answered Jan 14 at 18:34
germinatorgerminator
11
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Why two more vectors? Since $dim T=3$, you only need one more vector. So, apply Gram-Schmidt to $e_1=(-1,0,1,1)$, $e_2=(1,0,0,1)$, and $e_3=(0,1,1,1)$. The fact that $e_1$ and $e_2$ are orthogonal will make your computations easier.
answered Jan 14 at 18:27
José Carlos SantosJosé Carlos Santos
162k22128232
$endgroup$
$begingroup$
Thank you so much, I had tried this, done my calculations slightly wrong and ended up with something not orthogonal as the answer, so had presumed it was the wrong way to do it! Done again, and have an orthogonal basis! Worth asking for clarification..thanks again!
$endgroup$
– Andrea Brabrook
Jan 14 at 18:37
$begingroup$
Then you should have obtained$$frac1{sqrt3}left(-1,0,0,1right), frac1{sqrt2}(1,0,0,1)text{, and }frac1{sqrt{42}}(1,6,2,-1).$$
$endgroup$
– José Carlos Santos
Jan 14 at 18:43
$begingroup$
actually i had a different answer but presuming that yours is the orthonormal basis?
$endgroup$
– Andrea Brabrook
Jan 14 at 18:51
$begingroup$
This basis is the one that is obtained by applying Gram-Schmidt to the basis from my answer.
$endgroup$
– José Carlos Santos
Jan 14 at 18:54
add a comment |
$begingroup$
Why two more vectors? Since $dim T=3$, you only need one more vector. So, apply Gram-Schmidt to $e_1=(-1,0,1,1)$, $e_2=(1,0,0,1)$, and $e_3=(0,1,1,1)$. The fact that $e_1$ and $e_2$ are orthogonal will make your computations easier.
answered Jan 14 at 18:27
José Carlos SantosJosé Carlos Santos
162k22128232
$endgroup$
$begingroup$
Thank you so much, I had tried this, done my calculations slightly wrong and ended up with something not orthogonal as the answer, so had presumed it was the wrong way to do it! Done again, and have an orthogonal basis! Worth asking for clarification..thanks again!
$endgroup$
– Andrea Brabrook
Jan 14 at 18:37
$begingroup$
Then you should have obtained$$frac1{sqrt3}left(-1,0,0,1right), frac1{sqrt2}(1,0,0,1)text{, and }frac1{sqrt{42}}(1,6,2,-1).$$
$endgroup$
– José Carlos Santos
Jan 14 at 18:43
$begingroup$
actually i had a different answer but presuming that yours is the orthonormal basis?
$endgroup$
– Andrea Brabrook
Jan 14 at 18:51
$begingroup$
This basis is the one that is obtained by applying Gram-Schmidt to the basis from my answer.
$endgroup$
– José Carlos Santos
Jan 14 at 18:54
add a comment |
$begingroup$
Why two more vectors? Since $dim T=3$, you only need one more vector. So, apply Gram-Schmidt to $e_1=(-1,0,1,1)$, $e_2=(1,0,0,1)$, and $e_3=(0,1,1,1)$. The fact that $e_1$ and $e_2$ are orthogonal will make your computations easier.
answered Jan 14 at 18:27
José Carlos SantosJosé Carlos Santos
162k22128232
$endgroup$
Why two more vectors? Since $dim T=3$, you only need one more vector. So, apply Gram-Schmidt to $e_1=(-1,0,1,1)$, $e_2=(1,0,0,1)$, and $e_3=(0,1,1,1)$. The fact that $e_1$ and $e_2$ are orthogonal will make your computations easier.
answered Jan 14 at 18:27
José Carlos SantosJosé Carlos Santos
162k22128232
answered Jan 14 at 18:27
José Carlos SantosJosé Carlos Santos
162k22128232
answered Jan 14 at 18:27
José Carlos SantosJosé Carlos Santos
162k22128232
answered Jan 14 at 18:27
José Carlos SantosJosé Carlos Santos
162k22128232
162k22128232
$begingroup$
Thank you so much, I had tried this, done my calculations slightly wrong and ended up with something not orthogonal as the answer, so had presumed it was the wrong way to do it! Done again, and have an orthogonal basis! Worth asking for clarification..thanks again!
$endgroup$
– Andrea Brabrook
Jan 14 at 18:37
$begingroup$
Then you should have obtained$$frac1{sqrt3}left(-1,0,0,1right), frac1{sqrt2}(1,0,0,1)text{, and }frac1{sqrt{42}}(1,6,2,-1).$$
$endgroup$
– José Carlos Santos
Jan 14 at 18:43
$begingroup$
actually i had a different answer but presuming that yours is the orthonormal basis?
$endgroup$
– Andrea Brabrook
Jan 14 at 18:51
$begingroup$
This basis is the one that is obtained by applying Gram-Schmidt to the basis from my answer.
$endgroup$
– José Carlos Santos
Jan 14 at 18:54
add a comment |
$begingroup$
Thank you so much, I had tried this, done my calculations slightly wrong and ended up with something not orthogonal as the answer, so had presumed it was the wrong way to do it! Done again, and have an orthogonal basis! Worth asking for clarification..thanks again!
$endgroup$
– Andrea Brabrook
Jan 14 at 18:37
$begingroup$
Then you should have obtained$$frac1{sqrt3}left(-1,0,0,1right), frac1{sqrt2}(1,0,0,1)text{, and }frac1{sqrt{42}}(1,6,2,-1).$$
$endgroup$
– José Carlos Santos
Jan 14 at 18:43
$begingroup$
actually i had a different answer but presuming that yours is the orthonormal basis?
$endgroup$
– Andrea Brabrook
Jan 14 at 18:51
$begingroup$
This basis is the one that is obtained by applying Gram-Schmidt to the basis from my answer.
$endgroup$
– José Carlos Santos
Jan 14 at 18:54
$begingroup$
Thank you so much, I had tried this, done my calculations slightly wrong and ended up with something not orthogonal as the answer, so had presumed it was the wrong way to do it! Done again, and have an orthogonal basis! Worth asking for clarification..thanks again!
$endgroup$
– Andrea Brabrook
Jan 14 at 18:37
$begingroup$
Thank you so much, I had tried this, done my calculations slightly wrong and ended up with something not orthogonal as the answer, so had presumed it was the wrong way to do it! Done again, and have an orthogonal basis! Worth asking for clarification..thanks again!
$endgroup$
– Andrea Brabrook
Jan 14 at 18:37
$begingroup$
Then you should have obtained$$frac1{sqrt3}left(-1,0,0,1right), frac1{sqrt2}(1,0,0,1)text{, and }frac1{sqrt{42}}(1,6,2,-1).$$
$endgroup$
– José Carlos Santos
Jan 14 at 18:43
$begingroup$
Then you should have obtained$$frac1{sqrt3}left(-1,0,0,1right), frac1{sqrt2}(1,0,0,1)text{, and }frac1{sqrt{42}}(1,6,2,-1).$$
$endgroup$
– José Carlos Santos
Jan 14 at 18:43
$begingroup$
actually i had a different answer but presuming that yours is the orthonormal basis?
$endgroup$
– Andrea Brabrook
Jan 14 at 18:51
$begingroup$
actually i had a different answer but presuming that yours is the orthonormal basis?
$endgroup$
– Andrea Brabrook
Jan 14 at 18:51
$begingroup$
This basis is the one that is obtained by applying Gram-Schmidt to the basis from my answer.
$endgroup$
– José Carlos Santos
Jan 14 at 18:54
$begingroup$
This basis is the one that is obtained by applying Gram-Schmidt to the basis from my answer.
$endgroup$
– José Carlos Santos
Jan 14 at 18:54
add a comment |
$begingroup$
If "S is basis for T" and S has only 3 vektor in it, then you can assume, that dim(T) = 3 so you don't need two more vektors.
Just use the Gram-Schmidt process on those 3 basevektors in S.
But as we know that two of them are orthogonal, you just need to sove the last one.
answered Jan 14 at 18:34
germinatorgerminator
11
$endgroup$
add a comment |
$begingroup$
If "S is basis for T" and S has only 3 vektor in it, then you can assume, that dim(T) = 3 so you don't need two more vektors.
Just use the Gram-Schmidt process on those 3 basevektors in S.
But as we know that two of them are orthogonal, you just need to sove the last one.
answered Jan 14 at 18:34
germinatorgerminator
11
$endgroup$
add a comment |
$begingroup$
If "S is basis for T" and S has only 3 vektor in it, then you can assume, that dim(T) = 3 so you don't need two more vektors.
Just use the Gram-Schmidt process on those 3 basevektors in S.
But as we know that two of them are orthogonal, you just need to sove the last one.
answered Jan 14 at 18:34
germinatorgerminator
11
$endgroup$
If "S is basis for T" and S has only 3 vektor in it, then you can assume, that dim(T) = 3 so you don't need two more vektors.
Just use the Gram-Schmidt process on those 3 basevektors in S.
But as we know that two of them are orthogonal, you just need to sove the last one.
answered Jan 14 at 18:34
germinatorgerminator
11
answered Jan 14 at 18:34
germinatorgerminator
11
answered Jan 14 at 18:34
germinatorgerminator
11
answered Jan 14 at 18:34
germinatorgerminator
11
11
add a comment |
add a comment |
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